I have a problem in solving the following exercise:

Let $(\Omega, \mathcal{A},P)$ be a probability space and $(X_n)$ a submartingale with $\text{sup}_{n\geq 0} \mathbb{E}(|X_n|)< \infty $.

  1. Show that for each $n$ the sequence $(\mathbb{E}([X^+_m \mid \mathcal{F}_n])_{m\geq n}$ is increasing in $m$.
  2. Define $M_n:= \text{lim}_{m\rightarrow\infty}\mathbb{E}[X^+_m \mid \mathcal{F}_n]$. Show that $(M_n)$ is a positive, integrable martingale.

I understand that the precondition means that $X$ is $\mathcal{L}^1$-limited. But I don't see how this can help. Plugging into the martingale property also doesn't provide a good approach for the second question. So any help would be appreciated, thanks in advance.


1 Answer 1


For the first part, recall that if $X$ is a submartingale then so is $X^+$. So $\mathbb{E}[X_m^+ \mid \mathcal{F}_n] \geq X_{n}^+$ for all $m > n$. Now all that remains is to apply the tower property for conditional expectations and the fact that taking conditional expectations preserves ordering.

For the second point, it is clear that $M_n$ is non-negative since $X_m^+$ is non-negative for each $m$ (notice that $M_n$ is well-defined by the result from the first part).

By the monotone convergence theorem and part $1$ we have $$\mathbb{E}[M_n] = \lim_{m \to \infty} \mathbb{E}\big[\mathbb{E}[X_m^+ \mid \mathcal{F}_n]\big] = \lim_{m \to \infty} \mathbb{E}[X_m^+] \leq \sup \mathbb{E}[X_n] < \infty$$ so $M_n$ is integrable.

Checking the martingale property for $M$ is then similar. You want to show that for $n > r$ $$\mathbb{E} [M_n \mid \mathcal{F}_r] := \mathbb{E}\bigg[\lim_{m \to \infty} \mathbb{E}[X_m^+ \mid \mathcal{F}_n] \mid \mathcal{F}_r\bigg] = \lim_{m \to \infty} \mathbb{E}[X_m^+ \mid \mathcal{F}_r] = M_r.$$ To do this, you can use the conditional monotone convergence theorem to take the limit outside of the expectation. From there, the result is just an application of the tower law.

  • $\begingroup$ Now everything is clear to me. Thanks very much! $\endgroup$
    – d237
    Jun 18, 2018 at 8:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .