Question

Prove that, all odd order central moments are zero for symmetric distribution.​

Answer 1

Explanation:

Proposition 3.3.1.For all odd r≥3, any random variable which has a finite rt⁢h moment and for which the density is symmetric about the mean, μ, has an rt⁢h central moment of zero.

Proof.We will use the substitutions t=u-μ and s=μ-u. Symmetry means exactly that the density satisfies f⁢(μ+t)=f⁢(μ-t) for all t.

Firstly

I1 =∫μ∞(u-μ)r⁢f⁢(u)⁢du

=∫0∞tr⁢f⁢(μ+t)⁢dt

=∫0∞tr⁢f⁢(μ-t)⁢dt

by symmetry. Secondly

I2 =∫-∞μ(u-μ)r⁢f⁢(u)⁢du

=∫0∞(-s)r⁢f⁢(μ-s)⁢ds

=-∫0∞sr⁢f⁢(μ-s)⁢ds.

Hence μr=(I1+I2)/σr=0. ∎

3.2 The Law of the Unconscious Statistician3.4 The formula for a bivariate change of variable