Almost sure limit of a martingale process (Kakutani's theorem) 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:

*

*$N$ is bounded in $\mathcal{L}^2$ and that subsequently $M$ is uniformly integrable.

*$\prod_{i=1}^\infty r_k > 0 \implies \mathbb{E}M_\infty =1$.

*$\mathbb{E}M_\infty =1 \implies \prod_{i=1}^\infty r_k > 0$.

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.
 A: 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:


*

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

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


Combining both statements proves the assertion.


*

*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, $\mathbb{E}(M_{\infty})=1$.

*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.
