# For what value of $p$ is $(M_n)$ a martingale

Let $(X_i)_{i\ge 1}$ be i.i.d random variables, and let $(e_i)_{i \ge 1}$ be i.i.d random variables independent from the $(X_i)$. Assume $0\lt E(X_i^2) \lt +\infty$ and let $m = E(X_i), \sigma^2 = E(X_i^2)$. Assume as well that $e_1$ can only take values 1 and 2 with $P(e_1 = 1) = p, P(e_1 = 2) = 1-p$. Finally, define the product $M_n =X_1^{e_1}...X_n^{e_n}$. For what value of $p$ is $(M_n)$ a martingale?

Ok so for $(M_n)$ to be a martingale $E(M_{n+1}|X_1,...,X_n)= M_nE(X_{n+1}^{e_{n+1}})= M_n$

I think we have to use the limiting properties regarding the value of $p$ i.e. show that if $p<1/2$ then $M_n$ approaches $X_1^{2}...X_n^{2}$ or $M_n = X_1...X_n$ if $p>1/2$ but im not too sure.

Any help would be appreciated,

Thanks

-

You are nearly there. $M_n E(X_{n+1}^{e_{n+1}})=M_n\implies E(X_{n+1}^{e_{n+1}})=1$.
$$E(X_{n+1}^{e_{n+1}})=pE(X_{n+1})+(1-p)E(X_{n+1}^2)\\ =pm+(1-p)\sigma^2=1$$
Express $p$ in terms of $m$ and $\sigma^2$ and you are done!
It's much more simpler, you just stopped on a halfway. You've correctly mentioned that $$\mathsf E[M_{n+1}|X_1,\dots,X_n] = M_n\cdot\mathsf E[X_{n+1}^{e_{n+1}}]$$ thus you have to have $\mathsf E[X_{n+1}^{e_{n+1}}] = 1$ for $M_n$ to be a martingale. You have: $$\mathsf E[X_{n+1}^{e_{n+1}}] = p\cdot\mathsf E X_{n+1}+(1-p)\cdot\mathsf E X_{n+1}^2 = p(m-\sigma^2)+\sigma^2$$ and thus $$p = \frac{1-\sigma^2}{m-\sigma^2}.$$