Show that $\operatorname{Cov}(X,X^2)=0$ if X is a continuous random variable with symmetric distribution around the origin

Question

Let $X$ be a continuous random variable with symmetric distribution around the origin. All the moments of the variable exist. Show that $\operatorname{Cov}(X, X^2) = 0$.

What I've done: if $X$ is a continuous random variable with symmetric distribution around the origin, then $\mathbb{E}(X)=\mathrm{Median}=0$.

Coefficient of skewness = $\frac{μ^3}{σ^3}= \frac{\mathbb{E}(X^3)-3μσ^2-μ^3}{σ^3}=\frac{\mathbb{E}(X^3)}{σ^3}=0.$

So $\mathbb{E}(X^3)=0$

$$\operatorname{Cov}(X,X^2)=\mathbb{E}(XX^2)-\mathbb{E}(X)\mathbb{E}(X^2)=\mathbb{E}(X^3)-0*\mathbb{E}(X^2)=\mathbb{E}(X^3)=0.$$

What do you think?

Answer 1

There are probably a good amount of ways to show this. First of all $\operatorname{Cov}[X, X^2]=E[X^3]-E[X]E[X^2]$. Since the pdf is stated to be symmetric about the origin (i.e. it is an even function) $E[X^3]$ is determined by integrating an even function times an odd function ($x^3$) and so $E[X^3]=0.$ For the same reason $E[X]=0$, and therefore $\operatorname{Cov}[X, X^2]=0$.

Answer 2

Your proof is right and it is ok to use these formulas. I would not prove it in this way though. If you have already studied basic results on expectations (integral), I would prefer to prove it using the results for symmetric probability distributions. If $X$ is symmetric about $k$ ($P_{X-k}=P_{-(X-k)}$): then (1) $E(X)=k$ and $E[(X-k)^j]=0$ for $j$ odd, provided that moments exist. In particular, given the $3$rd moment exists, if $k=0$ and $j=3$, then $\operatorname{Cov}(X,X^2)=0$.