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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 $n\in\mathbb N$.

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  • $\begingroup$ @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.) $\endgroup$
    – 0xbadf00d
    Apr 8, 2021 at 8:51

2 Answers 2

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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

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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.

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