Skewness of the sample mean random varaiable

Question

We have X1 , ... , Xn being independent and identically distributed random variables with mean μ, variance σ² and skewness γ. Show that the skewness of the sample mean M is ^γ⁄_√n

Here is my incorrect reasoning:

We can assume that μ=0 and σ²=1 without loss of generality, that is if the Xj's are not standardised, we can standardise them. Then to determine the skewness of M we use the third standardised moment of M:

Skew(M) = E[^{(M - average.of.M)}⁄_stdev.of.M]³

average.of.M is unbiased so average.of.M = μ = 0.

stdev.of.M = sqrt[Var(M)] = sqrt[σ²/n] = ¹⁄_sqrt(n)

So Skew(M) = E[sqrt(n)*M]³ = n^3/2E[M]³

M³ = [^{(X1 + X2 + ... + Xn)}⁄_n]³ = ¹⁄_n³E[X1 + X2 + ... + Xn]³

[X1 + X2 + ... + Xn]³ is made up of terms with coefficients 1, 3, and 6 so that it can be written in a general form of [X1 + X2 + ... + Xn]³ = nX1³ + 3n(n-1)X1³ + 6(_nC₃)X1³

Plugging back into Skew(M), Skew(M) = n^3/2 $\times$ ¹⁄_n³E[nX1³ + 3n(n-1)X1³ + 6(_nC₃)X1³]

Simplifying by using Linearity of Expectation where E[X1³] = γ and applying the Choose formula we get Skew(M) = γ $\times$ n^3⁄₂

Where am I going wrong?

Answer 1

There are two errors. One is probably merely typographical: The third power should be inside the square brackets – the way you've written it, it's just $E[M]^3=0^3=0$.

The conceptual error is that you replaced all monomials by $X_1^3$. The idea to replace them is good and correct, but you can only replace them by applying the permutations under which the problem is invariant; there's no way to transform e.g. $X_1X_2^2$ into $X_1^3$ by a permutation. A correct expression would have been

$$ nX_1^3+3n(n-1)X_1^2X_2+6\binom n3X_1X_2X_3\;, $$

and the expectation values of the second and third terms vanish because they both contain a single variable.