Prove that $\{S_n^2-\sigma^2n\}$ is a martingale.

Question

Let $S=\{S_n\}$ be a mean zero random walk with $EX_1^2=\sigma^2$. Prove that $\{S_n^2-\sigma^2n\}$ is a random walk.

So we need that

1. $\{S_n^2-\sigma^2n\}$ is integrable, (easy)
2. $\{S_n^2-\sigma^2n\}$ is adapted, (again, easy), and
3. $E[S_{n_1}^2-\sigma^2(n+1)\mid\mathcal{F}_n] = S_n^2-\sigma^2n$ for all $n\in\mathbb{N}$, which is the part I'm stuck on.

So far I've been trying to just compute it directly and I have

\begin{align} E[S_{n_1}^2-\sigma^2(n+1)\mid\mathcal{F}_n] &=E[S_{n+1}^2\mid\mathcal{F}_n]-E[\sigma^2n+\sigma^2\mid\mathcal{F}_n] \\ &=E[S_{n+1}^2\mid\mathcal{F}_n]-E[EX_1^2n+EX_1^2\mid\mathcal{F}_n] \end{align} is this the way I should be approaching this? Is there a better "trick" to use?

Can I use $$E[S_{n+1}^2\mid\mathcal{F}_n]=S_n^2+\sigma^2+2E[S_nX_{n+1}\mid\mathcal{F}_n]?$$

Edit:

So the last display equation turns out to be $S_n^2+\sigma^2$ since $S_n$ and $X_{n+1}$ are independent, $S_n\in\mathcal{F}_n$, and $E[X_{n+1}\mid\mathcal{F}_n]=0$. But now I'm not sure about what to do with the term $-E[\sigma^2n+\sigma^2\mid\mathcal{F}_n]$.

Thank you!

Answer 1

Note that $$ S_{n+1}^2-\sigma^2(n+1)=(S_n+X_{n+1})^2-\sigma^2n-\sigma^2=S_n^2-\sigma^2n+2S_nX_{n+1}+X_{n+1}^2-\sigma^2 $$ So by linearity $$ \begin{align} E[S_{n+1}^2-\sigma^2(n+1)\mid \mathcal{F_n}]&=E[S_n^2-\sigma^2n\mid \mathcal{F_n}]+ 2E[S_nX_{n+1}\mid \mathcal{F_n}]+ E[X_{n+1}^2-\sigma^2\mid \mathcal{F_n}]\\ &=S_n^2-\sigma^2n+2S_nE[X_{n+1}\mid \mathcal{F_n}]+E(X_{n+1}^2)-\sigma^2\\ &=S_n^2-\sigma^2n+2S_nE[X_{n+1}]+E(X_{n+1}^2)-\sigma^2\\ &=S_n^2-\sigma^2n \end{align} $$ where in the second line we used the fact that $S_n^2-\sigma^2n\in \mathcal{F_n}$, $S_n\in \mathcal{F_n}$ and $X_{n+1}^2-\sigma^2$ is independent of $\mathcal{F_n}$. In the next line we used the fact that $X_{n+1}$ is independent of $\mathcal{F_n}$. In the final line we use the assumptions given in the problem.