For a multiplicative Martingale of independent, non-negative RV with mean 1, given $E(M_\infty)=1 \implies \prod E(\sqrt{X_k})>0$

Question

Problem: Given a sequence of independent, non-negative RV $(X_n)_{n\geq 1}$ with $E(X_n)=1$ and the martingale $M_n= \prod_{k=1}^n X_k$, $M_0:=1$ show that: $$E(M_\infty)=1 \implies \prod_{k=1}^\infty a_k>0, \text{ where } 0<a_n:= E(\sqrt{X_n}) \leq 1 $$

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My approach:

An infinite product is said to be convergent if the limit exists and is not zero. In the above situation the martingale $M_n$ is positive and therefore it converges with limit $M_\infty \in L^1$. That is $$ M_\infty= \prod_{k=1}^\infty X_k < \infty $$ Furthermore we must have $M_\infty >0$, because if $M_\infty =0$ then this would contradict the assumption that $E(M_\infty)=1$. Bringing this all together we have that the infinite product denoted by $M_\infty$ is indeed convergent given that $E(M_\infty)=1$

To show the implication now I want to relate the product $\prod a_k$ somehow with $M_\infty$.

I am uncertain about swapping the infinite product with expected value is allowed, but I will still do it below: $$ \prod_{k=1}^\infty a_k = \prod_{k=1}^\infty E\left(\sqrt{X_k)}\right) \overset{a)}= E\left( \prod_{k=1}^\infty \sqrt{X_k} \right)=E(\sqrt{M_\infty}) $$ a) I know that this is true in the finite case because my RVs are mutually independent, for the infinite case I would accept it to be a definition but this might be wrong

For $E(\sqrt{M_\infty})$ I can only use Jensen's Inequality for the concave function $\sqrt{.}$ that is $$E(\sqrt{M_\infty}) \leq \sqrt{E(M_\infty)} = 1 $$

Edit: I corrected the above, applied Jensen's Inequality correctly now. See Did's comment below

Hence I obtain the trivial statement that $$\prod_{k=1}^\infty a_k \leq 1 $$

Additionally: If I want to avoid the step a) I can use the finite case and that $\sqrt{M_n}$ is a super-martingale (because $\sqrt{.}$ is concave) that will then give me $$\prod_{k=1}^n a_k = E(\sqrt{M_n}) \geq E(\sqrt{M_0})=E(\sqrt{1})= 1 $$

I am looking for hints only to help me get unstuck or guide me in the right direction because the results above aren't fruitful.

Answer 1

Because $E(M_\infty)=1$, you have $P(M_\infty>0)>0$. Thus $E(\sqrt{M_\infty})>0$. Because $M_\infty=\lim_nM_n$ a.s., you can use Fatou's lemma to relate $E(\sqrt{M_\infty})$ to $\prod_{n=1}^\infty a_n$.

Answer 2

Jensen's Inequality is what I thought to do. Then

$$\prod_{k=1}^{n} a_k \ge 1 \to \lim_n \prod_{k=1}^{n} a_k \ge \lim_n 1 = 1 \ ?$$

LHS exists as you proved.

Also, why $\pm 1$?

$$\sqrt{E[M_{\infty}]} = \sqrt{1} = 1$$

Also

$$\sqrt{E[M_{0}]} = \sqrt{1} = 1$$