Download Value at Risk

Bank of England
Centre for Central Banking Studies
CEMLA 2013
Value at Risk.
David G. Barr∗
November 21, 2013
∗ Any
views expressed are those of the author and not necessarily those of the Bank of England.
1
Contents
1 Risk and volatility.
1.1 Illustrating a problem with volatility. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 Value at risk: Introduction.
2.1 VaR for a simple discrete distribution.
2.2 VaR and the Normal distribution. . .
2.2.1 Revision of hypothesis tests. . .
2.2.2 From hypothesis tests to VaR.
2.2.3 A general formula for Normally
2.3 VaR again, in words this time. . . . .
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distributed returns.
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3
3
7
. 7
. 8
. 8
. 10
. 11
. 13
3 Value at Risk: An example.
14
3.1 SP500: Daily returns. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
3.2 Longer time periods. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
4 Don’t forget....
16
5 Problems with VaR.
17
5.1 When the VaR represents a profit rather than a loss. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
5.2 VaR is ‘optimistic’. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
5.3 VaR may violate the basic rule of diversification. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
6 An alternative to VaR: Expected shortfall (ES).
20
6.1 Definition of expected shortfall. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
6.1.1 ES for a Normal distribution. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
6.2 The SP500 example again. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
A Volatility over several periods.
24
2
1. Risk and volatility.
• Can we sum up the riskiness of an asset, or portfolio, in a single-valued
measure?
– Obtaining a single-value measure is important if we want to rank
assets by risk.
• Traditionally this was achieved using volatility i.e. standard deviation.
• While volatility is still used, as a single summary measure of risk it
has some deficiencies:
– It’s symmetric i.e. upside and downside risks are measured in the
same way.
– It does not deal well with heavy-tailed distributions.
∗ ‘Heavy’ being defined in relation to the Normal distribution.
1.1. Illustrating a problem with volatility.
• The following time series of asset returns all have mean zero, and
volatility 1.
• But they come from three different distributions: N(0,1), t(3) and a
‘jump distribution’.
– The
√ variance of a t(n) is n/(n − 2), so we divide the t(3) data by
3.
3
√
Figure 1: N(0,1) and t(3)/ 3 variables.
4
Figure 2: A jump variable.
5
• Which is the most risky for a bank that will go bankrupt if the loss
hits -4%?
• Value at Risk (VaR) offers an alternative, also single-value, measure
of risk.
6
2. Value at risk: Introduction.
• A popular, but imperfect, alternative to volatility.
• Invented by JPMorgan and propagated in ‘RiskMetrics’,
– VaR is not symmetric (it looks only at losses).
– It treats heavy tails sensibly, up to a point.
• VaR analysis provides values for x and k in the following statement:
“We can be x% confident that we will not lose more than
$k.”
• But, before we can make use of the VaR measure we need to decide
on:
1. The length of the time period over which the risk is of concern.
2. The probability distribution of the risky variable (asset price etc).
– N.B. The VaR approach does not require Normality.
– We will see later that we don’t need the whole of the distribution.
3. A ‘confidence level’ i.e. x in the above statement.
2.1. VaR for a simple discrete distribution.
• But a discrete version makes the key VaR concept clear.
• Let a risky return have the following distribution:
Probability
0.02
0.03
0.06
0.89
7
Return
-12
-11
-10
≥ -9
• We can be 89% certain that we will not lose more than 9.
• We can be 95% certain that we will not lose more than 10, so...
• ...The Value at Risk at 5% is 10.
• The rest of the work involved in calculating VaR has to do with applying it to more realistic distributions.
2.2. VaR and the Normal distribution.
2.2.1. Revision of hypothesis tests.
• Calculating VaR is very similar to hypothesis testing in econometrics.
• Assume that an estimator for the equation y = xβ + β̂ =
Cov(x, y)
V ar(x)
has a Normal distribution with unknown mean β and standard deviation 1.
• I.e. β̂ ∼ N (β, 1).
• Assume that we want to test the null hypothesis that β = 0. Under
this hypothesis,
β̂ ∼ N (0, 1)
(1)
• We can then find 95% critical values for β̂ from
P r(β̂ > 1.96) = 2.5%
P r(β̂ < −1.96) = 2.5%
8
(2)
(3)
z
...
1.7
1.8
1.9
2.0
2.1
...
...
...
...
...
...
...
...
...
0.05
...
.9599
.9678
.9744
.9798
.9842
...
0.06
...
.9608
.9686
.9750
.9803
.9846
...
0.07
...
.9616
.9693
.9756
.9808
.9850
...
...
...
...
...
...
...
...
...
Table 1: Extract from cumulative standard Normal table, later to be called both Φ and N .
For (1.75, 0.9599), we have N (1.75) = 0.9599, and N −1 (0.9599) = 1.75.
2.5%
-1.96
Figure 3: Cumulative Normal, N(-1.96) = 0.025 = 2.5%, N(+1.96) = 0.975 = 97.5%.
• If our null hypothesis is true, the probability of getting β̂ outside the
range ±1.96 is only 5%.
• Similarly, the probability of getting an estimate outside the ranges...
±1.645 is 10%
±2.33 is 2%.
9
±2.57 is 1%.
• So if we get β̂ = 2.00:
– we can be 95% confident that the null hypothesis is wrong but...
– ...we cannot be 99% confident that it is wrong.
2.2.2. From hypothesis tests to VaR.
• We now move to considering financial risks instead of econometric
estimates.
• We start by looking at the risk of a financial asset.
• We continue to use the Normal distribution but:
– The Normal distribution extend to minus infinity.
– Very few assets have returns that can get this bad (a short call
option being an exception).
– E.g. an equity’s percentage return will lie in the range -100% to
+∞.
• From the examples we know that the probability of getting β̂ lower
than −1.645 is 5%.
• Now let β be the (unknown) future dollar return on an asset, let it
have distribution β ∼ N (0, 1), and let β̂ be the realisation of that
change.
• We get the probability that the asset price will fall by 1.645 or more
as 5%.
• This is the value at risk at 5% i.e.
VaR(5%) = 1.645.
10
and, similarly,
VaR(1%) = 2.33.
• Note that for these examples:
– We have assumed σ = 1.
– These numerical VaR results are for a N(0,1) distributed asset return, and not for any other distribution.
2.2.3. A general formula for Normally distributed returns.
• For the more general distribution r ∼ N (µ, σ), we transform r into
the standard normal variate z where
r−µ
z=
(4)
∼ N (0, 1)
σ
so that we can continue to use the above N (0, 1) table.
• For the lhs of the distribution r ∼ N (µ, σ) we have,
r−µ
< −1.645 = 0.05
Pr
σ
• We find the critical value r5% as
r5% − µ
= −1.645
σ
⇒ r5% = µ − 1.645σ
11
(5)
(6)
(7)
5%
-1.645
z = (r - mu)/sigma
Figure 4: Value at risk, in terms of z = (r − µ)/σ.
• In terms of z this would be
z5% =
r5% − µ
σ
and
r5% = µ + z5%σ
where z5% = 1.645.
• Note that this will typically give us a negative number, but by convention we treat losses as positive numbers.
• Hence the general formula for a Normally distributed return is,
V aR(5%) = − (µ − 1.645σ)
(8)
• For most of the data we use µ ≈ 0 so
V aR(5%) ≈ 1.645σ
which we often see expressed more generally as
12
(9)
V aR(S%) ≈ N −1(S%)σ
(10)
where N −1(S%) can be read off the standard Normal tables.
2.3. VaR again, in words this time.
• Given a distribution we can we find numbers such that we can state
that:
– ‘We can be x% confident that we will not lose more than $k.’
– Or, ‘The probability that we will not lose more than $k is x%.’
• VaR allows us to rewrite this as:
– ‘We can be 95% certain that we will not lose more than $VaR(5%).’
where VaR(5%) is a number to be calculated using the chosen distribution.
• Or, ‘We can be 99% certain that we will not lose more than $VaR(1%).’
• And, more generally,
– ‘We can be p% certain that we will not lose more than $VaR(100p)%).’
13
3. Value at Risk: An example.
3.1. SP500: Daily returns.
• The estimated logNormal distribution for SP500 daily returns is
N(0.0295%, 0.1399%).
• I.e.
ln(1 + r) ∼ N (0.000295, 0.001399)
• Find the 5% VaR for a $1m investment in the index.
• The 5% ‘critical value’ for ln(1 + r), is found from:
ln(1 + r)5% = − (µ − 1.645σ)
= − (0.000295 − 1.645 × 0.001399)
= 0.002006
(11)
(12)
(13)
• The equivalent return in natural units is found as:
1 + r5% = eln(1+r)5%
r5% = eln(1+r)5% − 1
= e0.002006 − 1
= 1.002008 − 1
= 0.002008
= 0.2008%
(14)
(15)
(16)
(17)
(18)
(19)
• So we can be 95% certain that we will not lose more than $2008 in any
one day.
• The 5% VaR here is referred to in some texts as its complement i.e.
as a 95% VaR.
• The time period is important. Here the data are for daily returns, and
are used to construct daily moments for the probability distribution.
14
• As a result, we get only a 1-day VaR.
3.2. Longer time periods.
• We may have data only for daily returns.
• If we want to calculate a 5-day VaR we need the moments for 5-day
returns.
• If the daily dollar returns are Normally distributed (µ, σ) and the
realisations are independent of each other we can scale the daily
mean and standard deviation up to 5 days as:
µ5 = 5 × µ1
√
σ5 = 5 × σ1
(20)
(21)
• We often approximate the mean by zero, in which case we get,
√
V aR(n days) ≈ n × V aR(1 day)
(22)
• So for the approximate value at risk over 30 days we get
√
V aR30(5%) ≈ 30 × V aR1(5%) = 5.48 × $2008 = $11004
(23)
• Adjusting the VaR for other distributions is more complicated than
this.
• Note that we have assumed the returns and in dollar terms here. If
we have proportionate returns, and
distributed,
√ these are logNormally
√
the equivalent scaling factor is e 30 instead of 30.
15
4. Don’t forget....
For Normally distributed returns, VaR(1%) is the return that is 2.33
standard deviations below the mean, multiplied by minus one.
and, more generally,
V aR(S%) = −σΦ(S%) for µ = 0
(24)
where Φ(S%) is the inverse Normal distribution function (the one in the
tables at the back of most statistics textbooks) and S% is typically 1% or
5% (as opposed to 99% or 95% as in Hull’s Risk Management text.)
16
5. Problems with VaR.
5.1. When the VaR represents a profit rather than a loss.
• If the mean is large in relation to the volatility, VaR(k%) may represent
a ‘small’ profit rather than a loss.
• The VaR(k%) measure makes no sense in this case.
Figure 5: ‘Positive’ Var.
5.2. VaR is ‘optimistic’.
• VaR(5%) = $1000 tell us that we can be 95% certain of not losing
more than $1000.
• But it tells us nothing about how much we should expect to lose in
the 5% of cases where things turn out badly.
• Consider the follow distribution of returns:
17
Figure 6: The VaR remains the same despite the increased conditional loss.
• The 5% Var is -2 for the disjointed distribution shown by the solid line.
• It remains at -2 if the left block moves further left. This would usually
be regarded as an increase in the risk, but VaR fails to capture it.
• This would be particularly important if the firm were to face bankruptcy
at -5.
5.3. VaR may violate the basic rule of diversification.
• There is general agreement that a sensible risk measure should satisfy
four ‘coherence’ characteristics.
• Coherence characteristics for any risk measure φ(X).
18
1. Monotonicity: If we have two asset returns, X and Y , and if in
all realisations X ≤ Y , then φ(X) ≥ φ(Y ).
2. Subadditivity: A diversified portfolio cannot be more risky than
its constituent assets, so φ(X + Y ) ≤ φ(X) + φ(Y ).
3. Homogeneity: If the portfolio value increases by n times then
so does the risk: φ(nX) = nφ(X).
4. Translation invariance: If we add $c of riskless cash to the
portfolio the risk should fall i.e. φ(X + c) = φ(X) − c.
• The second of these refers to portfolio diversification.
• VaR can fail to exhibit subadditivity if the tails of the distribution are
heavy enough (tail index > 2 - we look at tail indices in the Extreme
Value Theory session).
19
6. An alternative to VaR: Expected shortfall (ES).
6.1. Definition of expected shortfall.
• ES is the expected size of the loss given that the VaR has been exceeded
i.e.
(25)
ES(p) = −E[XkX ≤ V aR(p)] > 0
• For the discrete example we had earlier,
Probability
0.02
0.03
0.06
0.89
Return
-12
-11
-10
≥ -9
if V aR(5%) = 10 is exceeded, there are only two possible losses, -11
and -12.
• To get the expected value of the loss we have to scale the probabilities
to sum to 1, without changing their relative values.
• We do this as follows:
p−12
p−11 + p−12
0.02
=
0.02 + 0.03
= 0.4
p0−12 =
p−11
p−11 + p−12
0.03
=
0.02 + 0.03
= 0.6
p0−11 =
20
(26)
(27)
(28)
(29)
(30)
(31)
• Note that the denominator 0.02 + 0.03 = 0.05 = 5% is the VaR that
we are basing the ES on.
• We then get the expected loss as:
ES = −[0.4 × −12 + 0.6 × −11]
= 11.4
(32)
(33)
• As with VaR, the rest of the ES work deals with how to calculate ES
for more realistic distributions.
6.1.1. ES for a Normal distribution.
• For standard Normal (N(0,1)) Danielsson p87 gives the following VaR
and ES values for N (0, 1) profit/loss distribution:
p
0.5
0.1
0.05 0.025 0.01 0.0001
VaR(p) 0
1.282 1.645 1.960 2.326 3.090
ES(p) 0.798 1.755 2.093 2.338 2.665 3.367
• For example, for p = 0.05, we get V aR(5%) = 1.645 i.e. we can be
95% certain of not losing more than 1.645.
• If, however, we are told that this loss has been exceeded, we know that
the actual return must lie in the range (−∞, −1.645).
• Then the expected loss, given this information, is $2.093.
• Other things to note about ES:
– ES always satisfies the diversification rule.
– If the distribution of returns is Normal, then as p → 0 so V aR(p) →
ES(p). Look for this in the table above.
21
– The general formula for ES given rates of return distributed N (µ, σ),
is
φ(N −1(p))
(34)
ES(p) = − µ − σ
p
where φ(x) is the Normal density function, and Ψ(x) is the Normal
cumulative distribution function.
• So, for example, for µ = 0, σ = 1 and p = 0.01,
−1
φ(N (0.01)
ES(0.01) = − µ − σ
0.01
φ(−2.3263)
=
0.01
0.0267
=
0.01
= 2.6652
22
(35)
(36)
(37)
(38)
6.2. The SP500 example again.
• The SP500 estimated parameters were µ̂ = 0.0295%, σ̂ = 0.1399%
• To keep things simple we assume, incorrectly, that the returns are
Normally distributed (we assumed them to be log Normal earlier).
• We get,
V aR(0.05) =
=
=
=
−(µ − σN −1(0.05))
−0.0295 + 0.1399 × 1.645
−0.0295 + 0.2301
0.2006
(0.2008 when we assumed log Normality) and,
23
(39)
(40)
(41)
(42)
φ(N −1(0.05)
− µ−σ
0.05
φ(−1.645)
−0.0295 + 0.1399
0.05
0.1031
−0.0295 + 0.1399
0.05
−0.0295 + 0.1399 × 2.0699
−0.0295 + 0.2885
0.259
ES(0.05) =
=
=
=
=
=
(43)
(44)
(45)
(46)
(47)
(48)
A. Volatility over several periods.
• We are used to seeing the following relationship between variances for
multiples of random variables:
V (nx) = n2V (x)
(49)
σ5days = 5σ1day
(50)
or,
so why do we have
σ5days =
√
5σ1day
(51)
in equation (21)?
• This can be demonstrated in a 2-day example.
• Let the random return for the first day be x1, and for the second x2,
then
V (x1 + x2) = V (x1) + V (x2) + Cov(x1, x2)
(52)
24
• We assumed for (21) that the returns on the 2 days are independently
distributed, so Cov(x1, x2) = 0.
• We also assumed that they have the same variance, call it V (x).
• Hence
V (x1 + x2) = V (x1) + V (x2) = 2V (x)
and
σx1+x2 =
p
√
2V (x) = σx 2
(53)
(54)
.
• In equation (49) above we are multiplying a single realisation by n,
not adding two different realisations from the same distribution.
• The algebra behind (49), works as follows,
V (2x) =
=
=
=
V (x + x)
V (x) + V (x) + 2Cov(x, x)
V (x) + V (x) + 2V (x)
4V (x)
(55)
(56)
(57)
(58)
• So it all revolves around the Cov term, and whether or not it equals
zero.
25