Expected Value of a standard normal variable

Question

Suppose that Z has a standard normal distribution...

Evalutate $E[e^z]$

I am having some issues figuring out how to go about this problem.

I understand that $E[e^Z]$ is equal to the integral of $-\infty$ to $\infty$ of $ \dfrac{1}{(2\pi)^.5}e^\dfrac{-z^2}{2}e^z $

I have simplified the integral to $e^{z-z^2/2} $

but I am not sure to go about integrating this. I plugged it into wolframalpha and it gave me $e^.5$ but no explanation.

Answer 1

Note that \begin{align*} E[e^Z]&=\frac{1}{\sqrt {2\pi}}\int_{-\infty}^{\infty} e^ze^{-z^2/2}dz\\ &=\frac{1}{\sqrt {2\pi}}\int_{-\infty}^{\infty} e^{-z^2/2 +z}dz\\ &=\frac{1}{\sqrt {2\pi}}\int_{-\infty}^{\infty} e^{-(z-1)^2/2 +1/2}dz\\ &=\frac{\sqrt{e}}{\sqrt {2\pi}}\int_{-\infty}^{\infty} e^{-u^2/2}du\\ &=\sqrt{e}. \end{align*}

Answer 2

Just complete the square in the integral:

$$\exp(-\tfrac{1}{2}z^2 + z) = \exp(-\tfrac{1}{2}(z^2 - 2z)) = \exp(-\tfrac{1}{2}(z^2 - 2z + 1) + \tfrac{1}{2}) = \exp(-\tfrac{1}{2}(z - 1)^2) \sqrt{e}$$

Now you just need to substitute $z - 1 = t$.