Calculate $\operatorname{Cov}(X, X^2)$ for $X \sim N(\mu, \mu^2)$

Question

Calculate $\operatorname{Cov}(X, X^2)$ for $X \sim N(\mu, \mu^2)$.

I used the definition of $\operatorname{Cov}$ and $\operatorname{Var}$ to find $E(X^2)$:

$$\operatorname{Cov}(X, X^2) = E(X^3)- E(X)E(X^2) = E(X^3) - \mu(\mu^2+\mu^2)$$

I am not sure how to find $E(X^3)$. Can someone please help?

Answer 1

$X\sim N(\mu,\sigma^2)\Leftrightarrow Z=\frac{X-\mu}{\sigma}\sim N(0,1)$

$$\begin{align} E(X^3)&=\sigma^3E\left[\left(\frac{X-\mu+\mu}{\sigma}\right)^3\right]=\sigma^3E[(Z+p)^3],~~~p=\frac{\mu}\sigma\\ \\ E(X^3)&=\sigma^3(E(Z^3)+3pE(Z^2)+3p^2E(Z)+p^3)\\ \\ \text{Note:}~~~E(Z^3)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty z^3e^{\frac{-z^2}{2}}dz=0\\ \\ E(Z^2)&=1^2+0^2=1\\ \\ E(Z)&=0\end{align}$$

Therefore,

$$E(X^3)=\sigma^3(3p+p^3)$$

Answer 2

You can find $EX^3$ using characteristic function of normal random variable $X \sim \mathcal{N}(\mu, \sigma^2)$, $\phi_X(t)=e^{i\mu t-\frac{\sigma^2t^2}{2}}$.

It's well known fact, that $EX^k=(-i)^k\phi_X^{(k)}(0)$. If you don't want to find the third derivative of characteristic function explicitly, you can use Taylor expansion.