# Finding marginal distributions of a process

Suppose we have a process given by $S_t = S_0 \exp(\sigma W_t + (r - \frac{1}{2} \sigma^2 )t)$, and we wish to find the marginal distribution for $S_T$. (Note: $W_t$ is a $\mathbb{Q}$-Brownian Motion)

My book states that the marginal for $S_T$ is given by $S_0$ times the exponential of a normal with mean $(r - \frac{1}{2} \sigma^2)T$ and variance $\sigma^2 T$. How did they arrive at this conclusion?

I know that a random variable $X$ is a normal $N(\mu, \sigma^2)$ $\iff$ $\mathbb{E}[\exp(\theta X)] = \exp(\theta \mu + \frac{1}{2} \theta^2 \sigma^2)$. Is this the key observation, or am I missing something?

• Well $W_T$ is a normal random variable with mean zero and variance $T$, so shouldn't the distribution of $S_T$ follow directly from its definition? Commented Jun 2, 2016 at 23:47

$$dS_t=rS_tdt+\sigma\,S_tdW_t$$ by application of Ito's lemma for $X_t=\ln{S_t}$ we have $$dX_t=\left(r-\frac{1}{2}\sigma^2\right)dt+\sigma\,dW_t$$ as a result $$X_T=X_0+\left(r-\frac{1}{2}\sigma^2\right)T+\sigma\,W_T$$ so we can say $X_T$ has a normal distribution because $W_T$ is Guassian process with mean zero and variance $T$. Finally $$E[X_T]=X_0+\left(r-\frac{1}{2}\sigma^2\right)T$$ and $$Var[X_T]=\sigma^2T$$
Your book is being quite loose with terms. $S_t$ is lognormally distributed, meaning $\log S_t$ is normally distributed. To see this, take logs: $$\log S_t = \log S_0 + \left( r - \frac{\sigma^2}{2}\right)t + \sigma \sqrt{t} Z$$ where $Z \sim \mathcal{N}(0, 1)$. So, $$\log S_t \sim \mathcal{N}\left( \log S_0 + \left(r - \frac{\sigma^2}{2}\right)t, \sigma^2 t\right).$$ By the way, since I assume this is finance related, we can first subtract the $\log S_0$ from both sides of the first equation to see $$\log \left(\frac{S_t}{S_0}\right) \sim \mathcal{N}\left(\left(r - \frac{\sigma^2}{2}\right)t, \sigma^2 t\right).$$ In other words, this is what people mean when they say, "log returns are normal"
• Can you explain how you got $\sigma \sqrt{t} Z$ from $\sigma dW_t$? Commented Jun 3, 2016 at 2:52
• @Javier $W_t \sim N(0,t)$, so $\sigma W_t = \sigma\sqrt{t}Z$.