Download ISE Option Traders Statistics with Applications to Options

ISE Option Traders
Statistics with Applications to Options
Statistics with Applications to Options
Alan L. Tucker, Ph.D.
631-331-8024 (tel)
631-331-8044 (fax)
[email protected]
www.mtaglobal.com
March 2004
Copyright (c) 2001-2004 by Marshall, Tucker & Associates, LLC
All rights reserved.
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ISE Option Traders
Statistics with Applications to Options
Alan L. Tucker, Ph.D.
Alan Tucker was the Founding Editor of the Journal of Financial Engineering and is currently on the
editorial boards of the Journal of Derivatives and Global Finance Journal. He is the author of three textbooks and
numerous articles appearing in such journals as the Journal of Finance, Review of Economics and Statistics,
Journal of Financial and Quantitative Analysis, the Virginia Tax Review, and others. Dr. Tucker is a tenured
professor of finance at the Lubin School of Business, Pace University, and an adjunct professor of finance at the
Stern School of Business, New York University. As a consultant, he has worked for the US Treasury, the US
Department of Justice, JP Morgan, Morgan Stanley, UBS, Deutsche Bank, TIAA-CREF, Lazard, Merrill Lynch, LG
Securities, and others.
March 2004
Copyright (c) 2001-2004 by Marshall, Tucker & Associates, LLC
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Statistics with Applications to Options
Purpose:
The purpose of this presentation is refresh registrants about a number of
key statistical concepts.
Our approach is to impart said concepts within the context of a specific
application germane to option traders, namely, the computation of a
trader’s Value-at-Risk. Other applications are also addressed.
March 2004
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Statistics with Applications to Options
What is Value-at-Risk?
Value-at-risk (abbreviated either VAR or VaR) is a measure of risk that boils risk
down to a single easy-to-understand number.
Simply put, an option trader’s VAR is the maximum number of dollars that the
trader might lose over a specified period of time (called the risk horizon) at a
specified level of confidence.
For example:
Suppose that a trader’s one-day VAR is $25,000 at a 95% level of confidence.
This says that the trader expects to lose no more than $25,000 over a one-day
period 95 out of each 100 days. Of course, this also implies that the trader can
expect to lose more than $25,000 in a single day five days out of each 100 days.
March 2004
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Statistics with Applications to Options
How is VAR measured?
A trader’s VAR can be measured for different risk horizons and different levels of
confidence.
It can be measured for a single trader, a single group of traders (a desk), a
single division (group of desks), or for the trading firm as a whole.
VAR is used as a risk monitoring tool but also as a risk management tool.
For example, if the trader’s VAR turns out to be $35,000 and management
believes that a VAR above $25,000 is excessive, it can order a reduction in the
level of risk bearing to no more than a VAR of $25,000.
March 2004
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Statistics with Applications to Options
Management can set VAR limits for each trader, for each desk, for each division
and for the firm as a whole.
Thus, it is possible to monitor each trader’s VAR, aggregate these to a desk
VAR, and so forth. Presumably, if an option trader’s Greeks (delta, gamma and
vega) are within their limits, said trader’s VAR will be acceptable.
The aggregation of VARs is not linear. For example if a desk consists of two
traders and each trader has a VAR of $30,000, the desk VAR will likely be less
than $60,000. Indeed, it is possible that the desk VAR is less than $30,000 and
could even be zero!
Additionally, VAR is not linear with respect to risk horizons. For example, if a
trader’s one-day VAR is $10,000, his or her two-day VAR would likely be less
than $20,000.
March 2004
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Statistics with Applications to Options
Our Goals:
To get a good grasp on the basic statistical concepts associated with
measuring and interpreting an option trader’s VAR;
to understand some of the different ways that VAR can be measured;
to understand the aggregation properties and the horizon properties of
VAR;
to appreciate other applications of basic statistical concepts that are of
interest to option traders.
March 2004
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Statistics with Applications to Options
Basic Statistical Concepts
March 2004
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Statistics with Applications to Options
The Basics: Statistical Concepts and Risk Concepts
Risk measures, including VAR, are based on statistical concepts and tools.
We therefore need to develop the more important of these concepts and tools.
Later, we will see how they are applied to measure VAR. The concepts we will
need are:
random variable
probability
probability distribution
cumulative probability distribution
mean
variance
standard deviation
covariance
correlation
March 2004
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Statistics with Applications to Options
Random Variable:
In the language of statistics, a random variable is anything that will take on a
numeric value at some point in the future where the precise value that it will
take on is not presently known. At the same time, we might know the
probabilities associated with the different values the random variable can take
on.
We have to distinguish a random variable from a number. Once the random
variable takes on a numeric value it is no longer a random variable. It is now
a number.
We begin with the simplest type of random variable.
March 2004
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Statistics with Applications to Options
An experiment with a die:
If I throw a die, there are six possible numeric values that could result. Each of
these values has an associated probability.
As the die is rolling, it is a random variable. Once it stops, it is a number. The
number that results is called the “outcome.”
What are the values and the associated probabilities that this random variable
might take on?
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Statistics with Applications to Options
probabilities
1/6
1
2
3
4
5
6
values
A probability distribution: The complete set of values the random variable might
take on together with their associated probabilities. This is sometimes called a
probability density function (pdf).
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Statistics with Applications to Options
cumulative
probabilities
6/6
5/6
4/6
3/6
2/6
1/6
1
2
3
4
5
6
values
A cumulative probability distribution: The cumulative probability that the value of a
random variable will lie at or below each point.
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Statistics with Applications to Options
Mean: The mean is a measure of the center of a distribution in the sense that 50% of
the outcomes (weighted by their probabilities) will be at or above this central value and
50% will be at or below this value.
Where is the center of this distribution?
probabilities
1/6
1
2
3
4
5
6
values
The center, or mean, of the distribution is sometimes called the expected value.
It is also called the first moment of the distribution.
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Statistics with Applications to Options
Calculating a mean:
The calculation of a mean depends on whether we know the true probabilities
associated with the random variable.
For example, in the case of the die, if I tell you the die is “fair,” I am telling you
that the probability associated with each individual outcome is 1/6. But it is also
possible that I don’t know if the die is fair (that is, certain outcomes might have a
higher probability and others a lower probability).
The calculation of the mean of a random variable depends on whether we do or
do not know the probabilities.
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Statistics with Applications to Options
If we do know the true probabilities:
Denote the random variable by X and the different values the random variable might
take on by Xi . If we do know the true probabilities associated with the outcomes the
random variable might take on, then we can calculate the “true” mean (sometimes
called a population mean). The mean is denoted by the Greek letter  and calculated
as follows:
 =
N
 Xi × Prob[Xi]
i=1
For example, in the case of the die, the mean would be computed as:
 = (1 × 1/6) + (2 × 1/6) + (3 × 1/6) + (4 × 1/6) + (5 × 1/6) + (6 × 1/6)
=
March 2004
3.5
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If we do not know the true probabilities:
If we do not know the true probabilities, we cannot know the true mean. But, we
can formulate an estimate of it by conducting an experiment. Such an estimate is
called a sample mean.
The estimation process can also generate estimates of the probabilities of each
possible outcome. For example, we could roll the die 600 times. Denote the kth
outcome of this test by Ok, and the number of times the distribution is sampled by
Z. Then the sample mean is given by:
Z
 Ok
k=1
 = 
Z
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We tossed the die 600 times. We count the number of times each of the six
possible outcomes occurred and we plot them. This is called a histogram.
number of times
March 2004
103
106
1
2
90
102
3
4
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95
104
5
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From the histogram we can estimate the probabilities of each of the six possible
outcomes. For example, the estimated probability of getting a 1 is (103/600). The
estimated probability of getting a 2 is (106/600). This implies certain assumptions.
What are they?
number of times
March 2004
103
106
1
2
90
102
3
4
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95
5
All rights reserved.
104
6
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Statistics with Applications to Options
number of times
103
106
1
2
90
102
3
4
95
5
104
6
From this we can see that we can get the sample mean by adding up all the outcomes
and dividing by 600. Or, we can multiply each of the six possible outcomes by their
associated estimated probabilities. In either case we get 3.48667.
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Variance: Another important measure associated with a probability distribution is
called the variance. The variance, and a related measure called the standard
deviation, are both measures of dispersion. By measures of dispersion, we mean
measures of how far away from the mean a single outcome drawn at random from the
distribution (e.g., throw the die once) might be.
Calculating a Variance:
The calculation of a variance depends on whether we know the true probabilities
associated with the outcomes of the random variable. If we do, we can calculate a
true variance (sometimes called a population variance) If we don’t, the best we can
do is calculate a sample variance.
Note: Variance is also called the second moment of the distribution.
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Statistics with Applications to Options
If we know the true probabilities:
The variance of a random variable is usually denoted 2. To calculate a true
variance from the known probabilities, we employ the following formula:
2
N
=  (Xi - )2 × Prob[Xi]
i=1
Where X denotes the random variable, Xi denotes the ith possible outcome, 
denotes the mean of the random variable (assumed to have already been
calculated), and N is the number of different values that the random variable can
take on.
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2
N
=  (Xi - )2 × Prob[Xi]
i=1
Let’s use the variance formula above to calculate the variance associated with a
fair die. Again, there are only six possible outcomes (1,2,3,4,5,6) and each has
a probability of 1/6.
2 = [(1 - 3.5)2 × 1/6] + [(2 - 3.5)2 × 1/6] + [(3 - 3.5)2 × 1/6]
+ [(4 - 3.5)2 × 1/6] + [(5 - 3.5)2 × 1/6] + [(6 - 3.5)2 × 1/6]
= 2.91667
March 2004
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Statistics with Applications to Options
If we do not know the true probabilities:
If we do not know the true probabilities, we cannot know the true variance. But,
we can formulate an estimate of it by conducting an experiment. Such an
estimate is called a sample variance. We will denote the sample variance by
2.
Z
 (Ok - )2
k=1
2 = 
Z
Where Z is the number of times the sampling is repeated, and Ok is the
outcome from the kth sampling.
March 2004
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Example:
Consider again the die that we did not know was fair. We tossed the die 600 times.
From the 600 outcomes, we calculated a sample mean of 3.48667. We now use
this sample mean to help get the sample variance.
The sample variance formula produces the following result:
[103 × (1 - 3.48667)2] +[106 × (2 - 3.48667)2] + ….+ [104 × (6 - 3.48667)2]
2 = 
600
= 2.98982
March 2004
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Statistics with Applications to Options
Standard deviation:
The variance is a measure of dispersion, but it suffers from the fact that it is computed
from squared values. To eliminate this squaring effect, we compute a new measure
of dispersion called the standard deviation. This measure is simply the positive
square root of the variance.
 = +
In the case of the fair die:
 =
=
March 2004
2.91667
1.70783
2
In the case of the sampling experiment:
=
=
2.98982
1.72911
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Statistics with Applications to Options
•
March 2004
Alternative Methods of Estimating Volatility
–
Implied Volatility
–
Historic Volatility (RiskMetrics)
–
GARCH(1,1)
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Building an Implied Volatility Matrix
CSCO Calls, Closing Prices, 28 February 2001
Strike
Mar (16) Apr(41) July(132) October(223)
17.50
20.00
22.50
25.00
27.50
30.00
2 1/8
1 1/16
3/16
1/16
NA
NA
3 1/8
1 5/8
9/16
5/16
1/8
NA
6
3 7/16
2 1/4
1 9/16
15/16
9/16
6 1/4
5
2 3/4
2 1/8
1 1/2
1
Numbers in parentheses represent days until option expiration. CSCO
closing price on 28 February 2001 was $19.25. The interest rate term
structure was essentially flat at 4.5% with continuous compounding. A
calendar year is used.
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Statistics with Applications to Options
Implied Volatility Matrix
Strike
March
April
17.50
64.46% 85.14% 114.76% 91.49%
20.00
22.50
25.00
27.50
30.00
84.98
65.63
73.25
NA
NA
74.03
60.83
67.23
66.54
NA
July
78.84
71.28
69.80
65.10
62.44
(skew)
October
86.12 (term structure)
61.88
62.43
60.14
57.41
Example: 74.03% obtained by entering S = 19.25, X = 20.00, T - t = .1123
years (41 days), C = 1.625, and solving for implied volatility.
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•
March 2004
Term structure effects are principally occasioned by:
–
Mean Reversion
–
Scheduled Informational Events
–
Market Segmentations
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•
March 2004
Skews and Smiles are most often occasioned by:
–
Violations of the Assumption that Prices are Log-normally
Distributed
–
Leverage
–
Market Segmentation
–
Option Maturity
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Forward implied volatility is computed from spot implied
volatility much like a forward interest rate is computed from spot
rates. Keep in mind however that option volatility refers to the
annualized standard deviation of the continuously compounded
rate of return of the underlying asset, whereas most forward
rate computations involving interest rates involve a different
compounding frequency (such as a semi-annual periodicity in
the US Treasury bond market).
Question: What is the forward implied volatility of the nearest
at-the-money CSCO call between the third Friday of July and
the third Friday of October?
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Answer: The two relevant spot implied volatility measures are those of the
CSCO July 20 and CSCO October 20 calls. These are 78.84% and
86.12% respectively. Recall there are 132 days until the July expiry and
223 days until the October expiry, and thus we are looking for an estimate
of the 91-day forward vol that begins 132 days hence. Call this FV. The
answer is obtained as follows:
exp(.8612)(223/365) = exp(.7884)(132/365) x exp(FV)(91/365)
(.8612)(223/365) = (.7884)(132/365) + (FV)(91/365)
FV = .9668 = 96.68%.
Note that, just like with interest rate term structures, if the relevant segment
of the spot implied volatility term structure is upward sloping, then the
forward implied volatility term structure that it begets is even more steeply
upwardly sloped. Also note that it is possible to compute other forward
implied vols and therefore entire forward implied volatility term structures.
These may be useful, for example, for pricing forward-starting options such
as executive stock options that have a vesting period.
March 2004
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Statistics with Applications to Options
Measuring Historic Volatility. There are many ways to measure
volatility using historic data. We will focus on methods originally
introduced by Engle (Econometrica, 1982) and used today in
applications such as JP Morgan’s RiskMetrics.
Define V(n) as the variance of a stock, stock index, or other market
variable on day n, as estimated at the end of day n - 1. Let SD(n)
be its square root; SD(n) is commonly called “vol”.
Let the market variable at the end of day i be S(i). Let the variable
U(i) denote the continuously compounded return during day i
(between the end of day i - 1 and the end of day i):
U(i) = ln[S(i)/S(i - 1)].
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A model that estimates V(n) while giving more weight to more recent
data is:
(1) V(n) = E{a(i)[U(n - i)^2]},
where E is a summation operator where i = 1 to m, m is the total
number of observations (sample size) of the daily U(i), and the
variable a(i) represents the amount of weight given to the
observation i days ago. The a’s are positive and a(i) < a(j) when i >
j (because we want to assign less weight to older observations).
The weights must sum to one, that is, E[a(i)] = 1.
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An extension of equation (1) is obtained by assuming that there is a
long-run average volatility and that this should be given some
weight:
(2) V(n) = b(LV) + E{a(i)[U(n - i)^2]},
where LV is long-run volatility and b is the weight assigned to LV.
(Now the sum of b and the a’s must be one.) Equation (2) is known
as an ARCH(m) model, where the acronym ARCH stands for
“AutoRegressive Conditional Heteroscedasticity”. In practice, b(LV)
is replaced by a single variable, say w, when parameters are
estimated.
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The exponentially weighted moving average (EWMA) model is a
particular case of equation (1) where the weights, a(i) decrease
exponentially as we move back through time. Specifically, a(i + 1) =
k[a(i)] where k is a constant between zero and one. This weighting
scheme occasions the following simple formula for updating
volatility estimates:
(3) V(n) = k[V(n - 1)] + (1 - k)[U(n - 1)^2].
Here the estimate of the variance for day n, V(n), which is made at
the end of day n - 1, is calculated from V(n - 1) (the estimate that
was made one day ago of the variance for day n - 1) and U(n - 1)
(the most recent observation on changes in the market variable).
March 2004
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For example, suppose that k is 0.94, which is precisely the value
that JP Morgan uses to update daily volatility estimates in its
RiskMetric database. This value, being so close to 1, produces
estimates of daily volatility that respond relatively slowly to new
information provided by the U(i)^2. Also suppose that the
volatility estimate for day n - 1 is 1% per day, and that the
proportional change in the market variable during day n - 1 is
2%. So V(n - 1) = (0.01)^2 = .0001 and U(n - 1)^2 = (0.02)^2 =
.0004. Equation (3) gives V(n) = 0.94 x .0001 + 0.06 x .0004 =
.000118. The estimate of volatility is therefore (0.000118)^(0.50)
= 0.01086278, or about 1.086% per day, or about 17.24% per
annum using a trading day year (252 days).
March 2004
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A GARCH(1,1) model is a Generalized version of the EWMA model - itself a
particular ARCH model - just described (see Bollerslev, J. of Econometrics,
1986). Here V(n) is calculated from a long-run average variance, LV, as
well as from V(n - 1) and [U(n - 1)^2]. The model is:
(4) V(n) = b(LV) + a[U(n - 1)^2] + c[V(n - 1)].
Here c is a weight assigned to V(n - 1) and now a, b and c must sum to
one. The EWMA model is a nested version of the GARCH model of
equation (4) where b = 0, a = (1 - k) and c = k. [Under GARCH(1,1), V(n) is
based on the most recent observation of U^2 and V. A more general
GARCH(p,q) model calculates V(n) from the most recent p observations on
U^2 and the most recent q observations of V. For asymmetric GARCH
models and other variants, see Nelson (Econometrica, 1990) and Engle
and Ng (J. of Finance, 1993).]
March 2004
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To accommodate parameter estimation, the term b(VL) is usually replaced by a single
parameter w. Once w, a and c have been estimated, b is given by 1 - a - c. The long-term
variance LV is then calculated as w/b. For example, suppose that a GARCH(1,1) model is
estimated from daily data (using maximum likelihood estimation or variance targeting
techniques, c.f. Engle and Mezrich, RISK, 1996) as:
V(n) = .000002 + 0.13 x U(n - 1)^2 + 0.86 x V(n - 1).
This corresponds to w = .000002, a = 0.13, c = 0.86, b = 0.01 and LV = 0.0002. In other
words, the long-run average variance per day is 0.0002, corresponding to a vol of 1.4% per
day. Now suppose that the estimate of the vol on day n - 1 is 1.6% per day so that V(n -1) =
0.000256 and that the proportional change in the market variable on day n - 1 is 1% so that
[U(n - 1)]^2 is 0.0001. Then:
V(n) = 0.000002 + 0.13 x 0.0001 + 0.86 x 0.000256 = 0.00023516.
The new estimate of the volatility is therefore (0.00023516)^(0.50) = 0.0153 or 1.53% per
day.
March 2004
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Recall that the GARCH(1,1) model is similar to the EWMA model except that, in
addition to assigning weights that decline exponentially to past U^2, it also assigns
some weight to the long-run average volatility. Because of this added feature,
GARCH models can accommodate mean reversion in the volatility. Indeed, the
parameter c in the model is a type of “decay rate” similar to the parameter k in the
EWMA model. And for GARCH(1,1) per se, the variance V(n) exhibits mean reversion
with a reversion level of LV and a reversion rate of 1 - (a + c). (In the EWMA model,
(a + c) = 1 so the reversion rate is zero.)
March 2004
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Building an Historic Volatility Term Structure. Because GARCH models accommodate mean
reversion, they can be used to build forecasts of entire volatility term structures. These are
call “historic volatility term structures” because they are built using historic data samples and
are therefore not to be confused with implied volatility term structures. Still, one might
already envision trading strategies based on comparisons of the two term structures.
It can be shown that the GARCH(1,1) model described in equation (4) occasions the
following estimate of future variance:
(5) V(n + f) = LV + {[(a + c)^f] x (V(n) - LV)}
where V(n + f) is an estimate of the variance to occur on day n + f in the future. Notice that
when a + c < 1, the final term in equation (5) becomes progressively smaller as f increases.
(If a + c > 1 then the variance would not be reverting but would be “fleeing”.)
March 2004
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To build a historic volatility term structure suitable for options,
consider an option that lasting between day n and day n + N. One
can use equation (5) to compute the expected variance during the
life of the option as:
(6) (1/N) x S[V(n + f)],
where S is a summation operator for f = 0,…,N - 1.
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Suppose that for a particular stock, a + c = 0.9602 and LV = 0.00004422 (a
long-run daily vol of 0.66498%, or a trading-day yearly vol of 10.556%).
Also suppose that the current variance per day, V(n), is 0.00006. This
corresponds to a daily vol of 0.77460%, which is greater than LV, and an
annual vol of 12.30% based on a 252 trading-day year. The first table
below shows the historic volatility term structure (% per annum for a
calendar day year) based on these data and equation (6), while the second
table shows the impact on the term structure of a 1% change in the
instantaneous volatility. Of course, once one has a volatility term structure
like that in the first table below, one can always compute a forward vol and
indeed an entire forward vol term structure a la getting a forward implied vol
from an implied volatility term structure.
Notice in first table how the vol predicted in 500 days is closing in on the LV
value of 10.556%. Also notice that the term structure is downward sloping,
so the forward term structure would be even more steeply downwardly
sloped.
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Option Life (Days)
10
30
Option Volatility
12.03
11.61
(% per annum, 252-day trading year)
50
11.35
100
11.01
500
10.65
Option Life (Days)
Option Volatility Now
After 1% Change
Increase in Volatility
50
11.35
11.83
0.48
100
11.01
11.29
0.28
500
10.65
10.71
0.06
10
12.03
12.89
0.86
30
11.61
12.25
0.64
Note: 12.30% current daily vol moves to 13.30%. So the daily vol becomes
0.84% and the daily variance becomes 0.00007016. Applying equation (6)
to this new situation produces the second row in the second table.
March 2004
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Multiple Random Variables:
In situations involving statistics, we are often called upon to examine the
statistical behavior of multiple random variables. Specifically, we might be
interested in knowing how two random variables vary with respect to one
another.
For example, if we were to randomly select children at a local elementary school
and we were to measure their ages in months and their heights in inches, we
have in essence obtained joint observations on two different random variables.
What would we expect to find?
Do the two variables tend to move together?
How so?
Is the relationship perfect?
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What would it mean for the relationship to be perfect?
Height
Age
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What would it mean for the relationship to be perfect?
Height
Notice that the random variable called
height has a mean and a standard deviation.
Age
Is this what we are likely to find?
March 2004
Notice that the random variable called
age has a mean and a standard deviation.
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Height
Age
How could we describe this in
a statistical sense?
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Denote the two random variables X and Y
Y
Y
X
Positive Covariance
March 2004
Y
X
Negative Covariance
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X
Zero Covariance
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Measuring Covariance:
True Covariance:
N
CovX,Y =
M
  (Xi - µX) × (Yj - µY) × Prob[Xi and Yj]
i=1 j=1
Sample Covariance:
Z
 (Ox,k - uX) × (OY,k - uY)
k=1
CovX,Y = 
Z
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The problems with covariance:
As a measure of the degree to which two random variables move together, covariance has
two weaknesses.
The first of these concerns the choice of units of measure:
Age:
Years
Months
Weeks
Days
Seconds
Height:
Feet
Inches
Meters
Centimeters
Millimeters
Second, there is no theoretical upper or lower limit to a covariance. That is, covariances can
be extremely large positive numbers or extremely large negative numbers.
Both of these make it inconvenient to work with covariances.
March 2004
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Correlation Coefficients:
To address the weaknesses of covariance, statisticians developed the concept of a
correlation coefficient. A correlation coefficient is sometimes called a “standardized
covariance.” Correlation is usually symbolized by the Greek letter rho ():
X,Y =
CovX,Y

X × Y
True correlations are obtained from true covariances and true standard deviations
Sample correlations are obtained from sample covariances and sample standard
deviations.
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Correlation coefficients are bounded between +1 (called perfect positive correlation) and
–1 (perfect negative correlation).
CovX,Y
X,Y = 
X × Y
When the CovX,Y is positive the correlation is positive.
When the CovX,Y is negative, the correlation is negative.
When the CovX,Y is zero, the correlation is zero.
March 2004
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•
Simple Linear Regression
Two variables X and Y be said to be linearly related if their relationship can be
expressed by the following simple linear model:
yi = α + βxi + ei
Where yi is the value of the Y (“dependent”) variable for a typical unit of association
from the population for Y, xi is the value of the X (“independent” or “explanatory”)
variable for that same unit of association, α and β are parameters called the
regression constant and slope coefficients, respectively, and ei is a random
variable that is i.n.d. with mean 0 and variance equal to that of Y.
Under some classical assumptions regarding X and Y, we can estimate α and β,
estimate the strength of the linear relationship between X and Y, and conduct
certain significance tests.
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Using the least-squares criterion for best fit (whereby we estimate the intercept and slope
coefficients by minimizing the sum of the squared errors, where said errors represent the
distances between the X,Y coordinates and the fitted line), we have the following formulas
for a (our estimate of α) and b (our estimate of β):
a = y – bx
(1)
b = [nΣxiyi - ΣxiΣyi]/[nΣxi2 – (Σxi)2]
(2)
Where y is the mean value of yi (i = 1 to n, our sample size and the limits of the summation
operators), and x is the mean value of xi.
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For example, suppose that we have the following data relating production (X) and manufacturing
expenses (Y) for 10 firms:
X
Production (thousands of units)
40
42
48
55
65
79
88
100
120
140
March 2004
Y
Expenses (thousands of dollars)
150
140
160
170
150
162
185
165
190
185
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Then our estimates of a and b from equations 1 and 2 above are: a = 134.72, b= 0.3978. And
thus we have our estimated equation:
yc = 134.72 + 0.3978x,
Where yc is the calculated value of Y for a given X.
The “coefficient of determination” (usually denoted R2) provides an objective measure of the
goodness of the fit of the estimated equation. To understand how it is computed, first
consider how the “total deviation” for a particular value of yi from its calculated or “predicted”
value yc is decomposed partly into “explained deviation” and partly into “unexplained
deviation”:
(yi – y)
total deviation
March 2004
=
(yc – y)
explained deviation
+ (yi – yc)
unexplained deviation
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For example, in the case of our ninth observation (y9 = 190), we have 24.3 = 16.8 + 7.5.
Performing similar calculations for all ten observations in our sample, and squaring every
measure (so as to ensure that positive and negative total deviations do not offset), we get:
Σ(yi – y)2 =
Total Sum
of Squares
Σ(yc – y)2
Explained Sum
of Squares
+ Σ(yi – yc)2
Unexplained Sum
of Squares
For our sample, TSS = 2554.10, ESS = 1666.33, and USS = 887.77. In turn, the coefficient of
determination is simply the ratio of the ESS over TSS:
R2 = ESS/TSS = 1666.33/2544.10 = 0.6538 = 65.38%.
Here we would say that, subject to the assumptions invoked, about 65% of the total variability in
production expenses (Y) is due to/explained by the amount of units produced (X).
March 2004
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The correlation coefficient between Y and X is simply the square root of the coefficient of
determination, or here 0.8086 or about 81%. Here we say that production costs and output
are about 81% correlated.
Now we turn our attention to tests of significance. Such tests require a procedure known as
“analysis of variance” or ANOVA. You may recall the follow ANOVA table for simple linear
regression:
Source of
Variation
Regression
Error
SS
ESS
USS
df
1
n–2
F-statistic
ESS/[USS/(n – 2)]
For our sample, the F-statistic (with 1 and 8 degrees of freedom) is 1666.33/[(887.77/8)] = 14.99,
which is greater than 14.69, thus indicating that we can reject the null hypothesis of no linear
relationship between Y and X (R2 = 0) at the 0.01 level.
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We can similarly test for the significance of the estimated coefficients. For instance, the test
statistic for determining whether or not b is significantly different from zero is simply the
square root of the F-statistic, or 3.87. This figure is greater than 2.306 – the critical value of
t for a two-sided test with 8 degrees of freedom at the 5% level. Hence we conclude that b
is positive and statistically significant.
The t-statistic for b is also given by the estimated value of b (0.3978) divided by the standard
deviation of b. We can solve for the latter now: 3.87 = 0.3978/s, so s = 0.1028. This in turn
permits us to compute a confidence interval for b. For example, the 95% confidence interval
for b is 0.3978 +/- 2.306(0.1028) = [0.1608,0.6348].
We could go on and on here.
March 2004
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Types of distributions:
We started out our discussion of statistical distributions using one of the simplest types
of distributions.
This distribution had the properties that it was discrete and uniform.
By discrete distribution we mean that there are only a finite number of outcomes.
By uniform, we mean that every possible outcome has the same probability.
Some real life distributions do have these characteristics. But, most don’t.
March 2004
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As an example, define a new random variable as the number of dots facing up when a pair
of dice are thrown together. There are a total of twelve things that can happen and they do
not all have the same probability:
Possible Outcome
2
3
4
5
6
7
8
9
10
11
12
Probability
1/36
2/36
3/36
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36
[1,1]
[1,2
[1,3
[1,4
[1,5
[1,6
[2,6
[3,6
[4,6
[5,6
[6,6]
2,1]
2,2
2,3
2,4
2,5
3,5
4,5
5,5
6,5]
3,1]
3,2
3,3
3,4
4,4
5,4
6,4]
4,1]
4,2 5,1]
4,3 5,2 6,1]
5,3 6,2]
6,3]
What do the probability distribution and the cumulative probability distribution look like?
March 2004
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probability
6/36
5/36
This is discrete, but it
is not uniform!
4/36
3/36
2/36
1/36
2
March 2004
3
4
5
6
7
8
9
10 11
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12
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value
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Statistics with Applications to Options
cumulative
probability
36/36
30/36
24/36
18/36
12/36
6/36
2
March 2004
3
4
5
6
7
8
9
10 11
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12
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value
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Continuous distributions:
The distribution that we just described for the pair of dice is an example of a type of
discrete distribution called a binomial distribution. It is discrete, but the outcomes do
not have equal probability.
We now consider distributions that are not discrete.
Distributions that are not discrete are said to be continuous. A random variable is said to
be continuous if it can take on an infinite number of different values over some range of
values.
March 2004
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For example, the height of a child selected at random could have any value ranging
from the height of the smallest child on earth to the height of the tallest child on earth.
Suppose that the shortest child is 15 inches and the tallest child is 70 inches. How
many different heights are there between 15 inches and 70 inches if we measure with
absolute precision?
Answer: an infinite number of heights.
March 2004
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There are many types of continuous distributions but the most common type is one
called a normal distribution.
The normal distribution has several important characteristics:
It is continuous
It is symmetric
It can take on any value from negative infinity to positive infinity
It is fully described by its mean and its standard deviation (or variance)
It is extremely well defined mathematically
March 2004
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probability
(pdf)
X ~ N(µ, )
value
March 2004
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probability
(pdf)

µ
value
March 2004
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probability
(pdf)
X ~ N(µX, X)
Y ~ N(µY, Y)
X
X
Y
Y
µ
value
March 2004
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probability
(pdf)
Confidence Interval: 90%

µ
value
µ – 1.645
March 2004
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µ + 1.645
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Confidence Intervals: The role of z scores
probability
z = 1.645
z = 1.960
z = 2.326
z = 2.576
90% confidence interval
95% confidence interval
98% confidence
interval

99% confidence interval
Confidence Interval
Probability
µ
µ – z
March 2004
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µ + z
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value
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Example: A random variable that is normally distributed has a mean of 50 and a standard deviation
of 10. What is the 90% confidence interval?

probability
X ~ N(50, 10)
µ
µ – 1.645
33.55
March 2004
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µ + 1.645
66.45
All rights reserved.
value
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Cumulative
probability
1.2
1
0.8
0.6
0.4
0.2
0
March 2004
value
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Skewness (a measure of symmetry, called the third moment)
Skewed left
Symmetric (no skew)
Skewed right
Kurtosis (the tails, called the fourth moment)
Platykurtic: Data distributed heavily in midregion (thin tails)
Mesokurtic: Data is consistent with a normal distribution
Leptokurtic: Data is heavily distributed in the tails (fat tails)
March 2004
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Risk Profiles: Depicting an Exposure to a Single Market Risk
March 2004
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Commercial activity often exposes a corporation to one or more market risks.
Example:
Suppose that a farmer has 1000 acres of land on which he can plant wheat. Suppose
that the cost of growing a bushel of wheat is $4.00. Each acre of land will generate
30 bushels of wheat. Thus, his maximum production is 30,000 bushels.
Assume that there is no quantity uncertainty.
March 2004
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The farmer would plant the wheat in April and harvest it in October (six months later).
The spot price of wheat at the time of decision making (planting) is $4.90 a bushel.
The forward price of wheat (for October delivery) is $5.00 a bushel.
Which price counts?
If the spot price at harvest is equal to the current forward price for October,
how much profit does he make?
What risks does he run?
How can we depict this risk?
March 2004
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Profit = Quantity × Profit per Bushel
= Quantity × (Market Price - Cost of Production)
= 30,000 × (Price - $4.00)
What if the price is $5.00 in October? Profit =
What if the price is $2.00 in October? Profit =
What if the price is $7.00 in October? Profit =
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Profit = 30,000 × (Market Price - $4.00)
Risk Profile
Profit
$180,000
$120,000
$60,000
0
-$60,0000
$0
$1
$2
$3
$4
$5
$6
$7
$8
$9
Price in October
March 2004
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At the time of planting, the price of wheat in October is a random variable:
It has a mean, it has a standard deviation, and it has some type of probability
distribution.
Let’s assume that the distribution is normal.
The mean is $5.
The standard deviation is $0.75.
Can we depict this with a normal curve and show the 90% confidence interval?
March 2004
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Price ~ N(5, 0.75)
$2
$3
$4
$5
$6
$7
$8
value
µ
$3.766
$5 – 1.645($0.75)
March 2004
$6.234
$5 + 1.645($0.75)
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Notice that the translation from price per unit to profit is a linear transformation!
probability
Profit

$90,000
$60,000
$30,000
0
-$30,000
-$60,0000
$2
$3
$4
$3.766
$5 – 1.645($0.75)
March 2004
$5
µ
$6
$7
$8
value
$6.234
$5 + 1.645($0.75)
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How can we convert the mean price and the standard deviation of price to a mean
profit and a standard deviation of profit?
Mean profit = Quantity × (mean price - cost of production)
= 30,000 × ($5.00 - $4.00)
= $30,000
Standard deviation = Quantity × standard deviation of per unit profit
= 30,000 × $0.75
= $22,500
March 2004
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What is this farmer’s six-month VAR @ 95%?
probability

-$60000
$2
$3
$0
$30000
$4
$5
$30000 - 1.645($22,500)
March 2004
$90,000
$6
$7
Profit
$8
Price per bushel
30000 + 1.645($22,500)
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Would every wheat farmer have the same six-month VAR?
Suppose another farmer had 2000 acres on which he is planting wheat,
what does his risk profile look like relative to the first farmer?
What is the second farmer’s six-month VAR?
March 2004
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Farmer 2
Risk Profile
Profit
Farmer 1
$120,000
$60,000
0
-$60,0000
$0
March 2004
$1
$2
$3
$4
$5
$6
$7
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$8
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$9
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probability
-$60000
-$120000
$0
$0
$30000
$60000
$30000 - 1.645($22,500)
$60000 - 1.645($45,000)
March 2004
$90000
$180000
Profit Farmer 1
Profit Farmer 2
$30000 + 1.645($22,500)
$60000 + 1.645($45,000)
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The Roots of VAR: Portfolio Theory
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– Portfolio Theory is concerned with
• Risk = standard deviation of return
– for single assets
– for portfolios of assets
• Reward = mean return
– for single assets
– for portfolios of assets
– Portfolio theory was originally developed in the context of stock
portfolios and we will initially adopt this for convenience.
March 2004
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•
Measuring total return on a stock
– components of return
• dividend component
• gain component
– dollar returns vs percentage returns
R =
D + ( P E - PB)

PB
– annualizing the percentage return
• cumulative total return
• annual total return (we will call it the rate of return)
March 2004
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•
March 2004
Return as a random variable
– why is the rate of return a random variable?
– measuring expected (mean) return
– measuring the standard deviation of return
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Combining stocks to form portfolios:
–
–
–
–
–
March 2004
Harry Markowitz (1952)
return on a portfolio
mean return of a portfolio
variance of a portfolio
standard deviation of a portfolio
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•
Let there be N stocks.
•
•
•
•
•
Denote the return on stock i by Ri
Denote the mean return on stock i by i
Denote the standard deviation of return on stock i by i
Denote the covariance of returns of any two stocks i and j by covi,j
Denote the correlation coefficient for the returns on any two stocks by i,j
•
where
March 2004
i = 1,2,3,...,N and j = 1,2,3,...,N in all cases
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Let Rp denote the return on a stock portfolio
Let p denote the mean return on a stock portfolio
Let p2 denote the variance of return on a stock portfolio
Markowitz showed:
N
Rp =  wiRi
i=1
N
p =
 wii
i=1
N
p2
=
N
  wiwj covi,j
i=1 j=1
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Suppose that you plan to invest your money in a portfolio of three stocks as below.
By coincidence, each of the three stocks has an expected return of 15%. What is
the expected return on your portfolio?
Stock
1
2
3
Company
IBM
MSFT
INTC
Weight
30% (.3)
20% (.2)
50% (.5)
Mean Return
15%
15%
15%
N
p =
 wii
i=1
= (.3  15%) + (.2  15%) + (.5  15%)
= 15.0%
March 2004
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Suppose that after one year, the actual returns on the stocks in your portfolio are as
below.
Stock
1
2
3
Company
IBM
MSFT
INTC
Weight
.3
.2
.5
Actual return
14%
40%
10%
What was the actual return on your portfolio?
March 2004
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Stock
1
2
3
Company
IBM
MSFT
INTC
Weight
.3
.2
.5
Actual return
14%
40%
10%
What was the actual return on your portfolio?
N
Rp =  wiRi
i=1
= (.3  14%) + (.2  40%) + (.5  10%)
=
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17.2%
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Notice that the actual return on the portfolio deviated from the expected return on the
portfolio. That is, there was deviation from the expected. It is this potential for
deviation from expected return that we understand as risk.
Portfolio risk is measured as the variance (or its square root, the standard deviation) of
return.
N
p2 =
N
  wiwjcovi,j
i=1 j=1
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N
p2 =
N
  wiwj covi,j
i=1 j=1
Recall that a covariance is related to a correlation coefficient as follows:
covi,j
i,j = 
i × j
Therefore,
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covi,j = ij i,j
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Therefore:
p2
March 2004
N
=
N
  wiwj ij i,j
i=1 j=1
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p2
N N
=
  wiwj ij i,j
i=1 j=1
When i = j then ij = i2 , wiwj = wi2 , and i,j = 1.
p2
March 2004
N
=

wi2 i2
i =1
N
+
N
  wiwj ij i,j
i  j
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p2
Stock
1
2
3
March 2004
N
=
Company
IBM
MSFT
INTC

i=1
wi2 i2
Weight
.3
.2
.5
N N
+
  wiwj ij i,j
ij
Std. Dev.
19%
19%
19%
Copyright (c) 2001-2004 by Marshall, Tucker & Associates, LLC
Correlation
IBM MSFT INTC
1.0
0.5
0.4
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0.5
1.0
0.3
0.4
0.3
1.0
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p2
Stock
1
2
3
p2 =
N
=
Company
IBM
MSFT
INTC

i=1
wi2 i2
Weight
.3
.2
.5
N N
+
  wiwj ij i,j
ij
Std. Dev.
19%
19%
19%
Correlation
IBM MSFT INTC
1.0
0.5
0.4
0.5
1.0
0.3
0.4
0.3
1.0
(.32  .192) + (.22 .192) + (.52  .192)
+ (.3  .2  .19  .19  .5) + (.3  .5  .19  .19  .4)
+ (.2  .3  .19  .19  .5) + (.2  .5  .19  .19  .3)
+ (.5  .3  .19  .19  .4) + (.5  .2  .19  .19  .3)
March 2004
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Stock
Company
Weight
Std. Dev.
1
2
3
IBM
MSFT
INTC
.3
.2
.5
19%
19%
19%
p2 =
Correlation
IBM MSFT INTC
1.0
0.5
0.4
0.5
1.0
0.3
0.4
0.3
1.0
(.32  .192) + (.22 .192) + (.52  .192)
+ (.3  .2  .19  .19  .5) + (.3  .5  .19  .19  .4)
+ (.2  .3  .19  .19  .5) + (.2  .5  .19  .19  .3)
+ (.5  .3  .19  .19  .4) + (.5  .2  .19  .19  .3)
= 0.013718 + 0.008664
= 0.022382
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p2 =
0.022382
Therefore:
p =
=
March 2004
0.1496
14.96%
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Summarizing:
p = 15.00%
p = 14.96%
Conclusions:
What is the effect of diversification on expected return?
What is the effect of diversification on risk?
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p2
N
=

i=1
wi2 i2
N N
+
  wiwj ij i,j
ij
Diversification and risk
– unsystematic risk (diversifiable risk, also called specific risk)
• that risk which is company specific (e.g., a fire at a plant)
• caused by changes in those variables that impact a single company or
only a small group of companies (e.g., an explosion at a production
facility)
– systematic risk (non-diversifiable risk, also called market risk)
• that risk which is shared by all or most firms
• caused by changes in those variables that impact most companies
simultaneously (e.g., interest rates)
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Standard deviation of return
If a portfolio is well diversified, then the portfolio’s
standard deviation is all systematic risk. If a portfolio
is not well diversified, then the portfolio’s standard
deviation is partly unsystematic risk and partly
systematic risk.
unsystematic risk
systematic risk
10
March 2004
20
30
40
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Number of Stocks in Portfolio
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The Markowitz measures of portfolio risk and return can be used to develop a
confidence interval and a VAR-like measure.
Suppose we ask:
“What is the 90% confidence interval for the rate of return on a stock
portfolio if the portfolio’s annual rate of return has a mean of 15.00% and a
standard deviation of 14.96%?”
To answer this question, we must make some assumption about the nature
of the underlying distribution. We will assume that it is normal.
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probability
14.96%
90%
5%
5%
– 9.61%
15.00%
39.61%
90% confidence interval = 15.00% ± (1.645 × 14.96%)
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Annual rate of return
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From Portfolio Theory to Parametric VAR
(Variance-Covariance VAR)
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The first approach to measuring VAR was based on Markowitz’s measures of a
portfolio’s mean return and a portfolio’s standard deviation of return.
Because this measure was based on two statistical parameters (mean and
standard deviation) and the assumption that the probability distribution is
normal, the method is known as parametric VAR or Variance-Covariance VAR.
The calculation of parametric VAR requires two important adjustments to
Markowitz original measures of a portfolio’s parameters.
March 2004
1.
VAR is measured in number of monetary units (e.g., USD, JPY,
EUR, etc.). Markowitz measures are in terms of rates of return.
2.
VAR is measured over any desired risk horizon (1 day, 2 days, 3 days,
etc.). Markowitz assumed the investment horizon was one year in
length.
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The first adjustment requires that we re-state the mean and the standard
deviation in terms of monetary units, instead of percentage returns. We will
use dollars as the monetary units.
This adjustment is straightforward enough. Since percentage returns are
nothing more than the return per $1, we simply substitute the number of
dollars invested in each asset for the weight on that asset.
March 2004
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The expected profit becomes:
N
µ =
 Ii µi
i=1
where
µ denotes the expected (mean) dollar profit over the risk horizon
Ii denotes the number of dollars invested in asset i.
µi denotes the expected per dollar return on asset i over the risk
horizon. This is the same as the percentage return.
March 2004
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The variance of profit becomes:
N
2 =
N
  IiIjCovi,j
i=1 j=1
where
2 denotes the variance of dollar profit measured over the risk horizon
Ii and Ij denote the number of dollars invested in assets i and j respectively.
Covi,j denotes the covariance of the returns (measured over periods that
correspond to the length of the risk horizon) per dollar invested in asset i and
per dollar invested in asset j. This is the same as percentage returns.
March 2004
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The second adjustment is also straightforward. Markowitz assumed, for simplicity, that
percentage returns would be measured over a one-year period. This is why they are
called “rates of return.” But there is no reason that percentage returns need to be
measured over a period of one year. They could just as easily be measured over a period
of 1 hour, 1 day, 1 week, 1 month, and so forth.
When using VAR analysis, risk managers may use several risk horizons. That is, they
may state a one-day VAR, a one-week VAR, and a one-month VAR.
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It turns out that, if you know the statistical parameters to get a one-day VAR it is a simple
matter to get a VAR for any horizon one likes. For this reason, most publicly available
databases that can be tapped for the necessary covariance information use covariances for
$1 investments measured over one-day. This information is presented in matrix form and
is called a variance-covariance matrix.
The first firm to provide this data in a form that would feed into a parametric VAR model
was JP Morgan. The JP Morgan system together with its variance-covariance matrix
database is called RiskMetrics.
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Riskmetrics, and other systems like it, provide covariances that are based on some number
of days of data.
For example, suppose that the covariances are calculated using daily data for the past 45
market days. Each day one new day of data is added and the earliest day of data is deleted.
Thus, these are moving covariances.
1
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2
3
4
5
•
•
•
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44
45
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46
47
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Example of a Variance-Covariance Matrix (values measured in dollars per $1):
1-Day Variance-Covariance
2
3
Asset
1
1
0.000010
0.000008
–0.000002
0.000003
2
0.000008
0.000018
–0.000005
0.000004
3
–0.000002
–0.000005
0.000025
–0.000001
4
0.000003
0.000004
–0.000001
4
0.000015
Note: This variance-covariance matrix is a description of the market environment.
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1-Day Variance-Covariance
2
3
Asset
1
4
1
0.000010
0.000008
–0.000002
0.000003
2
0.000008
0.000018
–0.000005
0.000004
3
–0.000002
–0.000005
0.000025
–0.000001
4
0.000003
0.000004
–0.000001
0.000015
Suppose that a cash market trader has positions in these four assets only. He has a $10 mm long
position in Asset 1, a $15 mm short position in Asset 2, a $5 mm long position in Asset 3, and a
$10 mm short position in Asset 4. The expected returns on these positions are as follows:
Asset
1
2
3
4
March 2004
1-Day expected Return per $1 of investment (assumes long position)
$0.00027
firm’s estimates of the
–$0.00021
expected profits from its
$0.00035
positions (per $1 invested).
–$0.00030
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Calculate the Trader’s One Day VAR at 95%:
Step 1: Calculate the expected 1-day dollar return
Asset
1
2
3
4
Size of Position × Expected Return Per $1
$10,000,000
– $15,000,000
$5,000,000
–$10,000,000
$0.00027
–$0.00021
$0.00035
–$0.00030
=
Product
$2,700
$3,150
$1,750
$3,000
$10,600
March 2004
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Step 2: Calculate the Variance and Standard Deviation of 1-day profit:
Dollars Asset i
1,1
1,2
1,3
1,4
2,1
2,2
2,3
2,4
3.1
3,2
3,3
3,4
4,1
4,2
4,3
4,4
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10,000,000
10,000,000
10,000,000
10,000,000
-15,000,000
-15,000,000
-15,000,000
-15,000,000
5,000,000
5,000,000
5,000,000
5,000,000
-10,000,000
-10,000,000
-10,000,000
-10,000,000
×
Dollars Asset j
10,000,000
-15,000,000
5,000,000
-10,000,000
10,000,000
-15,000,000
5,000,000
-10,000,000
10,000,000
-15,000,000
5,000,000
-10,000,000
10,000,000
-15,000,000
5,000,000
-10,000,000
×
Covariance per dollar
=
0.000010
0.000008
-0.000002
0.000003
0.000008
0.000018
-0.000005
0.000004
-0.000002
-0.000005
0.000025
-0.000001
0.000003
0.000004
-0.000001
0.000015
Copyright (c) 2001-2004 by Marshall, Tucker & Associates, LLC
1000000000
-1200000000
-100000000
-300000000
-1200000000
4050000000
375000000
600000000
-100000000
375000000
625000000
50000000
-300000000
600000000
50000000
1500000000
2
=

=
All rights reserved.
6,025,000,000
$77,621
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Step 3: Calculate the 1-day VAR at the desired confidence level:
Note:
95% VAR z-score corresponds to 90% confidence interval z-score.
99% VAR z-score corresponds to 98% confidence interval z-score.
95% 1 day VAR z-score is 1.645, therefore 1-day VAR is calculated as follows:
VAR(1 day, 95%) =
µ – (1.645 × )
= $10,600 – (1.645 × $77,621)
= –$117,087
Thus, the maximum 1-day loss at a 95% level of confidence, is $117,087.
March 2004
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Note: Because a “loss” is understood to be a negative number, it would not be
appropriate to say:
The VAR is –$117,087, since VAR is generally understood to be a loss.
Thus, we usually drop the negative sign when quoting VAR.
An alternative way to avoid this sign confusion is to measure the VAR as follows:
VAR = (1.645 × ) – µ
This is the usual convention and we will use it from here on.
March 2004
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Note:
Some providers of VAR variance-covariance matrices assume that the VAR
will be measured at a specific level (usually 95%) and they pre-multiply each
covariance in the matrix by 1.6452.
When this is done, the variance-covariance matrix is a VAR-scaled
variance-covariance matrix. As a result, the standard deviation obtained is
a VAR-scaled standard deviation and has to be interpreted as such.
VAR =
 (VAR-scaled) – µ
It is important to read the technical document that accompanies variance-covariance matrix data to know
whether or not it has been scaled. This is available from the data supplier.
March 2004
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Adjusting VAR for longer risk horizons:
The procedure we just described provided a one-day VAR at a 95% level of confidence. We
know that we can get a one-day VAR at a higher or lower level of confidence by simply
adjusting the z-score.
But, what if the risk manager has been asked for say a 2-day VAR or a one-week (7-day) VAR or
a 1-month VAR.† How do we make these adjustments?
This turns out to be very simple. It will be approximately true that the T-day mean profit µ(T)
will be related to the one-day mean profit µ and the T-day standard deviation (T) will be
related to the one-day standard deviation  as follows:
µ(T) = T × µ
(T) =
T × 
†
It is important to know whether the variance-covariance matrix was generated on the basis of a trading day year
(251 days) or a calendar day year (365 days). For example, a week is 5 days in the former case and 7 days in the latter
case.
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µ(T) = T × µ
(T) =
T × 
Consider again the case we looked at earlier in which µ = $10,600 and  = $77,671
Risk Horizon
Mean
1 day
2 days
7 days
30 days
$10,600
21,200
74,200
318,000
Standard Deviation
VAR @ 95%
$77,621
109,773
205,366
425,148
$117,087
159,376
263,627
381,368
VAR = 1.645(T) – (T)
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VAR
1 2
March 2004
7
30
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Risk Horizon in days
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A note on daily expected profit:
Because daily expected profit is generally quite small relative to daily standard
deviation and daily expected profit is very hard to estimate, many users of VAR
simply assume that daily expected profit is zero.
March 2004
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Alternative Approaches to Measuring VAR
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Alternative Approaches to Measuring VAR:
The parametric approach to measuring VAR assumes that the risk manager has
available the relevant variance-covariance matrix and assumes that the distribution is
normal (or at least approximately so).
If either of these conditions is not satisfied, the parametric method would not seem to
fit the bill.
Another inherent problem with parametric VAR is that it assumes a linear
correspondence between the source of market risk and the size of the trader’s
exposure. This is not always the case. For example, if the trader’s portfolio
contains options or embedded options, this condition is violated.
There are several other methods for obtaining a VAR. Each has its strengths and
each its weaknesses. We will look at two of these: Historic VAR and Simulated VAR.
March 2004
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Historic VAR
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Historic VAR:
Historic VAR is a method of calculating VAR which assumes that the future will
be very much like the past.
In such a scenario, we can ask “What was the trader’s daily P&L each day for
the last 100 days?”
This information would be available to any trader that marks all of his or her
positions to market at the end of each day to generate a daily P&L.
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Once we have the daily P&Ls, we simply list them (rank them) from the highest to lowest:
Rank
Daily P&L
1
+$182,300
2
+$174,100
3
+$164,900
4
+$164,500
5
+$151,000
6
+$142,500
•
•
•
•
•
•
94
–$107,400
95
–$112,400
VAR @ 95%
96
–$115,300
97
–$120,200
98
–$127,000
99
–$134,100
100
–$142,500
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Historic P&L data can also be organized into a histogram. For example, we might say what
percentage of the time did the daily P&L lie between +$185,000 and $190,000 (answer 0)? What
percentage of the time did it lie between +$180,000 and $185,000, and so forth? This can be
plotted.
percentage
We can now ask “How does
this compare with a normal
distribution?”
daily P&L ranges
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Weaknesses of Historic VAR:
Historic VAR ignores important issues:
Has the scale of operations changed?
Has the trading strategy changed?
Has the use of leverage changed?
Have the instruments we trade changed?
Is the past a good indicator of the future?
shocks to the system
state of the economy
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Simulated VAR
(Monte Carlo VAR)
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Simulated Value-at-Risk:
Simulated VAR (also called Monte Carlo VAR) is most useful when the trader
whose VAR we are attempting to estimate holds positions that have non-linear
payoffs (also called non-linear risk profiles).
This will be the case whenever the trader holds positions in options, securities
with embedded options, or securities other than options that have non-linear
payoffs--such as bonds (e.g., convexity).
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To demonstrate why parametric VAR fails to work properly in cases involving nonlinear payoffs, let’s consider a trader with a long position in a straddle.
Straddles:
A long straddle involves a long position in both a call option and a put option on
the same underlying asset with the same expiration date and the same strike
price.
A short straddle involves a short position in both a call option and a put option on
the same underlying asset with the same expiration date and the same strike
price.
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Because a straddle is a combination of a call and a put, we need to take a moment
to look at the value diagrams of calls and puts.
A call option grants its owner the right but not the obligation to buy the underlying
asset from the option writer for the strike price written into the option contract. This
right is good for a limited time, called the time to expiry.
At expiration, the value of a call option on a single unit of the underlying is given by:
Value = max[S-X, 0]
where S = spot price of the underlying asset
X = strike price of the option
Prior to expiry, the value of the option is given by:
Value = max[S-X, 0] + time value
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A put option grants its owner the right but not the obligation to sell the underlying
asset to the option writer for the strike price written into the option contract. As
with a call, this right is good for a limited time called the time to expiry.
At expiration, the value of a put option is given by:
Value = max[X-S, 0]
where S = spot price of the underlying asset
X = strike price of the option
Prior to expiry, the value of the option is given by:
value.
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Value = max[X-S, 0] + time
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The valuation of an option at any time prior to expiry is complex and requires the
employment of an appropriate model.
Models are an attempt to mathematically capture the importance of the various
factors1 that determine an instrument’s value. In this case, the instrument is the
option. Often, there is less than universal agreement as to which of several
models is the “best” model for valuing a particular option.
The most widely known of option pricing models is the Black-Scholes-Merton
model.
1
The factors that determine the value of an instrument are sometimes called value drivers.
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Building a model requires that we make assumptions. For example, the BlackScholes-Merton model assumes:
The price of the underlying follows a random walk through time
The future price of the underlying asset is lognormally distributed
The volatility of the price of the underlying is constant
The risk-free rate of interest is constant
Trading is continuous (in both time and in the size of transactions)
There is no transaction cost associated with trading
Black/Scholes assumes no payout
Merton allows for a payout
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While the details will differ a bit, the value of an option, irrespective of which model is
employed, will be a function of five key variables, which constitute the value drivers:
the current price of the underlying (denoted S)
the strike price of the option (denoted X)
the annual volatility (denoted v)
the time to option expiration as a fraction of a year (denoted )
the rate of interest (denoted r)
This can be expressed in function form:
Value = f(S, X, v, , r)
If the underlying is a payout asset, then the payout must also be built into the model.
We assume that the underlying asset does not pay out anything.
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Statistics with Applications to Options
The Mechanics of Monte Carlo Simulation
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The Mechanics of Simulation:
In simulation analysis, the first thing we must do is determine the relevant
variables. Next, we must determine the type of distribution each variable has,
the parameters of the variables (e.g., , ), and the degree of correlation
between the random variables ().
Example:
Suppose that X and Y are two normally distributed random variables. X has a
mean of 5 and a standard deviation of 7. Y has a mean of 30 and a standard
deviation of 4. X and Y have a linear correlation of 0.6.
Simulate 100 joint outcomes on X and Y.
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Random Number Generator:
To simulate the outcomes on X and Y, we will employ a random number generator.
Excel has such a random number generator function built into it. It is called RAND().
RAND() generates observations on a continuous uniform distribution bounded
between 0 and 1. Nevertheless, the function can be used to generate observations
on a standard normal distribution as follows:
Let Ri be the ith value generated from RAND(). Now define E as follows
12
E =
Ri
– 6
i=1
E will have a standard normal distribution. That is
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E ~ N(0, 1)
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Suppose now that we generate two observations by this process. Let’s call the
first E1 and the second E2.
Each of these random variables is distributed as a standard normal:
E1 ~ N(0,1)
E2 ~ N(0,1)
While these two random variables are both standard normal, they are
independent of one another. That is, their correlation is 0.
We now want to add the appropriate degree of correlation.
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Adding the Correlation:
Define two new random variables as follows:
A = E1
and
B = ×E1 + E2×(1 - 2)½
A and B, like E1 and E2, are still standard normal random variables, but unlike E1 and
E2, which are independent of one another, A and B are correlated with one another
and the degree of correlation between them is given by .
In our example, the correlation is 0.6 so the relationship above would look as follows:
A = E1
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and
B = 0.6×E1 + E2×(1 - 0.62)½
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Adjusting the Means and Standard Deviations:
A and B have standard normal distributions with the appropriate degree of
correlation. We now want to adjust their means and their standard deviations so
that they have the parameters that X and Y are supposed to have. Define X and
Y as follows:
X = A×A + A
= A×7 + 5
Y = B×B + B
= B×4 + 30
At this point we have simulated the outcomes that we want:
X ~ N(5, 7)
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Y ~ N(30, 4)
X,Y = 0.6
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Consider the following:
Suppose that we have an option written on a stock. The option’s strike price is
$100. The stock’s current price is $102. The option covers 1 share. Suppose
that the option expires in 182 days, its annual volatility is 30%, and the risk-free
rate of interest is 5%.
Using the Black-Scholes-Merton Model
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What is the value of the option if it is a call?
Answer: $10.8333
What is the value of the option if it is a put?
Answer: $6.3710
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What value might the option have tomorrow if the option is a call?
The value of the option tomorrow will depend on the same value drivers as does the
value of the option today. The only difference is that we may not know today what the
inputs for those value drivers will be tomorrow.
The value drivers that might change between today and tomorrow are:
The amount of time remaining until the option expires
The price of the underlying asset
The volatility of the price of the underlying asset
The rate of interest
With the exception of the time to expiry, each of these “value-drivers” can change in an
unexpected way. Thus, they are each a source of “risk.” Simulation modeling of VAR
attempts to simulate input values for these value drivers in order to simulate a value for
the option.
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How does each value driver affect the value of an option?
We know that tomorrow the option will have only 181 days to expiry. Let’s suppose
that the volatility and risk-free rate of interest are constant, so that only the price of
the underlying is uncertain.
What might happen to the value of the call?
What might happen to the value of the put?
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Value
25
Long Call
20
Days = 181
r = 5%
vol = 30%
X = 100
15
10
5
11
3
11
0
10
7
10
4
10
1
98
95
92
89
86
0
Price of underlying
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Value
16
14
Long Put
12
Days = 181
r = 5%
vol = 30%
X = 100
10
8
6
4
2
11
3
11
0
10
7
10
4
10
1
98
95
92
89
86
0
Price of underlying
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A Closer Look at the Call:
Holding volatility and the interest rate constant, what happens with each passing
day?
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Value
25
181 days
20
r = 5%
vol = 30%
X = 100
120 days
60 days
15
0 days
10
5
11
3
11
0
10
7
10
4
10
1
98
95
92
89
86
0
Price of underlying
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A Closer Look at the Call:
Holding time to expiry constant, and the interest rate constant, what happens if
volatility increases or decreases?
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Value
25
vol = 40%
vol = 30%
vol = 20%
20
15
r = 5%
Days = 181
X = 100
10
5
11
3
11
0
10
7
10
4
10
1
98
95
92
89
86
0
Price of underlying
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A Closer Look at the Call:
Holding time to expiry and volatility constant what happens if interest rates increase
or decrease?
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25
Value
20
vol = 30
Days = 181
X = 100
15
10
5
11
3
11
0
10
7
10
4
10
1
98
95
92
89
86
0
Price of underlying
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A Closer Look at the Put:
Puts will exhibit similar behaviors:
As time grows shorter, the puts value will decline.
If volatility increases, the puts value will increase.
If interest rates rise, the value of a put goes down.
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The Straddle:
We are now ready to look at the straddle. The value of the straddle is the sum of
the values of the call and the put.
Holding volatility (30%), time to expiry (182 days), and the interest rate (5%)
constant, what does the value diagram of the straddle look like?
Suppose that the trader’s straddle covers 500,000 units.
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Now consider what happens to the value of our position between today and
tomorrow.
Today, the value of the call is $10.8333
Today, the value of the put is $6.3710
Thus, today the value of the straddle is $10.8333 + $6.3710 = $17.2043
And, the value of the position today is = $17.2043 × 500,000 = $8,602,150
The change in the value of the position = Value Tomorrow - Value Today
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We can graphically illustrate the change in the value of the position tomorrow but
only with respect to a change in at most two of the value drivers at a time.
Assume that volatility and the interest rate do not change.
We can then depict the change in value of the position with respect to a change
in the underlying’s price, remembering that one day will have passed.
This represents a “risk profile” with respect to one source of market risk (the price
of the underlying). That is, we are looking at the risk profile with respect to one of
the risk factors.
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Profit/Loss
3000000
2500000
2000000
vol = 30%
r = 5%
Days = 181
X = 100
1500000
1000000
500000
0
88 90
92 94 96
98 100 102 104 106 108
-500000
-1000000
Price of underlying
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Profit/Loss
3000000
2500000
What happens if we overlay this risk
profile with a normal distribution?
vol = 30%
r = 5%
Days = 181
X = 100
2000000
1500000
1000000
500000
0
88 90
92 94 96
98 100 102 104 106 108
-500000
-1000000
Price of underlying
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Profit/Loss
3000000
2500000
2000000
1500000
1000000
500000
0
88 90
92 94 96
98 100 102 104 106 108
-500000
-1000000
Price of underlying
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What happens if the underlying’s price changes, the volatility changes, and the
interest rate changes?
The multidimensionality of the problem and the non-linearities result in a situation
that is too complex to map into a parametric VAR.
This is where simulation analysis provides a solution.
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Simulating changing market conditions:
What we want to do is simulate vectors of possible outcomes for all the value
drivers that will influence the market value of the straddle. These outcomes
include: The price of the underlying, the volatility of the underlying, and the
interest rate.
This requires that we make decisions about:
1.
2.
3.
4.
5.
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The types of distributions describing these three value drivers
The mean of each value driver
The standard deviation of each value driver
The values of other possible parameters
The correlations among the different value drivers.
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Example:
Suppose that a DB trader is long a straddle on GTV stock. The straddle covers
500,000 shares. GTV is currently priced at $102, GTV has an annual vol of 30%,
the rate of interest is 5%. The straddle has a strike of $100 and 182 days to go.
Assuming that this is the only position this trader holds, determine the trader’s
one-day VAR at 95%.
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Under Black-Scholes-Merton, the future price of the underlying (GTV stock) is lognormally distributed, so that the natural logarithm of the future price is normally
distributed as follows:
lnST ~ N[lnS0 + (r – (σ2/2))T,σ√T]
Assuming a 365-day year, then the natural logarithm of the one-day future price of
GTV is distributed normally with mean 4.625 and standard deviation 0.0157:
lnS1 ~ N[ln(102) + (.05 – (.302/2))(1/365),.30√(1/365)] ~ N(4.625,0.0157)
Recall that the interest rate and volatility are constant in the Black-Scholes-Merton
framework. Thus there are no correlated variables in this particular illustration.
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Now, using the technique described above for sampling a standard normal
variate and then mapping it into one with mean 4.625 and standard deviation
0.0157, suppose that we obtain an estimate of lnS1 of 4.6407. Then the
estimate of the price of the stock in one day’s time is e4.6407 = 103.6169. Under
our new parameter set (S = 103.6169, r = .05, T = 181 days, σ = 0.30, and X =
100), the values of the call and put are $11.84 and $5.77, respectively. Thus the
estimate of the value of the straddle contract in one day is ($11.84 + $5.77) x
500,000 = $8,805,000. The one-day P&L is a gain of $202,850.
This entire process would be repeated over and over, perhaps 10000 times.
The profits and losses would be ranked (from greatest profit to greatest loss).
The 95% one-day VAR of the straddle would correspond to the outcome ranked
9501 out of the 10000.
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Suppose that we wanted an estimate of the long straddle position’s 95% VAR,
but for 5 days rather than 1 day. From the previous equation, the natural
logarithm of the underlying share price in five days would be normally distributed
with a mean of 4.625 and a standard deviation of 0.0351. So now we would
generate a random standard normal variate, and transform it to one with mean
4.625 and standard deviation 0.0351. This in turn would give us one estimate of
the share price in five days, and thus one measure of the 5-day P&L on the long
straddle position. Again, this process would be repeated 10000 times, with the
95% VAR being the 9501st ranked P&L.
If we wished, we could use a forward interest rate and a forward volatility
(instead of the 5% and 30% “spot” measures, respectively), in order to obtain
our estimate of a future stock value. Using forward values for r and σ may be
prudent if the VAR horizon is long and/or the interest rate and volatility term
structures are steeply sloped.
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While we have demonstrated the calculation of simulated VAR for a position
consisting of only one position (a straddle on one stock), there is no reason why
we could not have simultaneously generated potential values for all of the
positions in the trader’s book (where a book is defined as one wherein all
options have the same underlying stock, e.g., DELL).
Is it also possible to accommodate the calculation of simulated VAR when we
blend books, that is, when we have a portfolio of option positions and there is
more than one underlying stock (e.g., DELL and CSCO)? The answer is “yes”.
Here we must accommodate the correlations among all the stock prices. We
would do so for two stocks as described earlier, that is, when accommodating
two correlated and normally distributed variables. (Recall that we must work
with the natural logarithms of the two stock prices. That is, we must treat the
natural logs as normally distributed, and accommodate the correlation among
the natural logs of the prices.)
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What if we wanted to accommodate more correlated variables, for example, by permitting
multiple underlying stocks? Or, if we assume that the interest rate and volatility are lognormally distributed, how can we accommodate these variables and their possible
correlations with each other and with stock values? This can be accomplished as follows:
Consider the situation where we require n correlated samples from normal distributions
where the coefficient of correlation between sample i and sample j is ρi,j. We first sample n
independent variables xi (1 ≤ i ≤ n), from univariate standardized normal distributions. The
required samples are εi (1 ≤ i ≤ n), where
εi = Σαikxk
where Σ is a summation operator for k = 1 to i. For εi to have the correct variance and
correct correlation with εj (1 ≤ j < i), we must have
Σα2ik = 1 (where the summation factor is again for k = 1 to i),
and, for all j < i,
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Σαikαjk = ρij (where here the summation factor is for k = 1 to j).
The first sample, ε1, is set equal to x1. These equations for the α’s can be
solved so that ε2 is calculated from x1 and x2, ε3 is calculated from x1, x2, and x3,
and so on. The procedure is known as the Cholesky decomposition.
Note: Monte Carlo simulation tends to be numerically more efficient than other
simulation techniques when there are three or more stochastic variables,
because the time required to conduct a Monte Carlo simulation only
increases about linearly with the number of variables. It can also
accommodate products like path-dependent options.
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Potential Problems with Simulated VAR:
Don’t be fooled into believing that simulated VAR is non-parametric. It is.
Simulated VAR required knowledge of both the distributions and the parameters
of the distributions of the drivers (input values). Thus, simulated VAR is
parametric in nature. If parameters are misestimated or distributions are
incorrectly specified, simulation can generate faulty estimates of VAR.
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Model Risk:
Simulated VAR generates a change in value by deducting the potential value
of the option tomorrow from the value of the option today. The value today is
often observable in the market today, but the potential value tomorrow has to
be inferred from the simulated outcomes of the value drivers and the model
that takes the value driver inputs and produces an option value output.
This begs the question “What if we are using the wrong model?” The very
real possibility that we are using the wrong model is one form of a broader
class of problems called model risk.
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Blended VAR:
Some risk managers form a “blended” or “composite” VAR from VAR
estimates obtained using several different approaches.
For example, suppose that the following methods lead to the following 1-day
VARs at 95%:
Parametric VAR:
Historic VAR:
Simulated VAR:
$117,087
$115,300
$121,200
We could formulate a weighted average of these three VAR measures (either
equally weighted or by giving more weight to one method or another).
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What VAR is Not
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VAR is Not Traders
ISEWhat
Option
Statistics with Applications to Options
We have been trying to answer the questions: What is VAR and how is it measured?
We have concluded that VAR is a measure of the maximum loss a trader might
experience over a given risk horizon at a specified level of confidence.
It is equally important to understand what VAR is not.
VAR is not a statement of the absolute maximum loss the trader might suffer.
VAR does not attempt to address this question.
For example if the 1-day VAR at 95% is $1.2mm, we may conclude that the trader
will lose no more than $1.2 mm in a single day on 95 out of each 100 days. But, it
does not attempt to say how much might be lost on the other 5 days!
Thus, VAR is not a panacea for risk managers. It is merely one tool.
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Of course, one could use a 99% VAR or a 99.9% VAR in an effort to measure
the potential for loss under extreme conditions.
There are two problems with this. The first is that VAR estimates become
more and more unreliable the higher the level of confidence that is sought.
The reason for this has to do with departures of the underlying distributions
from the assumed normality. It is known, for example, that price distributions
often exhibit leptokurtosis (fat tails).
The one-day stock market declines in October 1929 and again in October
1987 are examples of this.
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Second, there is considerable risk that the relationships among the value
drivers (i.e., risk factors) might suddenly change. For example, under
conditions of stress in the market place (e.g. periods of flight to safety) the
correlations among prices sometimes rise sharply. When this happens, the
potential for loss can greatly exceed that implied by VAR.
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Stress Testing:
One way to measure the potential to experience a loss in excess of that
implied by VAR is to engage in stress testing.
In stress testing, we simulate VAR but we assume conditions of market stress.
By market stress, we mean we go back and look for periods when the markets
were under unusual stress (such as periods when there was a flight to quality).
We then measure the correlations among prices during those periods and the
standard deviations of prices during those periods. (It is important that we
employ values that are internally consistent.)
From these extreme-condition parameters, we can test the values produced by
our models and estimate VARs.
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