Expected Value of normally distributed r.v. is finite Let $X\sim \mathcal{N}(\mu, \sigma^2)$. How do I proof that $X\in L^p(\mathbb{P})$ for all $p\in[0,\infty)$?
Edit// First of all thank you, this is what I tried. Let $\mathcal{N}(0, 1)$
$$
E(|X|^n)=\int_{-\infty}^\infty |x|^n\displaystyle{(\frac{1}{\sqrt{2\pi} }e^{\frac{-x^2}{2}})} dx = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty|x|^{n-1}(|x|e^{\frac{-x^2}{2}})dx
$$
Partiall integration with
$$
u=|x|^{n-1};dv=|x|e^{\frac{-x^2}{2}}
$$
gives us
$$
du=(n-1)|x|^{n-1}; v=-e^{\frac{-x^2}{2}}
$$
We get
$$
E(X^n)=\frac{1}{\sqrt{2\pi}}(-|x|^{n-1}e^{\frac{-x^2}{2}}|_{-\infty}^{\infty}+(n-1)\int_{-\infty}^\infty|x|^{n-2}e^{\frac{-x^2}{2}}dx = \frac{n-1}{\sqrt{2\pi}}\int_{-\infty}^\infty|x|^{n-2}e^{\frac{-x^2}{2}}dx = (n-1)E(|X|^{n-2})
$$
We can repeat the same steps for $E(|X|^{n-2})$ again and so on. And $E(X^0)=\int_{-\infty}^\infty x^0 \frac{1}{\sqrt{2\pi} }e^{\frac{-x^2}{2}}=1$ so
$$
E(|X|^n)=(n-1)\cdot(n-3)\cdot(n-5)\cdot...<\infty
$$
 A: First, show that $E(e^X)$ is finite: the integral can be done by just completing the square. Then use the fact that $e^X$ dominates any polynomial.
A: It seems to me that it would be simpler to work with the explicit expression for $\mathbb{E}[\vert X\vert^p]$ and perhaps integration by parts. I think the following observations should do:
(1) $\mathbb{E}[\vert X\vert^p] =\int_{-\infty}^\infty \vert x\vert^p\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx$
(2) For $Z\sim N(0,\sigma^2)$ we have $\mathbb{E}[\vert Z\vert^p] =2\int_{0}^\infty z^p\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{z}{2\sigma^2}}dz$
(3) $\mathbb{E}[\vert Z\vert^p] $ is finite if and only if $\int_{1}^\infty z^p\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{z}{2\sigma^2}}dz$ is finite
(4) $\vert z\vert^p$ is increasing with respect to $p$ for $z>1$, and so it is enough to show that $\int_{1}^\infty z^{\lceil p \rceil}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{z}{2\sigma^2}}dz$ is finite.
(5) when $m$ is an natural number, integration by parts shows that $\int_{1}^\infty z^{m}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{z}{2\sigma^2}}dz$ is finite.
Perhaps there is a more elegant solution, but this seems like the easiest method to me.
