I have been struggling with the following exercise and I was wondering whether my solution is correct or not. I am pretty sure about the second part of the question (the martingale part) but not so sure about the first part of the question. Please tell me what you think.
Let $(X_n)_{n=1}^{\infty}$ be a sequence of i.i.d. r.v. on $(\Omega,\mathcal{F},\mathbb{P})$. Assume that $\mathbb{E}X_1=0$ and $\mathbb{E}X_1^2=1$, consider $\mathcal{F}_n=\sigma (X_1,...,X_n)$, and define $M_n=\sum\limits_{k=1}^n\frac{1}{k}X_1\cdot \cdot \cdot X_k$.
Calculate $\mathbb{E}(M_n^2)$ and show that $(M_n)_{n=0}^{\infty}$ is an $(\mathcal{F}_n)_{n=0}^{\infty}$ martingale.
Now my answer is the following:
$\mathrm{Var} (X_1)=(\mathbb{E}X_1^2)-(\mathbb{E}(X_1))^2=1$
Now: $\mathrm{Var}(M_n)=\mathrm{Var}(\sum_{k=1}^{n}\frac{1}{k}(X_1\cdot\cdot\cdot X_k))=\sum_{k=1}^{n}\frac{1}{k^2}\mathrm{Var}(X_1\cdot\cdot\cdot X_k)=\sum_{k=1}^{n}\frac{1}{k^2}$, since $\mathrm{Var}(X_1)=1$ and $(X_n)$ is an iid sequence.
Also we have that: $\mathbb{E}M_n=\mathbb{E}(\sum_{k=1}^{n}\frac{1}{k}(X_1\cdot\cdot\cdot X_k))=\sum_{k=1}^{n}\frac{1}{k}\mathbb{E}(X_1\cdot\cdot\cdot X_k)=0$ since $\mathbb{E}X_1=0$ and $(X_n)$ iid.
Now using that: $\mathbb{E}M_n^{2}=\mathrm{Var}(M_n)+(\mathbb{E}M_n)^2$, we get that: $\mathbb{E}M_n^2=0$.
Now to prove that $(M_n)_{n=0}^{\infty}$ is an $(\mathcal{F}_n){n=0}^{\infty}$ martingale, I proceeded as follows:
$\mathbb{E}(M_{n+1}|\mathcal{F}_n)=\mathbb{E}(M_n+\frac{1}{n+1}(X_1\cdot\cdot\cdot X_{n+1})|\mathcal{F}_n)=\mathbb{E}(M_n|\mathcal{F}_n)+\frac{1}{n+1}\mathbb{E}(X_1\cdot\cdot\cdot X_n|\mathcal{F}_n)=M_n+\frac{1}{n+1}\mathbb{E}(X_1|\mathcal{F}_n)\cdot\cdot\cdot\mathbb{E}(X_{n+1}|\mathcal{F}_n)=M_n+\frac{1}{n+1}\mathbb{E}X_1\cdot\cdot\cdot \mathbb{E}X_{n+1}=M_n+0=M_n.$
For the computation of $\mathbb E[M_n^2]$: the second equality sign has to be justified by the fact that the sequence $(P_j)_{j=1}^n= (X_1\dots X_j)_{j=1}^n $ is non correlated , that is, $\mathrm{Cov}(P_i,P_j)=0$ if $i\neq j$.
We actually have $\mathbb E[M_n^2]=\operatorname{Var}(M_n)+(\mathbb E M_n)^2= \operatorname{Var}(M_n) $.
In the proof that $(M_n)_{n\geqslant 1}$ is a martingale, the second equal sign does not hold, as it would imply that $$\mathbb E[X_1\dots X_{n+1}\mid\mathcal F_n ]=\mathbb E[X_1\dots X_{n}\mid\mathcal F_n ],$$ which is not true. Indeed, $$\tag{*}\mathbb E[X_1\dots X_{n+1}\mid\mathcal F_n ]= X_1\dots X_n\mathbb E[X_{n+1}\mid\mathcal F_n]= 0$$ because $X_{n+1} $ is independent of $\mathcal F_n$, while $$\mathbb E[X_1\dots X_{n}\mid\mathcal F_n ]=X_1\dots X_n.$$ (*) is sufficient to conclude that $\mathbb E[M_{n+1}\mid\mathcal F_n ]=M_n$.