Expectation of $\exp(X)$ when $X \sim N(\mu,\sigma^2)$

Question

Let be $X \sim N(2,4)$. What is the mathematical expectation of $e^{X}$?

My approach:

Let be $Z=\exp(X)$, then

\begin{align*} \mathbb{F}_{Z}(z)&=\mathbb{P}(Z\leq z) \\&=\mathbb{P}(\exp(X)\leq z)\\&=\mathbb{P}(X\leq \ln z)\\ &=\mathbb{F}_X(\ln z) \end{align*}

So,

\begin{align} \mathbb{E}(Z)&=\int_{- \infty}^{\infty} z \cdot f_{Z}(z)\\&=\int_{- \infty}^{\infty} z \cdot \underbrace{f_{X}(\ln z)}_{\text{Is this correct?}}\\&=\frac{1}{\sqrt{2 \pi}}\cdot \int_{- \infty}^{\infty} z \cdot \exp\left ( - \frac{(\ln z - 2)^{2}}{8} \right ) dz \end{align}

My doubts:

• Am I correct until this point?
• If it is necessary to compute that integral, do you have any hint?
• Is there another way, easier and faster, to compute that expectation?

I really appreciate your help! Thank you very much!

Answer 1

By definition of expectation,

$$\begin{align} \mathbb E[e^X] &= \int_{-\infty}^\infty e^x f_X(x)\,\mathrm dx \\[1ex] &= \frac1{4\sqrt{2\pi}} \int_{-\infty}^\infty e^{x-\frac1{32}(x-2)^2}\,\mathrm dx \\[1ex] &= \frac{e^{10}}{4\sqrt{2\pi}} \int_{-\infty}^\infty e^{-\frac1{32}(x-18)^2}\,\mathrm dx \end{align}$$

Can you see the PDF of another normal distribution here? (See LOTUS)