Determine $\mathbb{E}|X|^p$ for all $p > -1$ when $X$ has a normal distribution $\mathcal{N}(0, \delta^2)$ (I am stuck with the integral).

Question

How to determine $\mathbb{E}|X|^p$ for all $p > -1$ when $X$ has a normal distribution $\mathcal{N}(0, \delta^2)$?

I know that:

$$\mathbb{E}|X|^p = \int_{-\infty}^{\infty} |x|^p \cdot f(x)\mathrm{d}x$$ where $f(x)$ is the probability density function of $X.$

The probability density function of $X$ is $$f(x) = \frac{1}{\sqrt{2\pi}}e^{\frac{-x^{2}}{2}}$$ which is the density of the standard normal distribution.

$$\mathbb{E}|X|^p = \int_{-\infty}^{\infty} |x|^{p} \cdot \frac{1}{\sqrt{2\pi}}e^{\frac{-x^{2}}{2}}\mathrm{d}x = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} |x|^{p} \cdot e^{\frac{-x^{2}}{2}}\mathrm{d}x = $$

We put $z = \frac{-x^2}{2} \implies dz = -x \ dx \iff dx = \frac{dz}{-x}$ so we get:

$$= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{0} x^{p-1} \cdot e^{\frac{-x^2}{2}}dz -\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty} x^{p-1} \cdot e^{\frac{-x^2}{2}}dz =$$

$$= \frac{1}{ \sqrt{2 \pi} } \left( x^{p-1} \cdot e^{ \frac{-x^2}{2} } \Big|_{-\infty}^{0} \right) -\frac{1}{ \sqrt{2 \pi} } \left( x^{p-1} \cdot e^{ \frac{-x^2}{2} } \Big|_{0}^{\infty} \right) = \ ?$$

I don't know what to do next.

Answer 1

What is the density for a $N(0,\delta^{2})$ variate?. Is it not $f(x)=\frac{1}{\sqrt{2\pi}\sigma}\exp(-\frac{x^{2}}{2\sigma^{2}})$ . Then what would be $E(|X|^{p})$? Would it not be $\int_{-\infty}^{\infty}|x|^{p}f(x)\,dx$ ? . In particular what GEdgar has done is shown the moments for $N(0,1)$ variate. You can recover $N(0,\delta^{2})$ case by noting that if $X$ has $N(0,1)$ distribution then $\delta X$ has $N(0,\delta^{2})$ distribution.

Now you have that for a standard normal variate $X$,

$E(|X|^{p})=\int_{-\infty}^{\infty} |x|^{p} \cdot \frac{1}{\sqrt{2\pi}}e^{\frac{-x^{2}}{2}}\mathrm{d}x=2\int_{0}^{\infty}x^{p} \cdot \frac{1}{\sqrt{2\pi}}e^{\frac{-x^{2}}{2}}\mathrm{d}x$ .

Now substitute $\frac{x^{2}}{2}=z$ or $x=\sqrt{2z}$ to get that the above equals

$$2\cdot\int_{0}^{\infty} \frac{2^{\frac{p}{2}}}{\sqrt{2z}}z^{\frac{p}{2}}\frac{1}{\sqrt{2\pi}}e^{\frac{-x^{2}}{2}}\mathrm{d}x=2^{\frac{p}{2}}\frac{1}{\sqrt{\pi}}\int_{0}^{\infty}z^{\frac{p-1}{2}}e^{-z}\,dz$$

Now compare this to the definition of the Gamma Function to get that

$$\frac{2^{\frac{p}{2}}}{\sqrt{\pi}}\Gamma(\frac{p+1}{2})$$

Now these are for the standard normal variate. To get for $N(0,\delta^{2})$ variate notice that if $X$ has standard normal distribution then $Z=\delta\cdot X$ has mean $0$ and variance $\delta^{2}$ and hence is $N(0,\delta^{2})$ distributed.

So $\displaystyle E(|Z|^{p})=E(|\delta X|^{p})=\delta^{p}E(|X|^{p})=\delta^{p} \frac{2^{\frac{p}{2}}}{\sqrt{\pi}}\Gamma(\frac{p+1}{2})$ .

Answer 2

Your answer is not an elementary function (exception, $p$ is an integer). You can answer in terms of the Gamma function: $$ 2\,\int_{0}^{\infty }\!{\frac {{x}^{p}{{\rm e}^{-{x}^{2}/2} }}{\sqrt {2 \pi}}}\,{\rm d}x={\frac {1}{\sqrt {\pi}}{2}^{{ \rho/2}}\Gamma \left( {\frac {p}{2}}+{\frac{1}{2}} \right) } $$

I did the case where $X$ has distribution $\mathcal N(0,1)$.