How can I calculate the mean of the cube of random variable that follows chi-square distribuion? If $X + 1 \sim \chi^2_{1}$, what is the mean of the cube of $X$?
I'm confused between raw moment and central moment.
My idea is, since the mean of $X$ is 0, I should calculate the raw moment.
However, I'm not sure about that because it's mean.
 A: Expanding the cube out using the binomial theorem gives you:
$$\begin{equation} \begin{aligned}
\mathbb{E}(X^3) 
&= \mathbb{E}((X+1-1)^3) \\[6pt]
&= \mathbb{E}((X+1)^3 - 3(X+1)^2 + 3(X+1) - 1) \\[6pt]
&= \mathbb{E}((X+1)^3) - 3 \mathbb{E}((X+1)^2) + 3 \mathbb{E}(X+1) - 1. \\[6pt]
\end{aligned} \end{equation}$$
Since $X+1 \sim \chi_1^2$ we now need the raw moments of the chi-squared distribution, which are:
$$\mathbb{E}(X+1) = 1 \quad \quad \quad \mathbb{E}((X+1)^2) = 3 \quad \quad \quad \mathbb{E}((X+1)^3) = 15.$$
Substitution of these raw moments into the above equation gives:
$$\begin{equation} \begin{aligned}
\mathbb{E}(X^3) 
&= \mathbb{E}((X+1)^3) - 3 \mathbb{E}((X+1)^2) + 3 \mathbb{E}(X+1) - 1 \\[6pt]
&= 15 - 3 \cdot 3 + 3 \cdot 1 - 1 \\[6pt]
&= 15 - 9 + 3 - 1 \\[6pt]
&= 8. \\[6pt]
\end{aligned} \end{equation}$$
A: For any random variable $X$, central moment = raw moment if $E(X) = 0$, because
$E(X^k) = E[(X-E(X))^k]$. 
Let $Y=X+1$. Then $Y\sim \chi_1^2$, $E(Y)=1$, $X^3=(Y-1)^3$, So $E(X^3)$ = the third central moment of chi-square distribution with df = 1.  
