Deriving the Black–Scholes Probability of Expiring In the Money

Summary

The document derives the probability that a call finishes above its strike when the underlying follows geometric Brownian motion. At expiration, the asset price is lognormally distributed, so taking its logarithm converts the event that the price exceeds the strike into an inequality involving a normally distributed variable. Standardizing that variable gives a normal-tail probability.

By symmetry of the standard normal distribution, the probability is the cumulative normal function evaluated at d2. The derivation defines d2 using the initial asset price, strike, volatility, time to expiration, and drift. The drift must match the probability measure being used: under the risk-neutral measure it is the risk-free rate (with applicable carry adjustments), while a real-world probability uses the physical drift. This probability is distinct from the call’s risk-neutral pricing weight in the Black–Scholes formula, and the result assumes the model’s lognormal dynamics and constant volatility.

Key ideas

• Under geometric Brownian motion, the asset price at expiration is lognormally distributed.
• Taking logarithms turns the in-the-money condition into a threshold event for a normal variable.
• Standardizing the log-price produces the parameter d2, and normal-distribution symmetry yields the cumulative probability at d2.
• The drift used in d2 depends on whether the probability is measured under a risk-neutral or real-world model.

Tags

# How to get the probability of exercise call option in Black-Scholes model?


# How to get the probability of exercise call option in Black-Scholes model?












From Black-Scholes model, I'm trying to prove:

$p(S_t>K) = N(d_2)$

No luck yet!

Can anyone suggest a reference showing that how to obtain this equation?

All I get is:

$S_t = S_0e^{ (\mu-0.5 \sigma^2)t+\sigma B_t }$

And I looked for:

$E[S_t>K] $

Yet, could not make it to:

$N(d_2)$

## Answer by RRL (score 12)

https://quant.stackexchange.com/a/44233

With the underlying asset price $S_t$ following a geometric Brownian motion with drift $\mu$ (risk-neutral or otherwise) , we have at time $t = T$,

$$S_T = S_0e^{(\mu- \frac{\sigma^2}{2})T}e^{\sigma B_T} = S_0e^{(\mu- \frac{\sigma^2}{2})T}e^{\sigma\sqrt{T}\xi}$$

where $\xi \sim N(0,1)$ is a standard normal random variable. That is, $S_T$ is lognormally distributed.

The probability that a call option with strike price $K$ expires in the money is

$$P(S_T > K) = P(\log S_T > \log K) = P(\log\frac{S_T}{K} > 0),$$

since the natural logarithm is a monotone function and $S_T > K$ if and only if $\log S_T > \log K$.

Using

$$\log \frac{S_T}{K} = \log \frac{S_0e^{\mu T}}{K} - \frac{\sigma^2T}{2} + \sigma \sqrt{T} \xi,$$

we get after some rearrangement,

$$P(S_T > K) = P(\xi > -d_2)$$

where

$$d_2 = \frac{\log \frac{S_0e^{\mu T}}{K}}{\sigma \sqrt{T}} - \frac{1}{2}\sigma\sqrt{T}$$

By the symmetry of the normal distribution, we have $P(\xi > -d_2) = P(\xi < d_2) = N(d_2)$.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.