# 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.

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.

• Ah, thank you! I failed to recall Ito's Lemma, so I missed that explanation. – Eric Emer Apr 24 '13 at 7:15

Another way of seeing it is that the $-\frac12\sigma^2$ is just a correction term that comes from Jensen's inequality.

You need this when switching from supposedly symmetric returns (normal distribution) to the skewed price process (log-normal distribution).

The term 1/2 * sigma-squared arises through the application of Ito's Lemma. Keep in mind that the assumption is of a stock price that follows geometric BM with a constant drift and volatility. If you set up a delta-hedge portfolio and apply Ito calculus you will end up with an adjustment in the distribution by exactly above term. Another way of interpreting the term is that it represents the difference between the mean and the median of the log-normal distribution.

I do not have the book with me right now but Steven Shreve in his Stochastic Calculus II book has one of the most logical and easily understood derivations of Black Scholes through change of probability measure and it will make the shift in mean crystal clear.

I am not a huge fan of posting lengthy formulae nor am I a mathematician by heart so please take a look at the cited reference.

You can refer to Shreve's book, Volume II, Section 4.4.3 .

Assume that we have a generalized geometric Brownian motion $$dX_t = \sigma_t dW_t + (\alpha_t - \frac{1}{2} \sigma_t^2) dt ,$$ where the drift coefficient and the volatility are functions of $t$ also. $(dX_t)^2 = \sigma_t^2 dt + \mathcal{O}(dt^{3/2})$ .

Assume that the asset price is $$S_t = S_0 e^{X_t} = S_0 e^{\int_0^t \sigma_s dW_s + \int_0^t \alpha_s - \frac{1}{2} \sigma_s^2 ds} .$$

Let $S_t = f(X_t)$ and apply Ito's lemma to $df(X_t)$. Note that $f(x) = S_0 e^{x}$ and $\partial_x f = S_0 e^{x} = \partial^2_x f$. $$dS_t = df(X_t) = \partial_x f(X_t) dX_t + \frac{1}{2} \partial^2_x f(X_t) dX_t^2 = \alpha_t S_0 e^{X_t} dt + \sigma_t S_t dW_t ,$$ or $$\frac{dS_t}{S_t} = \alpha_t dt + \sigma_t dW_t .$$

In other words, for time-varying drift $\alpha_t$ and volatility $\sigma_t$, $S_t = S_0 \exp\left(\int_0^t \sigma_s dW_s + \int_0^t \alpha_s - \frac{1}{2} \sigma_s^2 ds \right)$ is the solution to the stochastic differential equation $\frac{dS_t}{S_t} = \alpha_t dt + \sigma_t dW_t$.

Note that, if $\alpha = 0$ and $\sigma_t = \sigma$, $Z_t = \exp \left( \sigma W_t - \frac{1}{2} \sigma^2 t\right)$ is a martingale.