Mgf of distribution to prove sample variance is equal to expectation of sample variance

Question

If X1,X2,...,Xn ∼ iid N(μ,σ2) and S2 is the sample variance, and E(S2) = σ2. Let Y = (n − 1)S2/σ2. Using the mgf of Y , how can you find the expected value of Y and conclude that E(S2) = σ2. What does the mgf of Y look like? How should I start the proof as well?

Answer 1

Since you said you know how to get the moment-generating function of $Y$, I'll start from there. We have, for the moment-generating function: $$M_Y(t) = (1-2t)^{-(n-1)/2}$$ Now, remember that $M_Y(t) = \Bbb E[e^{tY}]$, so that $\partial_t M_Y(0) = \Bbb E[Y]$. The derivative is: $$\partial_t M_Y(t) = (n-1)(1-2t)^{-(n-1)/2}$$ So we know that $$\Bbb E[Y] = n-1$$ But $\Bbb E[Y] = \frac{n-1}{\sigma^2}\Bbb E[S^2]$, so $$\frac{n-1}{\sigma^2}\Bbb E[S^2] = n-1 \Longrightarrow \Bbb E[S^2] = \sigma^2$$ as desired!