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


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 $E[Y^3] = 0$.

  • $\begingroup$ Oh thanks. Yeah. I suppose that makes sense so I should never have tried to expand $E[Y^3]$ $\endgroup$ Apr 19 '15 at 13:38
  • $\begingroup$ @StephanCasey you are welcome. $\endgroup$
    – Calculon
    Apr 19 '15 at 13:43
  • $\begingroup$ @Calculon How to deduce that $E[X^3] = 0?$ $\endgroup$
    – Idonknow
    Feb 16 '20 at 7:22
  • 1
    $\begingroup$ @Idonknow That is not necessarily true. $\endgroup$
    – Calculon
    Feb 16 '20 at 9:28

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