Any guidance on this topic is much appreciated. I solved the first part and am at the second part but got stuck a little solving the expression. In particular, I have no idea how to derive $\mathbb{E}[M_n^2]$. Thank you!
Suppose $X_1, X_2, \ldots$ are independent random variables with
$$
\Pr\{X_j = 1\} = \frac{1}{4}, \quad \Pr\{X_j = 0\} = \frac{1}{4}, \quad \Pr\{X_j = -1\} = \frac{1}{2}.
$$
Let $S_0 = 0$ and $S_n = X_1 + \cdots + X_n$. Let ${F}_n$ denote the information contained in $X_1, \ldots, X_n$. Then, let $M_n=2^{S_n}$. Is $\{M_n\}$ a square-integrable martingale? Find $E\left[\sum_{j=1}^{20} (M_j – M_{j-1})^2\right]$.
Solution So Far
Since $M_n$ is bounded ($M_n\leq 2^n$ and, thus, $\mathbb{E}[M_n^2]<\infty$), it is a square-integrable martingale.
Since $M_n$ is a square-integrable martingale:
$$\mathbb{E}[M_n^2]=\mathbb{E}[M_0^2]+\mathbb{E}\left[\sum_{j=1}^{n} (M_j – M_{j-1})^2\right]$$
Then, given $n=20$, we know that $\mathbb{E}[M_0^2]=1$
Best Answer
Note $$\begin{aligned}E[M^2_n]&=E[2^{2S_n}]\\ &=E[2^{2(S_n-S_{n-1})}2^{2S_{n-1}}]\\ &\stackrel{\textrm{Tower prop.}}=E[E[2^{2(S_n-S_{n-1})}|\mathscr{F}_{n-1}]2^{2S_{n-1}}]\\ &\stackrel{X_n\perp \mathscr{F}_{n-1},X_n\sim X_1}=E[2^{2X_1}]E[2^{2S_{n-1}}]\\ &=\bigg(\frac{2^2}{4}+\frac{1}{4}+\frac{2^{-2}}{2}\bigg)E[M_{n-1}^2]\\ &=\frac{11}{8}E[M_{n-1}^2] \end{aligned}$$ which implies $E[M^2_n]=(11/8)^n$ since $M_0=2^0=1$.