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Let $Z_1,Z_2, \ldots$ be independent non-negative random variables defined on a probability space $(\Omega,\mathcal{F},\mathbb{P})$ with $\mathbb{E} Z_n = 1$ for all $n \in \mathbb{N}$. The process $(M_n)_{n \in \mathbb{N}}$ defined by $M_n = \prod_{i=1}^n Z_i$ is a non-negative martingale. We know that $M_\infty$ exists as an almost sure limit of $M_n$.

We want to prove that $$\mathbb{E}M_\infty = 1 \iff \prod_{k=1}^\infty r_k > 0. \qquad (*)$$

First, we introduce $R_n = Z_n^\frac{1}{2}$ and subsequently, using Jensen's equality, we see that $$(\mathbb{E}R_n)^2 \leq \mathbb{E}R_n^2 = \mathbb{E} Z_n = 1.$$ It follows that $r_n := \mathbb{E}R_n \leq 1$ for all $n \in \mathbb{N}$, since $R_n = Z_n^\frac{1}{2}$ are non-negative random variables. Secondly, we let $N$ be the martingale defined by $N_n = \prod_{i=1}^n \frac{R_i}{r_i}$. In order to show that \begin{align} \mathbb{E}M_\infty = 1 \iff \prod_{k=1}^\infty r_k > 0, \qquad (*) \end{align} we first have to show that $N$ is bounded in $\mathcal{L}^2$ and that consequently $M$ is uniformly integrable.

However, I face difficulties regarding the elaboration of this last statement and the proof of the equivalence. Any help is appreciated!

EDIT: The three statements to prove:

  1. $N$ is bounded in $\mathcal{L}^2$ and that subsequently $M$ is uniformly integrable.
  2. $\prod_{i=1}^\infty r_k > 0 \implies \mathbb{E}M_\infty =1$.
  3. $\mathbb{E}M_\infty =1 \implies \prod_{i=1}^\infty r_k > 0$.

Proof 1. \begin{align} \mathbb{E}N^2 &= \mathbb{E}\bigg[ \prod_{i=1}^n \bigg( \frac{R_i}{r_i}\bigg)^2 \bigg]\\ &= \mathbb{E}\bigg[ \prod_{i=1}^n \bigg( \frac{Z_n}{\mathbb{E}[R_n]^2}\bigg) \bigg]\\ \end{align} How to show that the above expression is finite? Jensen's inequality does not seem to work on $\mathbb{E}[R_n]^2$. And if $N$ is bounded in $\mathcal{L}^2$, why does this imply that $M$ is U.I.?

For the proofs of 2 and 3 I have no suggestions.

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    $\begingroup$ If we consider the quintessential example $\Bbb{P}(Z_n = 0) = \Bbb{P}(Z_n = 2) = \frac{1}{2}$, then we have $M_{\infty} = 0$ a.s. Are you missing some extra condition? $\endgroup$ Jan 4 '17 at 18:58
  • $\begingroup$ Hmmpz. So the formulation of my questions seems to be incorrect then? $\endgroup$
    – iJup
    Jan 4 '17 at 19:03
  • $\begingroup$ That is what I am suspecting. $\endgroup$ Jan 4 '17 at 19:07
  • $\begingroup$ Yeah, okay. We want to prove that $\mathbb{E}M_\infty = 1 \iff (*)$. I will edit this in the initial post. $\endgroup$
    – iJup
    Jan 4 '17 at 19:19
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    $\begingroup$ @JohnDawkins: I don't agree. For example if $P(Z_k = 1) = 1-k^{-2}$ then by Borel-Cantelli we have $Z_k = 1$ for almost all $n$, almost surely, so the product can certainly have a nonzero limit. Note the $Z_k$ are not assumed to be identically distributed. $\endgroup$ Jan 4 '17 at 19:26
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Since the random variables $Z_n$, $n \in \mathbb{N}$, are by assumption independent and $\mathbb{E}(Z_n)=1$ for all $n$, we have

$$\begin{align*} \mathbb{E}(N_n^2) = \mathbb{E} \left( \prod_{i=1}^n \frac{Z_i}{r_i^2} \right) &= \prod_{i=1}^n \frac{\mathbb{E}(Z_i) }{r_i^2} = \frac{1}{\prod_{i=1}^n r_i^2}. \end{align*}$$

This shows that $(N_n)_{n \in \mathbb{N}}$ is bounded in $L^2$ if and only if $\prod_{i=1}^{\infty} r_i>0$. We are going to prove the following two statements:

  1. $\prod_{i=1}^{\infty} r_i> 0 \implies \mathbb{E}(M_{\infty})=1$.
  2. $\prod_{i=1}^{\infty} r_i=0 \implies \mathbb{E}(M_{\infty})=0$.

Combining both statements proves the assertion.

  1. Suppose that $\prod_{i=1}^{\infty} r_i>0$, i.e. $(N_n)_{n \in \mathbb{N}}$ is bounded in $L^2$. Since $0<r_i \leq 1$ for all $i$, we have $M_n \leq N_n^2$. Applying Doob's maximal inequality, we find $$\mathbb{E} \left( \sup_{1 \leq k \leq n} M_k \right) \leq \mathbb{E} \left( \sup_{1 \leq k \leq n} N_k^2 \right) \leq 4 \mathbb{E}(N_n^2),$$ and the right-hand side is uniformly bounded in $n$. Consequently, it follows from the monotone convergence theorem that $M^* := \sup_{n \geq 1} M_n \in L^1$. This implies that $(M_n)_{n \in \mathbb{N}}$ is uniformly integrable (since it is dominated by $M^* \in L^1$). Finally, since $M_n \to M_{\infty}$ almost surely and $(M_n)_{n \in \mathbb{N}}$ is uniformly integrable, it follows e.g. from Vitali's convergence theorem that $M_n \to M_{\infty}$ in $L^1$. Hence, in particular, $\mathbb{E}(M_{\infty})=1$.
  2. Suppose that $\prod_{i=1}^{\infty} r_i=0$. By the very definition of $N_n$, $$N_n \prod_{i=1}^n r_i = \sqrt{M_n}$$ which implies $$\sqrt{M_n} \xrightarrow[]{n \to \infty} N_{\infty} \prod_{i=1}^{\infty} r_i = 0$$ where we have used that $N_n$ converges almost surely to some random variable $N_{\infty}$ (since $(N_n)_{n \in \mathbb{N}}$ is a non-negative martingale, this follows from the martingale convergence theorem). Hence, $M_{\infty}=0$ which implies $\mathbb{E}(M_{\infty})=0$.

Remark: The statement is known as Kakutani's theorem.

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