I simply don't understand how to calculate the expected value of $X$, or $X^4$ for that matter. I tried doing the integral of $yf_x(y)dy$ from negative infinity to positive infinity but I don't know what y is supposed to be in this case or what $f_x(y)$ is supposed to be. Please help.
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5Sounds like you're supposed to calculate $\frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty x^4 e^{-x^2/2} \ dx$. – Randall Nov 28 '17 at 21:01
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Great, so how do I get to where you got? What do I plug in for sigma and mu? – Kyle Nov 28 '17 at 21:14
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1In $N(0,1)$, $\mu$ is zero and $\sigma$ is one. – Angina Seng Nov 28 '17 at 21:15
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For $\mathbb{E}(X^{4})$ see this: https://math.stackexchange.com/questions/990833/normal-distribution-ex4 – Nash J. Nov 28 '17 at 21:28
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For $\mathbb{E}(X)$ see this: https://math.stackexchange.com/questions/518281/how-to-derive-the-mean-and-variance-of-a-gaussian-random-variable – Nash J. Nov 28 '17 at 21:29
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Let $f(\cdot)$ be the PDF of $N(0,1)$. Note $$ f'(x)=-x f(x). $$ By integration by parts, $$ \int x^4f(x)dx=\int(-x^3)f'(x)dx=\underbrace{(-x^3)f(x)\Big|_{-\infty}^\infty}_0+3\times\underbrace{\int x^2f(x)dx}_{\text{var of }N(0,1)}=3. $$
Kim Jong Un
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Let $$g(t) = \int_{-\infty}^{\infty}e^{-tx^2} = \sqrt{\frac{2\pi}{t}}$$ $$g''(t) = \int_{-\infty}^{\infty}x^4e^{-tx^2} = ?$$
So we have $$g''(1/2) = \int_{-\infty}^{\infty}x^4e^{-x^2/2} = ? $$
Dionel Jaime
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First define a random variable $Y\sim\text{Gamma}(p)$ ($p>0$) if $Y$ has density $$ f_Y(y)=\frac{1}{\Gamma(p)}y^{p-1}e^{-y}\quad (y>0). $$ It is easy to see that $EY^d=\frac{\Gamma(p+d)}{\Gamma(p)}$ by the definition of the gamma function.
Now onto the problem. Let $X$ be a standard normal random variable. It is well known that $X^2\sim\chi^2_{(1)}$ or equivalently $X^2/2\sim\text{Gamma}(1/2)$. Write $X^2\stackrel{d}{=}2W$ where $W\sim\text{Gamma}(1/2)$. Then $$ EX^4=E(2W)^2=2^2EW^2=2^2\frac{\Gamma(1/2+2)}{\Gamma(1/2)}=2^2\left(\frac{1}{2}\right)\left(\frac{1}{2}+1\right)=1(3)=3$$ where we have used the fact that $\Gamma(p+1)=p\Gamma(p)$. We can generalize to compute all even moments of a standard normal. For $k\geq 1$ $$ EX^{2k}=E(2W)^k=2^kEW^k=2^k\frac{\Gamma(1/2+k)}{\Gamma(1/2)}=2^k\left(\frac{1}{2}\right)\left(\frac{1}{2}+1\right)\cdots \left(\frac{1}{2}+k-1\right). $$ But we can simplify further to get that $$ EX^{2k}=1(3)\cdots(2k-1)=\frac{(2k)!}{2^kk!}. $$
Sri-Amirthan Theivendran
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