Compute $L_p$ norm of exponential of standard Gaussian

Question

I am trying to compute the $L_p$ norm of $X$ such that $X$ is $e^{Y}$ and $Y$ is the standard Gaussian.

I started with the fact that

$$\|X\|_{Lp} = \left(E\left|e^{Y}\right|^{p}\right)^{1/p}$$

and that

$$E \left(e^{pY}\right) = \int_{-\infty}^\infty e^{pY} \frac{1}{\sqrt{2\pi}} e^{-\frac{Y^2}{2}} dy$$

Am I in the correct direction or did I make some mistake defining this integral?

Answer 1

You are in the good direction: complete the square to get $$ -\frac{y^2}2+py=-\frac 12\left(y^2-2py\right)=-\frac 12\left((y-p)^2-p^2\right) $$ and do the substitution $t=y-p$.