# 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:

$$N(d_2)$$

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 $$P(\xi > -d_2) = P(\xi < d_2) = N(d_2)$$.