# Expected value of transformed Gaussian random variable

I have a question regarding an equation from the first chapter in "Financial Calculus: An Introduction to Derivative Pricing" by Martin Baxter and Andrew Rennie.
The background is the computation of an expected value of a stock $$\mathbb{E} (S_t)$$, which in the setup described is nothing else than $$\mathbb{E} (S_0 \exp(X))$$ (continuously compounded). X is assumed to follow a normal distribution with $$\mu$$ and $$\sigma$$.
Now the author on recalls the "law of the unconscious statistician", which states $$\mathbb{E}(h(x)) = \int_{-\infty}^{\infty} h(x) \dot f(x) dx$$. So far so good.

The problem lies in the combination of this law with the fact that $$X$$ follows a normal distribution with density $$f(x) = \frac{1}{\sqrt{2 \cdot \pi \cdot \sigma^2}} \cdot \exp(\frac{-(x - \mu)^2}{2 \cdot \sigma^2})$$.
Now the expectation of the stock value derived out of the "law" is stated as $$\mathbb{E} (S_t) = S_0 \cdot \exp(\mu + \frac{1}{2}\sigma^2).$$
How can we deduce that out of the "law" and the integral? If I compute stuff inside the integral, I don't get anywhere. Undergrad here and my first question, would highly appreciate if you could tell me the trick applied here.

Grüße aus Deutschland.

• After substituting $f$ in the expectation, you need to complete the square for the power of exponential. Commented Oct 9, 2020 at 3:44

$$M_X(t)=\mathbb{E}(e^{tX})=\int_{-\infty}^{\infty}e^{tx}f(x)dx=\exp{[t^2\frac{\sigma^2}{2}+\mu t]}.$$
Now evaluate such result at $$t=1$$ (because we are looking for the expected value of $$e^X$$), and you obtain the result in your book.