Proving a normal random variable $ X \in L^p $ for all $ p \in [0, \infty ) $

Question

so I am struggling with the following task:

X ~ N($ \mu $, ${\sigma}^{2} $). Show that X $ \in $ $L^{p}($P$) $ for all p $ \in $ [1,$ \infty $).

So my idea (which I suppose is correct :D ) is to prove that the integral

$ \int_R x^p $ $\frac{1}{\sqrt{2\pi\sigma^2}}$ $ e^{-0.5 \frac{(x-\mu)^2}{\sigma^2}} dx $ is finite. I can further use the substitution $ X = \mu + \sigma Z $

(where E[Z] = 0, Var[Z] = 1) to make it bit simpler (or maybe not :). This leaves us with the integral

$\frac{1}{\sqrt{2\pi}} \int_R (\mu + \sigma z)^p e^{-0.5z^2} dz $ I need to show the integral is finite (the function that is inside is integrable), but I have no idea how to go further. Any tips?

Thanks in advance.

Answer 1

Since $$\lim_{x\to \infty }x^{p+2}e^{-x^2}=0,$$ for all $p\geq 0$, $$x^pe^{-x^2}=o\left(\frac{1}{x^2}\right),\quad \text{when }x\to \pm\infty, $$ and thus $(x\mapsto x^pe^{-x^2})\in L^1(\mathbb R)$. Therefore $X^p\in L^p(\Omega )$ as wished.

Answer 2

For every $p\geq 0$, the integral $$\int_{\mathbb{R}}(\mu+\sigma z)^pe^{-z^2/2}\,dz$$ converges absolutely. To see this, choose $M_p>0$ sufficiently large so that $$e^{-z^2/2}<\frac{1}{|\mu+\sigma z|^{p+2}}\quad\forall |z|>M_p$$ and then argue separately on the intervals $[-M_p,M_p]$ and $(M_p,\infty)$, $(-\infty,-M_p)$.

($M_p$ also may depend on $\sigma,\mu$, but it is not terribly important here).