Let $X_1,...$ be a sequence of idd r.v. with mean $0$ and variance $\sigma^2$. Show that $\{Z_n,n\geq1\}$ is a martingale when $Z_n=S_n^2-n\sigma^2$

Question

Let $X_1,...$ be a sequence of independent and identically distributed random variables with mean $0$ and variance $\sigma^2$. Let $S_n=\sum^n_{i=1}X_i$ and show that $\{Z_n,n\geq1\}$ is a martingale when $$Z_n=S_n^2-n\sigma^2$$

I know I need to show that $E(Z_{n+1}|\mathcal{F}_n)=Z_n$ to prove that $Z_n$ is a martingale. I know that if $Z_n=\sum^n_{i=1}X_i$, then $\{Z_n\}$ is a martingale. We show this by the following calculations: \begin{align} E(Z_{n+1}|\mathcal{F}_n)&=E(Z_n+X_{n+1}|\mathcal{F}_n)\\ &=E(Z_n|\mathcal{F}_n)+E(X_{n+1}|\mathcal{F}_n)\\ &=Z_n+E(X_{n+1})\\ &=Z_n+0=Z_n \end{align} I can see that this problem is analogous to that problem, but I am not sure how to go about it. Could I get some pointers? Thank you.

Answer 1

• Prove that $S_{n+1}^2 = S_n^2 + 2X_{n+1} S_n + X_{n+1}^2$.
• Show that $E[S_{n+1}^2 \mid \mathcal{F}_n] = S_n^2 + \sigma^2$.

Answer 2

Firstly, $S_{n+1}^2=(S_n+X_{n+1})^2=S_n^2+2S_nX_{n+1}+X_{n+1}^2$

Then \begin{align} E[Z_{n+1}-Z_n|\mathcal{F}_n]&=E[S_n^2+2S_nX_{n+1}+X_{n+1}^2-(n+1)\sigma^2-(S_n^2-n\sigma^2)|\mathcal{F}_n]\\ &=E[2S_nX_{n+1}+X_{n+1}-\sigma^2|\mathcal{F}_n]\\ &=E[S_nX_{n+1}|\mathcal{F}_n]+E[X_{n+1}^2|\mathcal{F}_n]-\sigma^2\\ &=2S_nE[X_{n+1}]+E[X_{n+1}^2]-\sigma^2\\ &=2S_n\cdot0+(\text{Var}(X_{n+1})+E[X_{n+1}]^2)-\sigma^2\\ &=0 \end{align} Last line follows from the fact that variance of $X$ is $\sigma^2$