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:



1 Answer 1


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 $\log S_T > \log K$.


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

we get after some rearrangement,

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


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

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


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