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I am trying to understand the martingale convergence theorem, specifically the following Levy's upward theorem:

Let $(\Omega, F, \mathbf{P})$ be a probability space and let $X$ be a random variable in $L^{1}$. Let $F_{*}=\left(F_{k}\right)_{k \in \mathbf{N}}$ be any filtration of $F$, and define $F_{\infty}$ to be the minimal $\sigma$-algebra generated by $\left(F_{k}\right)_{k \in \mathbf{N}}$. Then $$ \mathrm{E}\left[X \mid F_{k}\right] \rightarrow \mathrm{E}\left[X \mid F_{\infty}\right] \text { as } k \rightarrow \infty $$ both $\mathbf{P}$-almost surely and in $L^{1}$.

For positive $X$, can we make any statements about the monoticity of the martingale as well, for example, $\mathrm{E}\left[X \mid F_{s}\right] \le \mathrm{E}\left[X \mid F_{t}\right]$ for $s \le t$?

pineapple
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1 Answers1

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If $E(X|F_t)$ depends on $t$, the answer is negative. Indeed, suppose that $\xi = \mathrm{E}\left[X \mid F_{s}\right] \le \mathrm{E}\left[X \mid F_{t}\right] = \eta$ $ $ for $s \le t$. As $E\xi = E X = E\eta$ we have $E(\eta - \xi)=0$ and $\eta-\xi \ge 0$. Thus $\eta-\xi =0$ and $E(X|F_s) = E(X|F_t)$ for all $s\le t$. Thus $E(X|F_t)= \zeta$ for all $t$, where $\zeta = E(X|F_0)$. We got a contradiction.

Botnakov N.
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