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Options
An option, is a financial instrument whose value depends on the values of others, more
basic underlying variables. There are two basic types of options. A call option gives the
holder the right, but not the obligation, to buy the underlying asset by a certain date for a
certain price. A put option gives the holder the right, but not the obligation, to sell the
underlying asset by a certain date for a certain price. The price in the contract is known as
the exercise or strike price. The date in the contract is known as the expiration date or
maturity. American options can be exercised at any time up to the expiration date.
European options can be exercised only at the expiration date itself. Most of the options
that are traded on exchanges are American.
For the cost (premium) of an option, the holder locks in a specific exercise (strike) price for
the underlying. Obviously, the option will yield no payoff if the market price of the
underlying security exceeds the exercise price at the time of the option's expiration; the
option will simply be left to expire in this case and the holder will lose the cost paid for the
purchase of the option. However, the holder will realize a profit if the security's market
price falls below the exercise price. The gain grows linearly with the decline of the
security's market price at the option's expiration time. Hence, a put option can provide
coverage against potential declines in the price of the underlying security.
Clearly, the option will not yield a payoff, and will be left to expire, if the market price of
the security remains below the exercise price at maturity. But, in the event that the market
price of the security exceeds the strike price of the option at maturity, then the holder reaps
an immediate profit. By exercising the option, he will purchase the predetermined amount
of the security at the exercise price, which he can then solve in the market at the higher
price and reap the profit from the difference of the market price to the strike price. Hence, a
long position in a call option accentuates the upside potential in the event of market
upturns.
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Option positions
There are two sides to every option contract. On one side is the trader who has taken the
long position (i.e., he bought the option). On the other side is the trader (issuer- writer)
who has taken a short position (i.e., has sold or written the option). The writer of an
option receives cash up front but has potential liabilities later. The writer’s profit or loss is
the reverse of that for the purchaser of the option.
Four basic option positions are available:

A long position in a call option

A long position in a put option

A short position in a call option

A short position in a put option
It is often useful to characterize European option positions in terms of the payoff to the
trader at maturity. The initial cost of the option is then not included in the calculation. If X
is the strike price and ST is the final price of the underlying asset, the payoff from a long
position in a European call option is
Max(ST - X, 0)
This reflects the fact that the option will be exercised if ST > X and will not be exercised if
ST < X. The payoff to the holder of a short position in the European call option is
-Max(ST - X, 0)
or
Min(X - ST, 0)
The payoff to the holder of a long position in a European put option is
Max(X - ST, 0)
And the payoff from a short position in a European put option is
-Max(X - ST, 0)
or
Min(ST - X, 0)
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Factors affecting option prices
There are six factors affecting the price of a stock option:
1.
S0, The current stock price
2.
X, The strike price
3.
T, The time to expiration
4.
σ, The volatility of the stock price
5.
r, The risk-free interest rate
6.
d, The dividends expected during the life of the option,
Moneyness
- A call is in-the-money (ITM) if the asset price is greater than the exercise price (i.e. it
would be worth something exercised)
- A put is in-the-money if the asset price is less than the exercise price (i.e. it would be
worth something exercised)
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- A call is out-of-the-money (OTM) if the asset price is less than the exercise price (i.e. it
would be worth nothing exercised)
- A put is out-of-the-money if the asset price is greater than the exercise price (i.e. it
would be worth nothing exercised)
- A call is at-the-money (ATM) if the asset price is equal with the exercise price
- A put is at-the-money (ATM) if the asset price is equal with the exercise price
Strategies involving options
Certain strategies that concern specific combinations of options in hedging applications are
popular in practice. Here we consider the following option trading strategies: Straddle,
Strip, Strap, and Strangle, and Bearspread. All of these strategies involve combinations of
positions in call and put options. Each of them generates a different payoff profile. The
choice among them is based on the investor's view regarding potential movements in the
value of the underlying and his preferences for protection in the case of such movements.
Straddle strategy
This strategy suits investors who espouse the idea to ``exit the market in the face of a
storm. It is designed to provide protection in the event of increased volatility and yields
payoffs if there is a substantial movement in the price of the underlying security, either
downside or upside. A long straddle consists of long positions in one call and one put
option on the same underlying asset, at the same strike price, and with the same expiration
date - the length of the planning horizon in our case. An investor enters a long straddle
position when he anticipates an increase in volatility but is unsure about the direction of the
movement, yet he wishes to be covered in the event of sharp changes in the price of the
underlying in either direction. For a long straddle position to yield a profit, the underlying
stock index must swing sufficiently low or high so as to cover the total cost of the option
purchases. There is a linear profit against the movement of the stock index in either
direction once the cost of the option premiums is cleared. If the price of the underlying
stock index remains close to the strike price of the options the straddle results in a loss; the
maximum loss is the sum of the purchase prices of the call and put options that constitute
the straddle. However, in the event of a large movement of the stock index in either
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direction the straddle yields a substantial profit. Thus, the loss from a potential large
decrease in the stock index is recovered from a long position in the straddle.
Strip strategy
A long strip consists of a long position in one call and two puts with the same exercise
price and expiration date. An investor enters into a long strip position when he expects a
large move in the stock index and considers a decrease in the index more likely than an
increase. So, the investor again buys protection against large swings in the underlying stock
index but gives preferential weight towards coverage against downward moves. Again, for
a limited range of changes around the strike price the strip strategy results in a loss. But for
large swings in either direction it yields a positive payoff; the payoff is twice as steep for
downward moves than for upward moves.
Strap strategy
In a sense, the strap strategy is a mirror image of the strip strategy. A long strap consists of
a long position in two calls and one put with the same exercise price and expiration date.
An investor enters into a long strap position when he expects large moves in the stock
index but considers an increase in the index more likely than a decrease. Again the investor
reaps a profit in the event of a sharp movement of the stock index in either direction, but
gives preferential emphasis to gains from potential upside moves.
The constituent options in these three strategies are typically at-the-money options; that is,
their strike price is equal to the current value of the underlying index. Of course the closer
that the strike price of a call or a put option is to the current value of the underlying asset,
the more expensive the option is. Hence, these strategies are rather costly. Obviously, the
strip and strap are costlier than the straddle as they involve an additional
option.
Strangle strategy
Similar to the straddle, a long strangle consists of a long call and a long put option on the
same underlying asset and with the same expiration date; but in a strangle the two options
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have different exercise prices. The strike price of the put is lower than the current price of
the underlying, while that of the call is higher than the current price of the underlying.
Again, a long straddle yields a profit when there is a substantial move in the stock index in
either direction. The index must move farther in a strangle than in a straddle for the strategy
to yield a profit. But the downside risk if there is only a small change in the value of the
stock index is less with the strangle than with the straddle because the strangle is a cheaper
alternative than the straddle as the prices of its constituent options are lower than those in
the strangle. With a long strangle, an investor buys coverage against large movements in
the stock index in either direction; that is, he covers against volatility. The payoff pattern
resulting with a strangle depends on how close together the strike prices of the constituent
call and put options are; if both of these strike prices approach the current price of the
underlying stock index then the payoffs of the resulting strangle resemble those of the
straddle. The farther apart the strike prices of the constituent options are the lower the cost
of the strategy, but the farther the stock index must move for the strategy to realize a profit.
BearSpread strategy
A BearSpread is a strategy composed of two put options with the same expiration date.
The strategy involves a long ITM put and a short OTM put. The maximum payoff (in
downturns) of this strategy is the difference between the two strike prices, minus the net
premium for the two options. The break-even point is calculated as the difference between
the higher strike price of the ITM option and the net premium. The maximum loss is again
the net premium for the positions in the two options.
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Payoffs of option strategies
These are few of the ways in which options can be used to produce an interesting
relationship between profit and stock price.
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Upper and lower bounds for option prices
In this section we derive upper and lower bounds for option prices. If the option price is
above the upper or below the lower bound, there are profitable opportunities for
arbitrageurs.
Upper bounds
A call option gives the holder the right to buy one share of a stock for a certain price. No
matter what happens, the option can never be worth more than the stock price. Hence, the
stock price is an upper bound to the option price:
C <= S0,
If this relationship do not hold, an arbitrageur can easily make a riskless profit by buying
the stock and selling the call option.
A put option gives the holder the right to sell one share of a stock for X. No matter how
low the stock price becomes, the option can never be worth more than X. Hence,
P <= X,
We know that at time T the option will not be worth more than X. It follows that its value
today cannot be more than the present value of X:
P <= e-rT X,
If this were not true, an arbitrageur could make a riskless profit by selling the option and
investing the proceeds for the sale at the risk-free interest rate.
Lower bounds for call options
A lower bound for the price of a European call option on a non-dividend paying stock is
S0 - e-rT X
Consider the following two portfolios:
-
Portfolio A: one European call option plus an amount of cash equal to e-rT X
-
Portfolio B: one share
In portfolio A, if the cash is invested at the risk-free interest rate, it will grow to X at time
T. If ST > X, the call option is exercised at time T and portfolio A is worth ST.
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If ST < X, the call option expires worthless and the portfolio is worth X. Hence, at time T
portfolio A is worth
Max(ST , X)
Portfolio B is worth ST at time T. Hence, portfolio A is always worth at least as much as,
and is sometimes worth more than, portfolio B, at time T. If follows that it must be worth at
least as much as portfolio B today. Hence,
C + e-rT X >= S0,
Or
C >= S0 - e-rT X,
Because the worst than can happen to a call option is that it expires worthless, its value
must be positive. This means that C >= 0, and, therefore,
C >= max(S0 - e-rT X , 0)
Lower bounds for put options
A lower bound for the price of a European put option on a non-dividend paying stock is
e-rT X - S0
Consider the following two portfolios:
-
Portfolio C: one European put option plus one share
-
Portfolio D: an amount of cash equal to e-rT X
If ST < X, the put option in portfolio C is exercised at time T, and the portfolio becomes
worth X.
If ST > X, the put option expires worthless and the portfolio is worth ST. Hence, at time T
portfolio C is worth
Max(ST , X)
In portfolio D, if the cash is invested at the risk-free interest rate, it will grow to X at time
T.
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Hence, portfolio C is always worth at least as much as, and is sometimes worth more than,
portfolio D, at time T. If follows that it must be worth at least as much as portfolio D today.
Hence,
P + S0 >= e-rT X
Or
P >= e-rT X - S0,
Because the worst than can happen to a put option is that it expires worthless, its value
must be positive. This means that P >= 0, and, therefore,
P >= max(e-rT X - S0 , 0)
Put-Call Parity
We now derive an important relationship between C and P. Consider the following two
portfolios:

Portfolio A: One call option plus the amount of cash equal to e-rT X

Portfolio C: One put option plus one share
Both are worth
Max(ST , X)
At expiration of the option: Because the options are European, they cannot be exercised
prior to the expiration date. The portfolios must, therefore, have identical values today.
This means that
C + X e-rT = P + ST
This relationship is known as put-call parity.
Exercise:
A stock currently sells for $31. A 3-month call option with a strike price of $30 has a price
of $3. Assuming a 10% continuously compounded risk-free, what is the price of the
associated put option? What are the arbitrage possibilities when p = 2.25 ? when p = 1 ?
This problem requires the application of put-call-parity.We have:
p+ S0 =c + e−rT K or
p = 3 − 31 + e−rT 30,
or p=1.26
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a. Arbitrage when p=2.25
1.Buy call for $3
2.Short Put to realize $2.25
3.Short the stock to realize $31
4.Invest $30.25 for 3 months
If ST > 30, receive $31.02 from investment,
exercise call to buy stock for $30.
Net profit = $1.02
If ST < 30, receive $31.02 from investment,
put exercised: buy the stock for $30.
Net profit = $1.02
b. Arbitrage when p=1
1.Borrow $29 for 3 months
2.Short call to realize $3
3.Buy put for $1
4.Buy the stock for $31.
If ST > 30, call exercised: sell stock for $30.
Use $29.73 to repay the loan.
Net profit = $0.27
If ST < 30, exercise put to sell stock for $30.
Use $29.73 to repay the loan.
Net profit = $0.27
In case there are dividends, then the put-call parity is given by:
C + X e-rT = P + ST e-qT
Where q is the dividend yield.
Or, in case we have the exact amount D of dividends, the put-call parity is
C +D+ X e-rT = P + ST
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Exercises
1.
A stock currently sells for $32. A 6-month call option with a strike price of $35 has a
price of $2.27. Assuming a 4% continuously compounded risk-free rate and a 6% continuous
dividend yield, what is the price of the associated put option?
This problem requires the application of put-call-parity.We have:
p+ e−qT S0 =c+K+ e−rT K or
p = c − e−qT 32 + e−rT 35
p = $2.27 − e−0.06×0.532 + e−0.04×0.535 = $5.523.
2.
What are the arbitrage opportunities if the price of the put option in question 1 was $5?
What if this price was $6?
a. Arbitrage when p=6
1.Buy call for $2.27
2.Short Put to realize $6
3.Short the stock to realize $32
4.Invest $35.73 for 6 months
If ST > 35, receive $36.45 from investment,
exercise call to buy stock for $35. Pay dividends D=32*0.03=0.96
Net profit = $0.49
If ST < 35, receive $36.45 from investment,
put exercised: buy the stock for $35. Pay dividends D=32*0.03=0.96
Net profit = $0.49
b. Arbitrage when p=5
1.Borrow $34.73 for 6 months
2.Short call to realize $2.27
3.Buy put for $5
4.Buy the stock for $32.
If ST > 35, call exercised: sell stock for $35. Collect dividends D=0.96
Use $35.43 to repay the loan.
Net profit = $0.53
If ST < 35, exercise put to sell stock for $35. Collect dividends D=0.96
Use $35.43 to repay the loan.
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Net profit = $0.53
3.
What are the arbitrage opportunities if the price of the call option in question 1 was $2?
What if this price was $3? Suppose that the price of the put option is the one that satisfies the putcall-parity.
a. Arbitrage when c=2
1.Buy call for $2
2.Short Put to realize $5.523
3.Short the stock to realize $32
4.Invest $35.523 for 6 months
If ST > 35, receive $36.24 from investment,
exercise call to buy stock for $35. Pay dividends D=32*0.03=0.96
Net profit = $0.28
If ST < 35, receive $36.24 from investment,
put exercised: buy the stock for $35. Pay dividends D=32*0.03=0.96
Net profit = $0.28
b. Arbitrage when c=3
1.Borrow $34.523 for 6 months
2.Short call to realize $3
3.Buy put for $5.523
4.Buy the stock for $32.
If ST > 35, call exercised: sell stock for $35. Collect dividends D=0.96
Use $35.22 to repay the loan.
Net profit = $0.74
If ST < 35, exercise put to sell stock for $35. Collect dividends D=0.96
Use $35.22 to repay the loan.
Net profit = $0.74
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Option Pricing
Binomial Trees
A useful and very popular technique for pricing a stock option involves constructing a
binomial tree. This is a diagram representing different possible paths that might be
followed by the stock price over the life of the option.
A stock price is currently 20, and it is known that at the end of three months the stock
price will be either 22 or 18. We are interesting in valuing a European call option to buy
the stock for 21 in three months (the exercise price is 21).
Stock price =22
Option price =1
Stock price=20
Stock price =18
Option price =0
Consider a portfolio consisting of a long position in Δ shares of the stock and a short
position in one call option. We calculate the value of Δ that makes the portfolio riskless.
If the stock price moves up from 20 to 22, the value of the shares is 22Δ and the value of
the option is 1, so that the total value of the portfolio is 22Δ – 1. If the stock price moves
down to 18, the value of the shares is 18Δ and the value of the option is zero, so the value
of the portfolio is 18Δ. The portfolio is riskless if the value of Δ is chosen so that the final
value of the portfolio is the same for both alternatives. This means:
18Δ=22Δ – 1,
or
Δ=0.25
A riskless portfolio is, therefore,
Long: 0.25 shares
Short: 1 option.
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If the stock price moves up to 22, the value of the portfolio is
22*0.25-1=4.5.
If the value of the stock price moves down to 18, the value of the portfolio is
18*0.25=4.5.
Regardless of whether the stock price moves up or down, the value of the portfolio is
always 4.5 at the end of the life of the option.
Riskless portfolios must, in the absence of arbitrage opportunities, earn the risk free rate
of interest. Suppose that in this case the risk-free rate is 12% per annum. It follows that
the value of the portfolio today must be the present value of 4.5, or
4.5 e-0.12*0.25 =4.367
The value today of the stock price is 20. Suppose the option price is denoted by f. The
value of the portfolio today is
20*0.25 – f = 5-f
it follows that
5 – f = 4.367 and f = 0.633
In the absence of arbitrage opportunities, the current value of the option is 0.633. If the
value of the option were more than 0.633, the portfolio would cost less than 4.367 to set
up, and would earn more that the risk free rate. If the value of the option were less than
0.633, shorting the portfolio would provide a way to borrow money at less than the risk
free rate.
A generalization
We now generalize the argument by considering a stock whose price is initially S 0 and an
option on the stock whose current price is f. We suppose that the option lasts for time T
and that during the life of its the stock price can either move up to a new level, S0u, or
down to a new level S0d (u>1, d<1). If the stock price moves up to S0u, we suppose the
payoff from the option is fu. If the stock price moves down to S0d, we suppose the payoff
from the option is fd.
As before, we imagine a portfolio consisting of a long position in Δ shares and a short
position in one option. We calculate the value of Δ that makes the portfolio riskless. If
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there is an up movement in the stock price, the value of the portfolio at the end of the life
of the option is
S0uΔ – fu
If there is a down movement in the stock price, the value becomes
S0dΔ – fd
The two are equal when
S0uΔ – fu = S0dΔ – fd
Or

fu  f d
S0 u  S0 d
In this case, the portfolio is riskless and must earn the risk-free interest rate. The above
equation shows that Δ is the ratio of the change in the option price to the change in the
stock price as we move between the nodes.
If we denote the risk-free rate by r, the present value of the portfolio is
(S0uΔ – fu) e-rT
The cost of setting up the portfolio is
S0Δ – f
It follows that
S0Δ – f = (S0uΔ – fu) e-rT
Or
f = S0Δ – (S0uΔ – fu) e-rT
Substituting for Δ and simplifying, this equation reduces to
f = e-rT (p fu + (1-p) fd)
where
p
erT d
u d
Risk-Neutral Valuation
Although we do not need to make any assumptions about the probabilities of up and
down movements in order to derive the above equation, it is natural to interpret the
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variable p in the above equation as the probability of an up movement in the stock price.
The variable 1-p is then the probability of a down movement, and the expression
p fu +(1-p) fd
is the expected payoff from the option. With this interpretation of p, the above equation
states that the value of the option today is its expected future value discounted at the riskfree rate.
We now investigate the expected return from the stock when the probability of an up
movement is assumed to be p. The expected stock price at time T, E(ST), is given by
E(ST) = p S0u + (1-p) S0d.
Substituting for p, this reduces to
E(ST) = S0 erT
Showing that the stock price grows, on average, at the risk-free rate. Setting the
probability of the up movement equal to p is, therefore, equivalent to assuming that the
return on the stock equals the risk-free rate.
In a risk-neutral world, all individuals are indifferent to risk. They require no
compensation for risk, and the expected return on all securities is the risk-free interest
rate. We assume a risk-neutral world when we set the probability of an up movement to
p. Thus, the value of an option is its expected payoff in a risk neutral world, discounted at
the risk-free rate.
This result is an example of an important general principle in option pricing known as
risk-neutral valuation. The principle states that it is valid to assume the world is riskneutral when pricing options. The resulting option prices are correct not just in a riskneutral world, but in the real world as well.
Example.
We now return to the first numerical example. We define p as the probability of an
upward movement in the stock price in a risk-neutral world. In a risk-neutral world the
expected return on the stock must be the risk-free rate of 12%. This means that p must
satisfy
22p + 18 (1-p) = 20 e0.12*0.25
p = 0.6523
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At the end of the three months, the call option has a 0.6523 probability of being worth 1
and a 0.3477 probability of being zero. Its expected value is, therefore,
0.6523 * 1 + 0.3477 * 0 = 0.6523
In a risk-neutral world, this should be discounted at the riskless rate. The value of the
option today is, therefore,
0.6523 * e-0.12*0.25= 0.633
This is the same as the value obtained earlier, illustrating that no-arbitrage arguments and
risk-neutral valuation give the same answer.
Two-step Binomial Trees
We can extend the analysis to a two-step binomial tree, such as that shown in the
following Figure:
D 24.2
3.2
B
22
2.0257
E 19.8
0.0
20
A
1.2823
18
0.0 C
F
16.2
0.0
The objective of the analysis is to calculate the option price at the initial node of the tree.
This can be done by repeatedly applying the principles established earlier. The option
prices at the final node of the tree are easily calculated. They are the payoffs of the
option, when the exercise price is 21.
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At node C, the option price is zero, because C leads to either node E or node F and at
both nodes the option price is zero. We calculate the option price at node B. As before,
u=1.1 and d=0.9, and p=0.6523. Then, the price of the option at node B is given by
e-rT (p fuu + (1-p) fud) = e-0.12*0.25(0.6523 * 3.2 + 0.3477 * 0) = 2.0257
It remains for as to calculate the option price at the initial node, A. We do so, by focusing
on the first step of the tree:
e-rT (p fu + (1-p) fd) = e-0.12*0.25(0.6523 * 2.0257 + 0.3477 * 0) = 1.2823
The value of the option is 1.2823.
A Generalization
We can generalize the case of two time steps. The stock price is initially S0. During each
time step, it either moves up to u times its initial value or moves down to d times its
initial value.
S0u2
fuu
S0u
fu
S0ud
fud
S0
fu
S0d
fd
S0d2
fdd
Repeating application of the basic equation e-rT (p fu +(1-p) fd) gives:
fu = e-rT (p fuu + (1-p) fud)
fd = e-rT (p fud + (1-p) fdd)
f = e-rT (p fu + (1-p) fd)
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Substituting the first two equations to the last one we have
f = e-2rT [p2 fuu + 2p(1-p) fud + (1-p)2 fdd]
Thus, the option price is equal to its expected payoff in a risk-neutral world, discounted at
the risk-free rate.
Matching Volatility with u and d
In practice, when constructing a binomial tree to represent the movements in a stock
price, we choose the parameters u and d to match the volatility of the stock price. To see
how this is done, we suppose that the expected return on a stock is μ and its volatility is
σ. The step is of length Δt. The stock price either moves up by a proportional amount u or
moves down by a proportional amount d. The probability of an up movement (in the real
world) is assumed to be q.
The expected stock price at the end of the first time step is S0eμΔt. On the tree the
expected stock price at this time is
q S0u + (1-q) S0d
In order to match the expected return on the stock with the tree’s parameters, we must
therefore, have
μΔt
q S0u + (1-q) S0d = S0e
e t d
or q 
u d
The volatility of a stock price, σ, is defined so that,  t is the standard deviation of the
return on a stock price in a short period of time of length Δt. Equivalently, the variance is
σ2Δt. On the following tree (a) the variance of the stock price return is
S0u
S0u
p
q
S0
S0
1-q
1- p
S0d
(a) Real world
S0d
(b) Risk-neutral world
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qu2 + (1-q) d2 – [qu + (1-q)d]2
This uses the result that the variance of a variable Q equals E(Q2)-[E(Q)]2.
In order to match the stock price volatility with the tree’s parameters we must, therefore,
have
qu2 + (1-q) d2 – [qu + (1-q)d]2 = σ2Δt
e t d
and substituting q 
we have
u d
eμΔt (u+d) – ud – e2μΔt = σ2Δt
When terms in = Δt2 and higher are ignored, one solution to his equation is
u e
t
d e
t
The analysis shows that we can replace the tree (a) with the tree (b) where the probability
of an up movement is p, and then behave as though the work is risk neutral. The variable
p is given by
p
ert d
u d
It is the risk-neutral probability of an up movement.
21
Exercise
What is the value of a European call option with an exercise price of $40 and a maturity date
six months from now if the stock price is $28, the instantaneous variance of the stock price is
0.5 and the risk-free rate is 6%?
We have: So=$28, K=$40, r=0.06, σ2=0.5, Τ=0.5, and Τ/2=0.25 (for each step of the binomial
tree)
The risk-neutral probability is p 
ert d
=0.433, while u e
u d
t
=1.424 and d e
t
=0.702
The binomial tree, for a call option is the following
D
56.78
16.78
39.87
7.16
28
3.06
B
E
28
0.0
A
19.66
0.0
We know that:
fu = e-rT (p fuu – (1-p) fud)
C
F
13.8
0.0
0.0
fd = e-rT (p fud – (1-p) fdd)
f = e-rT (p fu – (1-p) fd)
Substituting, we have for the price of the call option, c=3.06.
From put-call parity, the price of the put option is p=13.87
22
Black-Scholes pricing model
We begin be considering the discrete-time random walk description:
W(t+1)=W(t) + e(t+1);
W(0)=W0
e: i.i.d. N(0,1)
The variable t represent time and is measured in discrete integer increments.
W(t) is the level of the cumulant of e(t); it is called a random walk because it appears that
W takes random steps up and down through time. We would like to describe a process
that has the same characteristics as the random walk but observed more frequently:
W(t+Δ)=W(t) + e(t+Δ);
W(0)=W0
e: i.i.d. N(0,Δ)
This newly defined process has the same expected drift and variance over n periods as the
first process has over one period (t=1/n). The change ΔW over Δt is ε(Δt)1/2.
Now consider the behavior of the process as Δ → dt
W(t+dt)=W(t) + e(t+dt);
W(0)=W0
e: i.i.d. N(0,dt)
And define dW(t) = W(t+dt)-W(t). We heuristically define dt as the smallest positive real
number such that dta=0, whenever a>1. Either of these processes, dW(t) or e(t+dt), is
referred to as a white noise.
Recall that dW may be thought of as a normally distributed random variable with mean
zero and variance dt.
The process W(t) is referred to as a standard Wiener process.
Arithmetic Brownian Motion
Let a(X,t)=a and σ(X,t)=σ, two constants. Then the process X is said to follow arithmetic
Brownian Motion with drift a and volatility σ, if:
dX = adt + σ dW
The process is an appropriate specification for economic variables that grow at a linear
rate and exhibit increasing uncertainty. The process X has the following properties
(among others):
1. X may be positive or negative
2. If u>t, then Xu is a future value of the process relative to time t. The distribution
of Xu given Xt is normal with mean Xt + a(u-t) and standard deviation σ(u-t)1/2.
3. The variance of a forecast Xu tends to infinity as u does.
23
The three properties indicate that arithmetic Brownian Motion is appropriate for variables
that can become positive or negative, have normally distributed forecast errors, and have
forecast variance that increases linearly in time. Usually, asset returns are assume to
follow arithmetic Brownian Motion.
Geometric Brownian Motion
Let a(X,t)=a and σ(X,t)=σ, two constants. Then the process X is said to follow geometric
Brownian Motion with drift a and volatility σ, if:
dX = aXdt + σX dW
The process is an appropriate specification for economic variables that grow
exponentially at an average rate of a and have volatility proportional to the level of the
variable. The process X has the following properties (among others):
1. If X starts with a positive value, it will remain positive.
2. X has an absorbing barrier at 0: Thus, if X hits 0, then X will remain at zero.
3. The conditional distribution of Xu given Xt is lognormal. The conditional mean of
ln(Xu) for u>t is ln(Xt)+ a(u-t) – (1/2) σ2(u-t) and the conditional standard
deviation of ln(Xu) is σ(u-t)1/2. ln(Xu) is normally distributed. The conditional
expected value of Xu is Xt exp[a(u-t)].
4. The variance of a forecast Xu tends to infinity as u does.
Geometric Brownian motion (GBM) is often used to model security values, since the
proportional changes in security price are independent and identically normally
distributed.
A variable has lognormal distribution if the natural logarithm of the variable is normally
distributed. If the stock price S follows a Geometric Brownian Motion,
dS = aSdt + σSdW
then
dlnS=(μ-σ2/2)dt+σdW
24
from this equation we see that the variable lnS follows a generalized Wiener process. The
change in lnS between time 0 and time T is normally distributes, so that
lnST – lnS0 ~ N[(μ-σ2/2)T, σ(T)1/2]
from this follows that
ln(ST/S0) ~ N[(μ-σ2/2)T, σ(T)1/2]
and
lnST ~ N[lnS0 + (μ-σ2/2)T, σ(T)1/2]
where ST is the stock price at a future time T, S0 is the stock price at time 0, and N(m,s)
denotes a normal distribution with mean m and standard deviation s. The above equation
shows that lnST is normally distributed so that ST has a lognormal distribution.
Example
Consider a stock with an initial price of 40$, an expected return of 16% per annum, and a
volatility of 20% per annum. From the above equation, the probability distribution of the
stock price, ST, in six months time is given by
lnST ~ N[lnS0 + (μ-σ2/2)T, σ(T)1/2],
lnST ~ N[ln40 + (0.16-0.22/2)0.5, 0.2(0.5)1/2]
lnST ~ N[3.759,0.141]
There is a 95% probability that a normally distributed variable has a value within 1.96
standard deviations of its mean. Hence, with 95% confidence interval,
3.759 - 1.96*0.141< lnST <3.759 + 1.96*0.141
This can be written
e(3.759-1.96*0.141) < ST < e(3.759+1.96*0.141)
or
32.55 < ST < 56.56
Thus, there is a 95% probability that the stock price in six months will lie between 32.55
and 56.56.
A variable that has a lognormal distribution can take any value between zero and infinity.
Unlike the normal distribution, it is skewed so that the mean, median, and mode are all
different. From the above equation and the properties of the lognormal distribution, it can
be shown that the expected value of ST is given by
25
E(ST)= S0eμT
This fits with the definition of μ as the expected return. The variance of ST, Var(ST), can
be shown to be given by
 2
Var(ST)= S02e2[e
1]
Example
Consider a stock where the current price is 20, the expected return is 20% per annum and
the volatility is 40% pre annum. The expected stock price in one year, E(S T), and the
variance of the stock price in one year, Var(ST), are given by
E(ST)= S0eμT = 20e0.2*1 = 24.43
 2
Var(ST) = S02e2[e
1] = 400e2*0.2*1[e
0.4 2 *1
1] 103.54
The lognormal property of stock prices can be used to provide information on the
probability distribution of the continuously compounded rate of return earned on a stock
between times 0 and T. Define the continuously compounded rate of return per annum
realized between times 0 and T as h. It follows that
ST= S0ehT
and
h=(1/T)ln(ST/S0)
It follows that
h ~ N[μ-σ2/2, σ(T)-1/2],
Thus, the continuously compounded rate of return per annum is normally distributed with
mean μ-σ2/2 and standard deviation σ(T)-1/2.
Example
Consider a stock with an expected return of 17% per annum and a volatility of 20% per
annum. The probability distribution for the actual rate of return (continuously
compounded) realized over three years is normal with mean
μ-σ2/2 = 0.17- 0.22/2 = 0.15
or 15% per annum and standard deviation
σ(T)-1/2 = 0.2(3)-1/2 = 0.1155
26
or 11.55% per annum. Because there is a chance that a normally distributed variable will
lie within 1.96 standard deviations of its mean, we can be 95% confident that the actual
return realized over three years will be between –7.6% and 37.6% per annum.
We show that μΔt is the expected return provided by S in a very short period of time, Δt.
It is natural to assume that μ is also the expected return on the stock over a relatively long
period of time when this return is expressed with continuous compounding. However,
this is not the case. The last equation for h shows that the expected return continuously
compounded is per μ-σ2/2 year. An important general result is that whenever there is
some uncertainty about the return, then the expected return on an asset per annum in a
period Δt is greater that the expected return on the asset expressed with a compounding
frequency of Δt.
Example
Suppose that the following is a sequence of returns per annum on a stock, measured
using annual compounding:
15%, 20%, 30%, -20%, 25%
Our best estimate of the expected return in one year is calculated by taking the sum of the
returns and dividing by five. It is 14%. To estimate the expected return realized over a
five years with annual compounding, we imagine investing 100 in the stock. Its value at
the end of the five year period would be
100*1.15*1.20*1.30*0.80*1.35=179.40
This corresponds to a return of
(1.7940)1/5 – 1= 0.124
or 12.4% per annum with annual compounding. The expected return per annum over five
years with annual compounding is, therefore, less than the expected return in one year.
Example
A stock price has a volatility of 30% per annum, and provides an expected return of 15%
per annum. The current stock price is 100. What is the expected return and the volatility
of the change ΔS of the stock price, when the time interval is one week?
The process for the stock price is
dS = aSdt + σSdW
27
dS/S = adt + σdW=0.15dt+0.30dW
If S is the stock price at a particular time and ΔS is the increase in the stock price in the
next small interval of time,
dS/S = 0.15Δt+0.30ε(Δt)1/2
where ε is a random drawing from a standardized normal distribution. Consider a time
interval of one week or 0.0192 years and suppose that the initial stock price is 100. Then,
Δt=0.0192 and S=100, and
ΔS=100(0.00288+0.0416ε)
Or
ΔS=0.288+0.0416ε
Showing that the price increase is a random drawing from a normal distribution with
mean 0.288 and standard deviation 4.16
Derivation of the Black-Scholes Differential Equation
Consider a continuous and differentiable function, f, of two variables S and t. If Δt is a
small change in S and Δt is a small change in t, and Δf is the resulting small change in f, a
well known result in ordinary calculus is that
f 
f
f
S  t
S
t
In other words, Δf is approximately equal to the rate of change of f with respect to S and
t, multiplied by ΔS and Δt respectively. If more precision is required, a Taylor series
expansion of Δf can be used:
df 
2 f
f
f
dS  dt  1 2 dS 2 ...
S
t
2 S
The stock price process S is GBM:
dS = μSdt + σS dW
Suppose f is the price of a call option or other derivative contingent on S. The variable f
must be some function of S and t.
Since
dS = μSdt + σS dW,
we have that
28
df (
2 f
f
f
f
S   1 2  2 S 2)dt  SdW
S
t 2 S
S
The discrete versions of the above equations are
ΔS = μSΔt + σS ΔW,
and
f (
2 f
f
f
f
S   1 2  2S 2)t  SW
S
t 2 S
S
where ΔS and Δf are the changes in f and S in a small time interval Δt. It follows that by
choosing a portfolio of the stock and the derivative, the Wiener process can be
eliminated.
The appropriate portfolio is
-1 : derivative
+
f
: shares
S
The holder of this portfolio is short one derivative and long an amount
f
of shares.
S
Define Π as the value of the portfolio. By definition:
Π= - f +
f
S
S
The change ΔΠ in the value of the portfolio in the time interval Δt is given by
ΔΠ=-Δf+
f
ΔS
S
Substituting the above equations in to the last one, we have
(
f 1  2 f 2 2

 S )t
t 2 S 2
Because the equation does not involve ΔW, the portfolio must be riskless during time Δt.
Thus, the portfolio must instantaneously earn the same rate of return as other short-term
risk-free security.
It follows that:
ΔΠ = rΠ Δt
Where r is the risk-free interest rate. Substituting in the last equation, we have
29
(
f 1  2 f 2 2
f

 S )t r(f  S)t
t 2 S 2
S
so that
2 f
f
f
rS  1 2  2 S 2 rf
t
S 2 S
This is the Black-Scholes differential equation. It has many solutions, corresponding to
all the different derivatives that can be defined with S as the underlying variable. The
particular derivative that is obtained when the equation is solved depends on the
boundary conditions that are used.
In the case of a European call option, the key
boundary condition is:
f=max(S-X,0)
when t=T
In the case of a European put option, it is:
f=max(X-S,0)
when t=T
Solving the above differential equation, using the boundary conditions just mentioned,
we have the Black-Scholes formulas for the prices of a European call and put options on
a non dividend-paying stock:
cS0 N(d1 ) XerT N(d2 )
and
p XerT N(d2 )S0 N(d1 )
where
d1 
ln( S0 / X)(r  2 / 2)T
 T
d2 d1  T
and Ν(Χ) is the cumulative probability distribution function for a variable that is
normally distributed with a mean of zero and a standard deviation of 1:
X
 y2 /2
N(X) 1 e
dy
2 
Example
Consider the situation where the stock price six months from the expiration of an option
is 42, the exercise price of the option is 40, the risk-free interest rate is 10% per annum,
30
and the volatility is 20% per annum. This means that S0=42, X=40, r=01, σ=0.2 and
T=0.5. Then
d1 
ln( S0 / X)(r  2 / 2)T ln( 42/ 40)(0.10.22 / 2)0.5

0.7693
 T
0.2 0.5
d2 d1  T =0.7693-0.2 0.51/2=0.6278
Xe-rt =40 Xe-0.05 =38.049
Hence, if the option is a European call option, its value, c, is given by
c = 42 N(0.7693) - 38.049 N(0.6278)
And if the option is a European put option, its value, p, is given by
p = 38.049 N(-0.6278) – 42 N(-0.7693)
Using the polynomial approximation,
N(0.7693) = 0.7791
N(0.7693) = 0.2209
N(0.6278) = 0.7349
N(-0.6278) = 0.2651
So that,
c=4.76 and
p=0.81
Hedging with put options
Portfolio managers Portfolio managers can use index options to limit their downside risk.
Suppose that the value of an index today is S0.
When the portfolio’s beta is 1
Consider a manager in charge of a well diversified portfolio whose beta is 1,0. A beta of
1,0 implies that the returns from the portfolio mirror those from the index. If the dividend
yield from the portfolio is the same as the dividend yield from the index, the percentage
changes in the value of the portfolio can be expected to be approximately the same as the
percentage changes in the value of the index. Each contract on the S&P 500 is on 100
times the index. It follows that the value of the portfolio is protected against the
31
possibility of the index falling below K if, for each 100S0 dollars in the portfolio, the
manager buys one put option contract with strike price K.
Suppose that the manager's portfolio is worth $500.000 and the value of the index is
1.000. The portfolio is worth 500 times the index. The manager can obtain insurance
against the value of the portfolio dropping below $450.000 (Rp =-10%) in the next three
months by buying five put option contracts with a strike price of 900. The strike price is
given by: K= S0(1+ RM), where RM is the market return associated with the allowed
portfolio return Rp (as given by the CAPM) . Suppose that the risk-free rate is 12%, and
the volatility of the index is 22%. The parameters of the option are:
S0=1000,
K=900,
r=0.12,
σ=0.22,
T=0.25,
Using the Black-Scholes formula the price of this put option is $6,48. The cost of the
insurance is therefore 5*100*6,48=3.240.
To illustrate how this works, consider the situation where the index drops to 880 in three
months. The portfolio will be worth about $440.000. The payoff from the options will be
5 x (900 - 880) x 100 = $10.000, bringing the total value of the portfolio up to the insured
value of $450.000.
When the portfolio’s beta is Not 1
If the portfolio's returns are not expected to equal those of an index, the capital asset
pricing model can be used. This model asserts that the expected excess return of a
portfolio over the risk-free interest rate equals beta times the excess return of a market
index over the risk-free interest rate. Suppose that the $500.000 portfolio just considered
has a beta of 2,0 instead of 1,0. Suppose further that the current risk-free interest rate is
12% per annum, and the dividend yield on both the portfolio and the index is expected to
be 4% per annum. As before, we assume that the S&P 500 index is currently 1.000.
Table 1. shows the expected relationship between the level of the index and the value of
the portfolio in three months.
Table 1. Relationship between value of index and value of portfolio for beta = 2,0
32
Value of index
in three months
1.080
1.040
1.000
960
920
880
Value of portfolio
in three months ($)
570.000
530.000
490.000
450.000
410.000
370.000
Suppose that S0 is the value of the index. It can be shown that, for each 100S0 dollars in
the portfolio, a total of beta put contracts should be purchased. The strike price should be
the value that the index is expected to have when the value of the portfolio reaches the
insured value. Suppose that the insured value is $450.000, as in the case of beta = 1,0.
Table 1 shows that the appropriate strike price for the put options purchased is 960. In
this case, 100S0 = $100,000 and beta = 2,0, so that two put contracts are required for each
$100.000 in the portfolio. Because the portfolio is worth $500.000, a total of 10 contracts
should be purchased.
Alternatively, to find the strike price we use the CAPM:
Rp + q = Rf + [RM + q – Rf]b where we find that for Rp =-10% the Rm is -4%.
Then,
K= S0(1+ RM)=960
To illustrate that the required result is obtained, consider what happens if the value of the
index falls to 880. As shown in Table 1, the value of the portfolio is then about $370.000.
The put options pay off (960 — 880) x 10 x 100 = $80.000, and this is exactly what is
necessary to move the total value of the portfolio manager's position up from $370.000 to
the required level of $450.000.
33