# If $\tau_n,\tau$ are stopping times with $\tau=\inf_n\tau_n$, then $\text E[X\mid\mathcal F_{\tau_n}]\to\text E[X\mid\mathcal F_\tau]$

Let $$(\Omega,\mathcal A,\operatorname P)$$ be a probability space.

Lemma 1: Let $$Y\in\mathcal L^1(\operatorname P)$$, $$I\subseteq\mathbb R$$ be countable and $$(\mathcal G_t)_{t\in I}$$ be a filtration on $$(\Omega,\mathcal A)$$. Then $$M_t:=\operatorname E\left[Y\mid\mathcal G_t\right]\;\;\;\text{for }t\in I\cup\{\operatorname{sup}I\}$$ is an $$(\mathcal G_t)_{t\in I\cup\{\operatorname{sup}I\}}$$-martingale, where $$\mathcal G_{\sup I}:=\sigma(\mathcal G_t:t\in I)$$. Moreover, $$M_t\xrightarrow{t\to\operatorname{sup}I}M_{\operatorname{sup}I}\;\;\;\text{almost surely}\tag1.$$

Now let $$X\in\mathcal L^1(\operatorname P)$$, $$(\mathcal F_t)_{t\ge0}$$ be a filtration on $$(\Omega,\mathcal A)$$ and $$\tau$$ be a $$\mathcal F$$-stopping time. Then $$\tau_n:=\frac{\lceil 2^n\tau\rceil}{2^n}$$ is an $$(\mathcal F_{k2^{-n}})_{k\in\mathbb N_0}$$-stopping time for all $$n\in\mathbb N$$, $$(\tau_n)_{n\in\mathbb N}$$ is nonincreasing and $$\tau_n\xrightarrow{n\to\infty}\tau.\tag2$$ Assuming that $$\mathcal F$$ is right-continuous, it holds $$\mathcal F_\tau=\bigcap_{n\in\mathbb N}\mathcal F_{\tau_n}\tag3.$$

How can we use Lemma 1 and $$(3)$$ to conclude that $$\operatorname E\left[X\mid\mathcal F_{\tau_n}\right]\xrightarrow{n\to\infty}\operatorname E\left[X\mid\mathcal F_\tau\right]\tag4$$ almost surely?

I think the conclusion should be easy given I already wrote, but I'm not able to finish. My idea is to take $$Y=\operatorname E\left[X\mid\mathcal F_\tau\right]$$ and $$\mathcal G_n:=\mathcal F_{\tau_n}$$ for $$n\in\mathbb N$$ in Lemma 1. Then $$(M_{\tau_n})_{n\in\mathbb N}$$ is an $$(\mathcal F_{\tau_n})_{n\in\mathbb N}$$-martingale and $$M_\tau=\operatorname E\left[M_\tau\mid\mathcal F_{\tau_n}\right]$$ for all $$n\in\mathbb N$$.

• @KaviRamaMurthy You mean strictly decreasing? (Note that $(\tau_n)$ is nonincreasing. The claim is made in Theorem 6.29 in the book (1st edition) of Kallenberg.) Apr 8, 2021 at 8:51

Suggested modification: "My idea is to take $$Y=\operatorname E\left[X\mid\mathcal F_\tau\right]$$":

Why don't you consider $$Y=X$$ in lemma 1?

Rest of Proof:

Then, as you point out, Consider, $$\mathcal G_n:=\mathcal F_{\tau_n}$$ for $$n\in\mathbb N$$ in Lemma 1. Then $$(M_{\tau_n})$$ is an $$(\mathcal F_{\tau_n})$$ martingale and $$M_{\tau_n}=\operatorname E\left[X\mid\mathcal F_{\tau_n}\right]$$ for all $$n\in\mathbb N$$.

Now let us calculate $$\mathcal G_{\sup I}$$, given (3) : \begin{align*} \mathcal G_{\sup I}&=\sigma(\mathcal G_t:t\in I) \end{align*} Since in our case $$I$$ is actually $$\mathbb{N}$$, $$\mathcal G_{\sup I}=\sigma(\mathcal G_n:n\in \mathbb N) = \sigma(\mathcal F_{\tau_n}:n\in \mathbb N)$$. But from (3) we have that: $$\mathcal F_\tau=\bigcap_{n\in\mathbb N}\mathcal F_{\tau_n}$$. So $$G_{\sup I}=\mathcal F_\tau$$

Now from the second part of Lemma 1, we get that: \begin{align*} M_t &\xrightarrow{t\to\operatorname{sup}I} &M_{\operatorname{sup}I}\;\;\;&\text{almost surely}\\ \operatorname E\left[Y\mid\mathcal G_t\right] &\xrightarrow{t\to\operatorname{sup}I}&\operatorname E\left[Y\mid\mathcal G_{\sup I}\right]\;& \text{almost surely}\\ \end{align*} We now substitute in the appropriate values: \begin{align*} \operatorname E\left[X\mid\mathcal F_{\tau_n}\right] &\xrightarrow{n\to\operatorname{sup}\mathbb{N}} \operatorname E\left[X\mid \mathcal F_\tau \right]\;& \text{almost surely}\\ \operatorname E\left[X\mid\mathcal F_{\tau_n}\right] &\xrightarrow{n\to\infty} \operatorname E\left[X\mid \mathcal F_\tau \right]\;&\text{almost surely}\\ \end{align*}

which is the desired result

The conditional expectation $$\Bbb E[X\mid\mathcal F_{\tau_n}]$$ converges a.s. and in $$L^1$$ to some random variable $$Z$$, by the martingale convergence theorem. Because $$(\mathcal F_t)$$ is right continuous, you can take $$Z$$ to be $$\mathcal F_\tau$$-measurable. Now check that $$\Bbb E[Z\cdot 1_A]=\Bbb E\left[\Bbb E[X\mid\mathcal F_\tau]\cdot 1_A\right]$$ for each $$A\in\mathcal F_\tau$$. This will prove that $$Z=\Bbb E[X\mid\mathcal F_\tau]$$ a.s.