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If a set of $k$ random variables $x$ is drawn from the same normal distribution with mean $μ$ and standard deviation $σ$, then:

  • the mean of the distribution of $Σx$ is $kμ$
  • the mean of the distribution of $Σx²$ is $k(μ²+σ²)$

Is there a similar closed-form expression for the mean of the distribution of $Σx³$ ?

Sorry if I've got the terminology wrong!

Richard Collins
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  • Technically you should mention that the observations are i.i.d. check out the proposed duplicate that Patrick suggests. – Michael R. Chernick Oct 19 '19 at 00:40
  • The first answer gives you one way to arrive at the result except that 1 is plugged in for the mean & 2 for the standard deviation. The second answer gives you the way to solve it generally using moment generating functions. – Michael R. Chernick Oct 19 '19 at 00:47

1 Answers1

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Suppose $X$ is $N(\mu,\sigma^2)$, then we can agree that \begin{align*} \text{E}[X] & = \mu\,,\\ \text{E}[X^2] & = \mu^2+\sigma^2\,,\\ \text{E}[(X-\mu)^3] & = 0\,, \end{align*} the last because the distribution of $X$ is symmetrical around 0. Now, \begin{align*} 0 & = \text{E}[(X-\mu)^3] = \text{E}[X^3-3\mu X^2+3\mu^3X-\mu^3] \\ & = \text{E}[X^3] -3\mu \text{E}[X^2] + 3\mu^3-\mu^3 \\ & = \text{E}[X^3] -3\mu(\mu^2+\sigma^2)+2\mu^3 = \text{E}[X^3] -\mu^3-3\mu\sigma^2 \end{align*} which means that $\text{E}[X^3] = \mu^3+\mu\sigma^2$. If $X_i$, $i=1,\ldots,k$ are all $N(\mu,\sigma^2)$ then $$ \text{E}[\sum_{i=1}^k X_i^3] = k(\mu^3+\mu\sigma^2) $$ You could easily extend this to 4th moment, 5th moment and so on, using the fact that for even $n$: $$ \text{E}[(X-\mu)^n] = \sigma^n (n-1)(n-3)(n-5)\cdot\ldots\cdot1 $$

StijnDeVuyst
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