# 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

## 1 Answer

The trick is the use of the moment generating function of a normal random variable:

$$M_X(t)=\mathbb{E}(e^{tX})=\int_{-\infty}^{\infty}e^{tx}f(x)dx=\exp{[t^2\frac{\sigma^2}{2}+\mu t]}.$$

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.

• Thank you, great job! So basically we have the "coincidence" of our function h(x) to be exactly constructed in the way we get the moment generating function, right? Makes perfect sense. Commented Oct 9, 2020 at 7:58