Prove that a symmetric distribution has zero skewness.
Okay so the question states : First prove that a distribution symmetric about a point a, has mean a.
I found an answer on how to prove this here: Proof of $E(X)=a$ when $a$ is a point of symmetry
Of course I used method 2
But now for the rest of this proof I'm struggling.
$\mu_{2X}$ and $\mu_{3X}$ are the $2^{nd}$ and $3^{rd}$ moments about the mean, respectively (w.r.t X).
$$E[X] = \mu = a$$
$$\mu_{2X}= E[(X-a)^2] = E[((a+Y)-a)^2] = E[Y^2]$$
$$\mu_{3X} = E[Y^3]$$
Please just check my notation. I always use subscripts to show which distribution is in queston but especially with Skewness, can this be done?
$$\text{Skewness} = \sqrt{\beta_{1X}} = \frac{\frac{1}{n}\sum_{x \in S}y^3}{(\sqrt{\frac{1}{n}\sum_{x \in S}y^2})^3}$$
So I thought it just needs to be shown that the numerator is always equal to 0. I don't know if I approached it correctly but it seemed to make sense to me and now I'm stuck
Best Answer
I have a simpler proof. I hope this is ok. Let $Y = X - a$ be a random variable. Now note that due to symmetricity $Y$ and $-Y$ have the same distribution. That implies $$E[Y^3] = E[(-Y)^3]$$ This implies $E[Y^3] = 0$.