3

I was wondering if the property exists where if you have $\ E[(Y- \mu)^3]$ you can write it as $\ E[(Y- \mu)^2] E[(Y- \mu)] $ ?

MathStudent
  • 149
  • 2
    Well, it CAN happen that E(Y^3)=E(Y)E(Y^2), for example if Y is symmetric, but this is not a general fact at all. – Did Oct 29 '13 at 20:05
  • My question says that when $\ \dfrac{E[(Y - \mu)^3]}{ \sigma ^3} = 0 $ then Y is symmetric. So does that mean I can't use it?? Since it may not be zero all the time? – MathStudent Oct 29 '13 at 20:11
  • 1
    No, $E[(Y-\mu)^3]=0$ does not imply that $Y$ is symmetric around $\mu$. – Did Oct 29 '13 at 21:53

2 Answers2

6

Unfortunately, this is not generally true.

It is true that if $X$ and $Y$ are independent random variables, then $$ \mathbb{E}[XY]=\mathbb{E}[X]\cdot\mathbb{E}[Y]. $$

There are certainly pairs of non-independent random variables for which you can do this; two random variables $X$ and $Y$ so that this property holds are called uncorrelated.

Unfortunately, it is not generally true that $Y-\mu$ and $(Y-\mu)^2$ are uncorrelated.

Nick Peterson
  • 32,430
2

Is $Y$ normally distributed by any chance? If it is then Stein's Lemma can be useful for calculating higher order moments.

Stein's Lemma: Let $Y\sim N(\mu,\sigma^{2})$ and let $g$ be a differentiable function satisfying $\mathbb{E}|g\prime(Y)|<\infty$ then $\mathbb{E}[g(Y)(Y-\mu)]=\sigma^{2}\mathbb{E}[g\prime(Y)]$

in your case $g(Y)=(Y-\mu)^{2}$ and so $g\prime (Y)=2(Y-\mu)$, for which $\mathbb{E}[...]=0$

and so $\mathbb{E}[(Y-\mu)^3]=0$

!but only works for normally distributed Y.

Mael
  • 741