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!
Best Answer
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)