Given: X ~ Gaussian($\mu$, $\sigma^{2}$) and Y = $e^{X}$. Compute $E(Y)$ and $Var(Y)$.

Question

Not sure how to start this problem at all. Could someone help?

Answer 1

• For $\mathbb{E}[Y]$, we have

\begin{align} \mathbb{E}[Y]\ &=\ \int_{-\infty}^{\infty} e^x\cdot \frac{1}{\sqrt{2\pi}\sigma}e^{-(x - \mu)^2 / 2\sigma^2} dx \\ &=\ \int_{-\infty}^{\infty}e^{\mu + y}\cdot \frac{1}{\sqrt{2\pi}\sigma}e^{-y^2 / 2\sigma^2}dy \\ &=\ e^\mu\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}\sigma}e^\frac{-y^2 + 2\sigma^2y}{2\sigma^2} dy \\ &=\ e^\mu\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}\sigma}e^\frac{-(y - \sigma^2)^2 + \sigma^4}{2\sigma^2} dy \\ &=\ e^\mu e^{\sigma^2/2} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}\sigma}e^\frac{-(y - \sigma^2)^2}{2\sigma^2} dy \\ &=\ e^{\mu + \frac{\sigma^2}{2}} \end{align} For the second equality, we use transform $y = x - \mu$.

• For $\mathbb{E}[Y^2]$, you can derive it similarly and the result is $\mathbb{E}[Y^2] = e^{2\mu + 2\sigma^2}$.

• Use $Var[Y] = \mathbb{E}[Y^2] - \mathbb{E}[Y]^2$ to obtain $Var[Y]$ then.

Note: Given a normal random variable $X$, the random variable $Y = e^X$ follows log-normal distribution.