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7/28/2019 Black Scholes Notes

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shorter. One minus the probability that someone is shorter than me is the probability that theyre

taller than me. Because of symmetry, being unusually tall is the same as being unusually short. So

we can also use N(z) = 1 N(z) to calculate probabilities. From the normal table, there is a0.1587 probability that someone is one standard deviation below the mean or shorter. Once again,

I get that 15.87% of people are taller than me. Both approaches always work.

How unusual is someone 205cm? For this problem, A is 205cm. The z-score is

z=205 175

10= 3. (3)

I look that up on a normal table and find that the probability of being 205cm or below is 0.9986.

Alternatively, you have only a 0.0014 probability of falling three standard deviations or more below

the average height. Either way you look at the problem, there is almost a 100% probability that a

person is shorter than 205cm. If you are 205cm, you expect only one person in a thousand taller

than you.

2 We Assume Returns are Normal Random Variables

We assume that returns are normally distributed. We assume that the mean return using risk-

neutral probabilities is Rf 2

2 T. So z-scores will use that as the mean. We assume the

standard deviation is T. So, really we have enough information to figure out z-scores.

Heres an example. Suppose the risk-free rate is 5% and the standard deviation of Telstra returns

is 20% per year. Whats the risk-neutral probability that the return was 10% or greater over the

next six months?

The mean return is

(0.05 0.202

2) 0.5 = 0.015, (4)

so the z-score for a 10% return is

z=0.10 0.015

0.2

0.5= 0.60. (5)

Looking that up on a normal table, N(0.60) = 0.7257. That means theres a 73% chance that a

return is 10% or below given our average return. If theres a 73% chance that the return is below

10%, there is a 27% chance that the return is above 10%. Alternatively, we could look up N(0.60)to find a probability of 0.2743. Once again, either approach gives the same answer.

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3 Exercising a Call Option Requires a High Enough Return

over the Life of the Option

Suppose that the price of Telstra stock is currently $3.00. You have a six month call option onTelstra stock with a strike price of $3.05. Suppose the risk-free rate is 5% and the standard deviation

of Telstra returns is 20%. Whats the risk-neutral probability that you exercise the option?

Well, exercise depends on the stock return. If the current price is $3.00, we need a continuously-

compounded return of at least ln

3.05

3.00

= ln[3.05] ln[3] = 0.0165 before wed exercise. If the

return is not at least that big, the stock price ends up below $3.05 and we dont exercise. So what

is the probability that the return is at least that high?

The mean return is(0.05 0.20

2

2) 0.5 = 0.015, (6)

so the z-score for a 1.65% return is

z=0.0165 0.015

0.2

0.5= 0.01. (7)

Looking that up on a normal table, N(0.01) = 0.5040. There is about a 50.4% chance that the

return is lower than that number, which means theres about a 49.6% chance that the return is

higher than that number. Alternatively, we could look up N(0.01) to find a probability of 0.4960.Once again, either approach gives the same answer.

In general, the z-score we looked up is:

z=

A ln[K] ln[S0]

Mean Rf

2

2

T

T

Standard Deviation. (8)

while d2 is defined as

d2 =

ln[S0] ln[K] +Rf

2

2

T

T

= z. (9)

Thus, d2 is the z-score wed look up to answer the following question: What is the probability that

a return big enough to make our call option end up above the strike price? That means N(d2) tells

us the probability that well exercise the call option because the return was high enough.

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