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

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

2 Answers 2

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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

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

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  • $\begingroup$ 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) $\endgroup$
    – GAGA
    Mar 14, 2019 at 3:56
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    $\begingroup$ 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 $\endgroup$
    – jlewk
    Mar 14, 2019 at 4:04
  • $\begingroup$ (My previous comment treats the case mu=1, the answer now handles any mu). $\endgroup$
    – jlewk
    Mar 14, 2019 at 5:14
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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 $(\frac{s_1}{n})^3$:

enter image description here

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

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