The question is: "Let $(X_n)$ be a sequence of independent nonnegative random variables. Show that $M_n = \prod_{j=1}^{n} X_j$ is a submartingale (supermartingale) with respect to $\mathcal{F}_n = \sigma(X_1, X_2, \ldots, X_n)$, if $E(X_n) \geq 1$ ($E(X_n) \leq 1$) holds for all $n$."
Is the following proof correct? In the following proof, how did we get this step: \begin{align*} E(M_n | \mathcal{F}_{n-1}) &\leq E(X_n) \cdot \prod_{j=1}^{n-1} E(X_j) \\ &\leq E(X_n) \cdot \prod_{j=1}^{n-1} 1 \end{align*}
====
Let $(X_n)$ be a sequence of independent nonnegative random variables. We want to show that $M_n = \prod_{j=1}^{n} X_j$ is a submartingale (supermartingale) with respect to $\mathcal{F}_n = \sigma(X_1, X_2, \ldots, X_n)$ if $E(X_n) \geq 1$ ($E(X_n) \leq 1$) holds for all $n$.
To prove this, we consider the conditional expectation $E(M_n | \mathcal{F}_{n-1})$. By the properties of conditional expectation, we have:
\begin{align*} E(M_n | \mathcal{F}_{n-1}) &= E\left(\prod_{j=1}^{n} X_j \bigg| \mathcal{F}_{n-1}\right) \\ &= E\left(\prod_{j=1}^{n} X_j \bigg| X_1, X_2, \ldots, X_{n-1}\right) \quad \text{(independence of } X_1, X_2, \ldots, X_n) \\ &= E\left(E\left(\prod_{j=1}^{n} X_j \bigg| X_1, X_2, \ldots, X_{n-1}\right)\right) \quad \text{(tower property)} \\ &= E\left(X_n \cdot \prod_{j=1}^{n-1} X_j \bigg| X_1, X_2, \ldots, X_{n-1}\right) \\ &= E(X_n) \cdot E\left(\prod_{j=1}^{n-1} X_j \bigg| X_1, X_2, \ldots, X_{n-1}\right) \\ &= E(X_n) \cdot \prod_{j=1}^{n-1} E(X_j) \quad \text{(independence of } X_j \text{ for } j > n-1) \end{align*}
Now, if $E(X_n) \geq 1$ for all $n$, we can bound the conditional expectation as follows:
\begin{align*} E(M_n | \mathcal{F}_{n-1}) &\leq E(X_n) \cdot \prod_{j=1}^{n-1} E(X_j) \\ &\leq E(X_n) \cdot \prod_{j=1}^{n-1} 1 = E(X_n) \cdot (n-1) \end{align*}
Since $E(X_n) \geq 1$, it follows that $E(X_n) \cdot (n-1) \geq 1 \cdot (n-1) = n-1$. Therefore, we have:
\begin{align*} E(M_n | \mathcal{F}_{n-1}) &\leq n-1 \end{align*}
This inequality holds for all $n$, which satisfies the submartingale property.
To show that $M_n$ is a supermartingale when $E(X_n) \leq 1$ holds for all $n$, we can reverse the inequality. Starting from:
\begin{align*} E(M_n | \mathcal{F}_{n-1}) = E(X_n) \cdot \prod_{j=1}^{n-1} E(X_j) \end{align*}
Since $E(X_n) \leq 1$, we have:
\begin{align*} E(M_n | \mathcal{F}_{n-1}) \geq E(X_n) \cdot \prod_{j=1}^{n-1} E(X_j) \end{align*}
Again, using the assumption $E(X_j) \leq 1$ for all $j$, we have:
\begin{align*} E(M_n | \mathcal{F}_{n-1}) \geq E(X_n) \cdot \prod_{j=1}^{n-1} 1 = E(X_n) \cdot (n-1) = n-1 \end{align*}
Therefore, we have:
\begin{align*} E(M_n | \mathcal{F}_{n-1}) \geq n-1 \end{align*}
This inequality holds for all $n$, satisfying the supermartingale property.
In conclusion, the sequence $M_n = \prod_{j=1}^{n} X_j$ is a submartingale when $E(X_n) \geq 1$ holds for all $n$, and it is a supermartingale when $E(X_n) \leq 1$ holds for all $n$.
To see that $(M_n)_{n\in\mathbb{N}}$ is a submartingale w.r.t. the filtration $\mathcal{F}_n$, when $E(X_n)\geq 1$, we need to show that $E(M_n|\mathcal{F}_{n-1})\geq M_{n-1}$. We have, using independence and measurability on the second equality:
\begin{align} E(M_n|\mathcal{F}_{n-1})=E(X_nM_{n-1}|\mathcal{F}_{n-1})=E(X_n)M_{n-1}\geq M_{n-1}, \end{align} where in the last inequality we used the hypothesis of the expectation.
The supermartingale case goes just as direct: if $E(X_n)\leq 1,\forall n$, then $E(M_n|\mathcal{F}_{n-1})=E(X_n)M_{n-1}\leq M_{n-1}$.
Recall that if $X$ is independent of $\mathcal{G}$, then $E(X|\mathcal{G})=EX$ and if $X$ is measurable w.r.t. $\mathcal{G}$, then $E(X|\mathcal{G})=X$.