options - In Black-Scholes, why is $log{frac{S_{t+triangle t}}{S_t}} sim phi{((mu - frac{1}{2}sigma^2)triangle t, sigma^2 triangle t)}$?

I don't understand why in the formula

$$\log{\frac{S_{t+\triangle t}}{S_t}} \sim \phi{\left((\mu - \frac{1}{2}\sigma^2)\triangle t, \sigma^2 \triangle t\right)}$$ the mean is $(\mu - \frac{1}{2}\sigma^2)\triangle t$ and not just $\mu \triangle t$. I am aware that it is supposed to represent a lognormal distribution, but I guess I'm missing something, or that explanation isn't simple enough.

Answer

So we have the BS-Model

$$dS_t=S_t(\mu dt +\sigma dW_t)$$

W.l.o.g we assume $S_0=1$. Itô's lemma implies that

$$S_t=\exp{(\sigma W_t+(\mu-\frac{1}{2}\sigma^2)t)}$$

We know that $W_t$ is normally distributed with mean $0$ and variance $t$. Now have a look at the r.v.

$$X_t=\sigma W_t+(\mu-\frac{1}{2}\sigma^2)t$$

$\sigma W_t$ is the random part and $\gamma:=(\mu-\frac{1}{2}\sigma^2)t$ is deterministic. Hence $E[X_t]=\sigma E[W_t]+\gamma=\sigma\cdot 0+\gamma=\gamma$. We also have the rule $Var(Y+a)=Var(Y)$, for constants $a$ and a r.v. $Y$. Hence the variance of $X_t$ is given by $\sigma^2t$.

By properties of the $\exp(x)$ function, we have

$$\frac{S_{t+\Delta t}}{S_t}=\exp{(\sigma(W_{t+\Delta t}-W_t)+(\mu-\frac{1}{2}\sigma^2})(t+\Delta t-t))=\exp{(\sigma(W_{t+\Delta t}-W_t)+(\mu-\frac{1}{2}\sigma^2})\Delta t)$$

You can apply the same argument as for $X_t$, using that $W_{t+\Delta t}-W_t\sim\mathcal{N}(0,\Delta t)$.

Why it should be the lognormal distribution should be clear. Let me know if something is not clear to you.