# 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 $$E\left[\overline X^3\right]$$?

• Think moment generating functions. Mar 14, 2019 at 2:00
• How do you think $E(\bar{X}^3)$ with $X\sim N(\mu,1)$ relates to $E((X-\mu)^3)$ where $X-\mu \sim N(0,1)$ ? Mar 14, 2019 at 2:00

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

$$E[\bar X^3] = 3\mu/n +\mu^3.$$

• I think when you start $Z= (\bar X - \mu )$ you should divide with the Sd of $\bar X$ which is equal to 1/ (root of n)
– GAGA
Mar 14, 2019 at 3:56
• You may as well divide by sqrt(n) but it's not needed. The scaling you mention appears at the end, where I wrote that $E[Z^2] =1/n$. The final answer $3+3/n$ is correct, see wolframalpha.com/input/?i=E%5BX%5E3%5D+where+X+is+N(1,1%2Fn) for instance Mar 14, 2019 at 4:04
• (My previous comment treats the case mu=1, the answer now handles any mu). Mar 14, 2019 at 5:14

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: $$\frac{3 \mu}{n} + \mu^3$$