Justify $\lim_{r \to \infty} E[M_r 1_F] = E[\lim_{r \to \infty} M_r 1_F]$

Question

Probability with Martingales

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I think we have

$$M_r 1_F \le |M_r 1_F|$$

and

$$E[|M_r 1_F|] \le E[|M_r|] \le \sup_r E[|M_r|] < \infty$$

Hence by DCT

$$\lim_{r \to \infty} E[M_r 1_F] = E[\lim_{r \to \infty} M_r 1_F] = E[M_{\infty} 1_F]$$

Is that wrong?

If not how is $\color{blue}{\text{blue}}$ used for $\color{red}{\text{red}}$?

Answer 1

1. To apply DCT, you need an RV $X$ in $L^1$ so satisfying $|M_r|{\bf 1}_F \le X$ for all $r$. The fact that $\sup_{r} E[ |M_r|]< \infty$ does not guarantee the existence of such an $RV$.
2. The argument goes as follows.

Fix $n$.

We want to show that $E [ M_\infty |{\cal F}_n] = M_n$, or, equivalently, that for any $F \in {\cal F}_n$, $E[M_n {\bf 1}_F ] = E[M_\infty {\bf 1}_F]$.

To do that first observe that for any $r$:

$$(1)\quad | E[ M_r{\bf 1}_F] - E[M_\infty {\bf 1}_F] | \le E [|M_r-M_\infty|{\bf 1}_F] \le E[|M_r-M_\infty|],$$

and, for any $r\ge n$, the martingale property gives

$$(2)\quad E[ M_r {\bf 1}_F ] = E[ E [M_r |{\cal F}_n] {\bf 1}_F]= E[M_n {\bf 1}_F].$$

Therefore, plugging $(2)$ into $(1)$ gives:

$$(3)\quad |E[M_n {\bf 1}_F ] - E[M_\infty {\bf 1}_F] |\underset{r\ge n}{\le} E[|M_r-M_\infty|].$$

Recall that $n$ is fixed. Now take $r\to\infty$, and by $(*)$ the RHS of $(3)$ tends to $0$. Therefore the LHS of $(3)$ is zero.