How can I compute $\mathbb{E}[Z^4]$ where $Z\sim N(0,1)$

Question

Let $Z\sim N(0,1)$ and $Y=a+bZ+cZ^2$. I want to compute the variance of $Y$. This is what I did: $$\operatorname{Var}(Y)=0+b^2\operatorname{Var}(Z)+c^2\operatorname{Var}(Z^2)=b^2+c^2\operatorname{Var}(Z^2)$$ To get $\operatorname{Var}(Z^2)$, I tried to use the definition $\operatorname{Var}(Z^2)=\mathbb{E}[Z^4]-\mathbb{E}[Z^2]^2$ But im having with this part. If this was a odd for example $\mathbb{E}[Z^3]$ you can say that because of the symmetry of the normal distribution $\mathbb{E}[Z^3]=0$, but in this is pair.

Thank you

Answer 1

Integrate $\int_0^\infty x^4 e^{-x^2/2}\; dx$ by parts using $u = x^3$, $dv = x e^{-x^2/2}\; dx$. Or change variables with $x = \sqrt{t}$ and use properties of the Gamma function.

Answer 2

If you have seen the Moment Generating Function, we can solve this by using it.

We have $$ \mathbf{E}(e^{tZ}) = e^{\frac{t^2}{2}}.$$ By equating the coefficient of $t^4$, we have $$\frac1{24}\mathbf{E}(Z^4) = \frac18.$$

This gives $\mathbf{E}(Z^4)=3$.

Answer 3

$$\mathbb{E}[Z^4]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}z^4e^{\large-\frac{z^2}{2}}dz=\frac{2}{\sqrt{2\pi}}\int_{0}^{\infty}z^4e^{\large-\frac{z^2}{2}}dz$$ set $\frac{z^2}{2}=u$, we have $z\,dz=du$ and $z^3=2\sqrt{2}u\sqrt{u}$, thus $$\mathbb{E}[Z^4]=\frac{4}{\sqrt{\pi}}\int_{0}^{\infty}u\sqrt{u}\,e^{-u}du=\frac{4}{\sqrt{\pi}}\Gamma\left(\frac{5}{2}\right)=\frac{4}{\sqrt{\pi}}\times \frac32 \times \frac 12\Gamma\left(\frac{1}{2}\right)=3$$ please check this link

Note $$\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$$ Generally $$\mathbb{E}[Z^{2k}]=\frac{(2k)!}{2^k k!}$$

Answer 4

It can be shown that $$\lim_{z\to \pm \infty} g(z)\exp\left(-\frac12z^2\right) = 0 \implies \mathsf{E}[g'(Z)] = \mathsf{E}[Zg(Z)]$$

This can be done with integration by parts, letting $u = g(z)$. Thus,

$$\mathsf{E}[Z^4] = \mathsf{E}[Z \cdot Z^3] = \mathsf{E}[3Z^2] = \mathsf{E}[Z \cdot3Z] = \mathsf{E}[3] = 3$$