Prove that $S_n^2-s_n^2$ is martingale

Question

Let $(X_i)$ be iid such that $EX_i = 0$ and $\operatorname{Var}X_i = \sigma_i^2$. Let $s_n^2 = \sum_{i=1}^n \sigma_i^2$ and $S_n = \sum_{i=1}^n X_i$. Prove that $S_n^2 - s_n^2 $ is martingale.

My attempt:

I would show that for $Y_n = S_n^2 - s_n^2$ I have $E[Y_{n+1} - Y_n \mid \mathcal{F}_n] = 0$

Next I can write: $E[Y_{n+1} - Y_n \mid \mathcal{F}_n] = E[2S_nX_{n+1} + X_{n+1}^2 - \sigma_{n+1}^2 \mid \mathcal{F}_n]$

Of course $E[2S_nX_{n+1}] = 0 $ so I have $E[Y_{n+1} - Y_n \mid \mathcal{F}_n] = E[X_{n+1}^2 - \sigma_{n+1}^2 \mid \mathcal{F}_n] = E[X_{n+1}^2 \mid \mathcal{F}_n] - \sigma_{n+1}^2$.

Wat can I do now?

Answer 1

I assume that $\mathcal F_n$ is the $\sigma$-algebra generated by $X_1,\dots,X_n$. It remains to show that $\mathbb E[X_{n+1}^2\mid\mathcal F_n]=\sigma_{n+1}^2$. Since $X_{n+1}$ is independent of $\sigma(X_1,\dots,X_n)$, we have $\mathbb E[X_{n+1}^2\mid\mathcal F_n]=\mathbb E[X_{n+1}^2]$.