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Let

  • $\mathbb{F}$ be a filtration
  • $(X_t)_{t\ge 0}$ be a right-continuous $\mathbb{F}$-supermartingale
  • $\sigma,\tau$ be bounded $\mathbb{F}$-stopping times with $\sigma\le \tau$ and $\sigma^n:=2^{-n}\lceil 2^n\sigma\rceil$, $\tau^n:=2^{-n}\lceil 2^n\tau\rceil$

I want to show, that we've got $$\operatorname{E}\left[X_{\tau^m}\mid\mathcal{F}_{\sigma^n}\right]\stackrel{n\to\infty}{\to}\operatorname{E}\left[X_{\tau^m}\mid\mathcal{F}_\sigma\right]\;\;\;\text{and}\;\;\;X_{\sigma^n}\stackrel{n\to\infty}{\to}X_\sigma\tag{1}$$ almost surely and in $L^1$.


Maybe it's not that hard to prove $(1)$, but I absolutely don't find a starting point. So, what do we need to do to prove $(1)$?

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  • $\begingroup$ Why do people always use $$2^{-n}\lceil 2^n\alpha\rceil$$ instead of $$\frac 1n\lceil n\alpha\rceil\;?$$ Does that make any difference? $\endgroup$ – 0xbadf00d Jul 2 '15 at 9:55
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    $\begingroup$ The first one gives a decreasing sequence; the second one is not necessarily decreasing. $\endgroup$ – saz Jul 2 '15 at 16:02
  • $\begingroup$ @saz Why? I don't see that. $\endgroup$ – 0xbadf00d Jul 3 '15 at 20:15
  • $\begingroup$ Which one? And do you assume anything on the filtration? One would expect $$\mathbb{E}(X_{\tau^m} \mid \mathcal{F}_{\sigma^n}) \to \mathbb{E}(X_{\tau^m} \mid \mathcal{F}_{\sigma \color{red}{+}});$$ I would say. $\endgroup$ – saz Jul 5 '15 at 11:22
  • $\begingroup$ consider $x>0$ such that $0<3x<1$ and $1<4x$ therefore $\frac{1}{3}\lceil3x\rceil = 1/3$ and $\frac{1}{4}\lceil4x\rceil = 2/4=1/2 >1/3$ $\endgroup$ – Conrado Costa Jul 6 '15 at 20:32
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So let's see that $X_{\sigma^n} \to X_{\sigma}$ in $L^1$. Since $\sigma^n \downarrow \sigma$ right continuity gives us that $X_{\sigma^n} \to X_{\sigma}$ almost surely. Now call $X(n) = X_{\sigma^n}$ and $\mathcal{F}_n = \mathcal{F}_\sigma$.

Note that

1) $X(n)$ is $\mathcal{F}_n$ adapted

2) $X(n) \in L^1$ for every $n$

3)$\Bbb{E}[X(n+1) - X(n)\vert \mathcal{F}_n] \geq 0$

4) $\Bbb{E}[X_n]\leq\Bbb{E}[X(n+1)]\leq \Bbb{E}[X_\sigma] <\infty$

Let $\Delta_{n} = \Bbb{E}[X(n) - X(n-1)\vert \mathcal{F}_{n-1}] \geq 0$.

Let $A_{n} = \sum_{i=2}^{n} \Delta_i$

$$\Bbb{E}[A_n] = \Bbb{E}[X(n) - X(1)]\leq \Bbb{E}[X_\sigma - X(1)]< \infty$$

$A_{\infty}:= \sum_{i=2}^\infty \Delta_i$.By the monotone convergence theorem $\Bbb{E}[A_\infty]< \infty$.

Define now, $M(n) = X(n) - A(n)$. Note that $M_n$ is a martingale. Indeed $$\Bbb{E}[M(n+1) - M(n)\, \vert \mathcal{F}_n] = \Bbb{E}[X(n+1) - X(n) - \Delta_{n+1}\, \vert \mathcal{F}_n] = 0 $$

Since $\mathcal{F}_n \subset \mathcal{F}_{n-1}$, $$\Bbb{E}[M(1)- M(n)\vert \mathcal {F}_n] = \Bbb{E}[\Bbb{E}[M(1)- M(n)\vert \mathcal {F}_1]\vert\mathcal {F}_n]=0$$ $M(n)$ is uniformly integrable.

Moreover since $A_n \leq A_\infty \in L^1$ $A_n$ is also uniformly integrable. Therefore $X_n = M_n + A_n$ is also uniformly integrable.

Thus, almost sure convergence (from right continuity) implies convergence in law which combined with Uniform integrability yields convergence in $L^1$.


we now want to prove $$\operatorname{E}\left[X_{\tau^m}\mid\mathcal{F}_{\sigma^n}\right]\stackrel{n\to\infty}{\to}\operatorname{E}\left[X_{\tau^m}\mid\mathcal{F}_\sigma\right]$$ for $T = \tau^m$ $$f_n(\omega)=\Bbb{E}[X_T \vert \mathcal{F}_{\sigma^n}](\omega)$$ Is almost an uniformly integrable backward martingale therefore it converges almost surely and in $L^1$.

To see that where is the backward martingale remember that $\mathcal{F}_{\sigma^n} \supset \mathcal{F}_{\sigma^m}$ when $m>n$ $$\Bbb{E}[f_n \vert \mathcal{F}_{\sigma^m}] = f_m $$

Then consider $Y_{n} = f_{-n}$ and $\mathcal{F}_{n} = \mathcal{F}_{\sigma^{-n}}$ now this sequence is a an uniformly integrable backward martingale.

Then the limit $Y_{-\infty}= \lim_{n\to -\infty} X_n$ exists almost surely and in $L^1$ define $f = Y_{-\omega}$ and note that $\lim_{n\to \infty} f_n(\omega) = f(\omega)$ is our natural candidate for $\Bbb{E}[X_T \vert \mathcal{F}_{\sigma}]$ since $\sigma^n \downarrow \sigma$.

We need to check two properties

A) $\int_A f(\omega)\,d\Bbb{P}(\omega) =\int_A X_T(\omega)\,d\Bbb{P}(\omega) $ for every $A \in \mathcal{F}_\sigma$

Since $A \in \mathcal{F}_\sigma \Rightarrow A \in \mathcal{F}_{\sigma^n}$ we have for each $n > 0$: $$\int_A f_n(\omega)\,d\Bbb{P}(\omega) =\int_A X_{T}\,d\Bbb{P}(\omega) $$

Therefore since $X_{\sigma^n} \xrightarrow[]{L^1} X_{\sigma} $ taking limits in both sides we obtain (use monotone convergence for the left side) $$\int_A f(\omega)\,d\Bbb{P}(\omega) =\int_A X_{T}(\omega)\,d\Bbb{P}(\omega) $$

B) the last condition we need to check is that $f(\omega)$ is $\mathcal{F}_{\sigma}$ measurable. Since $f_n$ is $\mathcal{F}_{\sigma^n}$ measurable and $\mathcal{F}_{\sigma^n}\supset\mathcal{F}_{\sigma^{n+1}}$we conclude that $f$ is $\mathcal{F}_{\sigma^n}$ measurable for every $n$. since $\sigma^n \downarrow \sigma$ $forall \delta>0$ we choose $n$ such that $\mathcal{F}_{\sigma^n} \subset \mathcal{F}_{\sigma + \delta}$ This implies that $f is \mathcal{F}_{\sigma}^+$ measurable. If your filtration satisfies the usual conditions (see https://almostsure.wordpress.com/2009/11/08/filtrations-and-adapted-processes/) then you conclude that $f$ is $\mathcal{F}_{\sigma}$ measurable and therefore that

$$\lim_n \Bbb{E}[X_T \vert \mathcal{F}_{\sigma^n}] = \lim_n f_n(\omega) = f(\omega) = \Bbb{E}[X_T \vert \mathcal{F}_{\sigma}]$$

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