Finding $E[\overline{X}^3]$ where $X_1,\ldots,X_n$ are i.i.d $N(\mu,1)$ Suppose $X_1, X_2, \ldots, X_n$ are  i.i.d. Normal$(\mu , 1)$ random variables with $μ \in$ $\mathbb{R}$. 
How can we calculate   $E\left[\overline X^3\right]$?
 A: Let $Z=\bar X - \mu$ so that $Z$ has mean $0$.
Since $(a+b)^3 = a^3+3a^2b + 3ab^2 +b^3$ you obtain
$$
E[\bar X^3] = E[(Z+\mu)^3] = E[Z^3] + 3 E[Z^2]\mu + 3 E[Z] \mu^2 + \mu^3.
$$
Because $Z$ is normal with mean 0, it is symmetric about 0 so that $E[Z^3]=0$ and $E[Z]=0$. We are left with
$$E[\bar X^3] = 3 E[Z^2]\mu +\mu^3.$$
If remains to compute $E[Z^2]$ which is also the variance of $\bar X$, equal to $1/n$. The final answer is
$$
E[\bar X^3] = 3\mu/n +\mu^3.
$$ 
A: There is a much more general solution to this problem that holds for any distribution whose moments exist. It is a problem known as finding moments of moments. The modus operandi for solving such problems is to work with power sum notation $s_r$, namely: $s_r = \sum_{i=1}^n X_i^r$. 
We seek $\mathbb{E}[ (\frac{s_1}{n})^3]$ ... which is just the $1^\text{st}$ raw moment of $(\frac{s_1}{n})^3$:

where:


*

*$\mu_r$ denotes the $r^{th}$ central moment of $X$ i.e. $\mu_r = E[(X-\mu)^r]$

*${\hat\mu}_1$ denotes the 1st raw moment (i.e. the population mean), and

*RawMomentToCentral is a function from the mathStatica package for Mathematica.
In the special case of $X \sim N(\mu, 1)$, the third central moment $\mu_3$ is zero (by symmetry), and the variance $\mu_2 = 1$, and so the solution simplifies to: $\frac{3 \mu}{n} + \mu^3$
