Find a moment of a distribution, given another moment

Question

Let $X$ be a random variable with the following density function:

$$f(x)= {1\over\sqrt{18\pi}}e^{a(c-x)^2}$$ where $-\infty<x<\infty$.

Given $\mathbb{E}[X^2] = 9$, find $\mathbb{E}[X^8]$.

Now I know the formula to find $\mathbb{E}[X^8]$ but I could not understand the distribution of $X$. It seems to have standard normal distribution with $\sigma = 3$, $c=0$ and $a=1/18$. But I am not sure. If it is normally distributed how can I show it?

I tried $\int{x^2f(x)} = 9$ to find $a$ and $c$ but couldn't do it. Any help is appreciated.

Answer 1

Hint: Use $\int_{-\infty}^{\infty} f(x) dx = 1$ to find $a$. Use $E(X^2)=9$ to compute $c$. You will realize that the distribution is a Gaussian distribution with $c=0$ and $a = -1/18$. Afterwards, use $E(X^n)=1\cdot 3 \cdots (n-1) \sigma^n$ for even $n$ to compute $E(X^8)$.

Answer 2

First of all $f$ is a density function hence $$\int_{\Bbb R} f(x) dx = 1$$ should hold.

That gives $\alpha = -\frac{1}{18}$.

Then you have $E[X^2] = 9$ what leads to $c=0$ if you plug in $a = -\frac{1}{18}$.

Then use that density to calc $E[X^8]$

Answer 3

The PDF of a normal distribution with mean $\mu$ and standard deviation $\sigma$, abbreviated $\mathcal{N}(\mu, \sigma^2)$, is $$\dfrac{1}{\sigma\sqrt{2\pi}}\exp\left[\frac{-1}{2}\left(\frac{x-\mu}{\sigma}\right)^2\right]$$ for $x \in \mathbb{R}$. Thus, considering your PDF, you have $$f(x) = \dfrac{1}{\sqrt{18\pi}}e^{a(c-x)^2}=\dfrac{1}{\sqrt{9}\sqrt{2\pi}}e^{a(x-c)^2}=\dfrac{1}{3\sqrt{2\pi}}e^{a(x-c)^2}\text{.}$$ Hence, $X \sim \mathcal{N}(c, 9)$.

It follows that $$\mathbb{E}[X^2]=\text{Var}(X)+\left(\mathbb{E}[X]\right)^2 =9+c^2=9$$ so that $c^2 = 0$, or $c=0$.

Thus, $$f(x) = \dfrac{1}{3\sqrt{2\pi}}\exp\left[\dfrac{-1}{2}\left(\dfrac{x}{3} \right)^2 \right]$$

for $x \in \mathbb{R}$. Now $$\mathbb{E}[X^8] = \int_{0}^{\infty}\dfrac{1}{3\sqrt{2\pi}}x^8\exp\left[\dfrac{-1}{2}\left(\dfrac{x}{3} \right)^2 \right]\text{ d}x$$ Perform the substitution $z = \dfrac{x}{3}$, so that $\dfrac{\text{d}z}{\text{d}x} = \dfrac{1}{3}$. Then $$\begin{align}\int_{0}^{\infty}\dfrac{1}{3\sqrt{2\pi}}x^8\exp\left[\dfrac{-1}{2}\left(\dfrac{x}{3} \right)^2 \right]\text{ d}x &= \int_{0}^{\infty}\dfrac{1}{\sqrt{2\pi}}(3z)^8\exp\left[\dfrac{-1}{2}z^2 \right]\text{ d}z\\&=\dfrac{6561}{\sqrt{2\pi}}\int_{0}^{\infty}z^8e^{-z^2/2}\text{d}z\\ &=6561\int_{0}^{\infty}\dfrac{1}{\sqrt{2\pi}}z^7ze^{-z^2/2}\text{d}z\text{.}\end{align}$$ One can notice that the integral above is $\mathbb{E}[Z^8]$, where $Z \sim \mathcal{N}(0, 1)$.

Now we integrate by parts. Let $u = z^{7}$ and $\text{d}v = ze^{-z^2/2}$, so that $\text{d}u=7z^{6}\text{ d}z$ and $v=-e^{-z^2/2}$. Then $uv$ (when evaluated from $0$ to $\infty$) will yield $0$, and $$-\int_{0}^{\infty}v\text{ d}u=\int_{0}^{\infty}\dfrac{1}{\sqrt{2\pi}}7z^{6}e^{-z^2/2}\text{ d}z=7\int_{0}^{\infty}\dfrac{1}{\sqrt{2\pi}}z^6e^{-z^2/2}\text{ d}z=7\cdot\mathbb{E}[Z^6]\text{ d}z\text{.}$$ If we were to repeat the above procedure, we would obtain $$\mathbb{E}[Z^8]=6561(7)\mathbb{E}[Z^6]=6561(7\cdot 5)\mathbb{E}[Z^4]=6561(7\cdot 5 \cdot 3)\mathbb{E}[Z^2]=6561 \cdot 7\cdot 5 \cdot 3$$ since $\mathbb{E}[Z^2]=\text{Var}(Z)+\left(\mathbb{E}[Z]\right)^2=1+0^2=1$, giving $$\mathbb{E}[Z^8]=688905$$