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When is $\lim\limits_{t \rightarrow \infty} \mathbb{E}[X|\mathcal{F}_{t}] =\mathbb{E}\left[X|\lim\limits_{t\rightarrow\infty}\mathcal{F}_{t}\right]$?

Is there a theorem like monotone convergence or dominated convergence for a problem of this sort?

One specific case of interest would be when $\{\mathcal{F}_{t}\}$ is a sequence of sub-sigma algebras such that $\forall s<t[\mathcal{F}_{s}\subseteq \mathcal{F}_{t}]$ (that is, it is non-decreasing).

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  • $\begingroup$ Presumably the $t$ in the title should be $n$. You may also want to re-copy the question from the title in the main body. I assumed the $F_n$ was already a filtration as that makes $F_{\infty} = \cup_{i=1}^{\infty} F_i$ a natural definitoin. In general, how were you thinking $F_{\infty}$ to be defined? $\endgroup$ – Michael Mar 1 '19 at 20:33
  • $\begingroup$ Correct, I will do such. Thanks! With respect to the edited comment, $\mathcal{F}_{\infty}$ should be defined as the limit of $\mathcal{F}_{t}$ as $t\rightarrow \infty$ $\endgroup$ – BayesIsBae Mar 1 '19 at 20:34
  • $\begingroup$ I changed the notation as it is indeed usually the case that $\mathcal{F}_{\infty}$ is reserved for that specific case. $\endgroup$ – BayesIsBae Mar 1 '19 at 20:42
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    $\begingroup$ @Michael $\mathcal{F}_{\infty}$ should not be defined in that way. It may happen that $\cup_{n=1}^{\infty}\mathcal{F}_{n}$ is not a $\sigma$-algebra even if $\mathcal{F}_{1}\subseteq\mathcal{F}_{2}\subseteq\ldots$. In fact, $\mathcal{F}_{\infty}$ is defined as $\mathcal{F}_{\infty}=\vee_{n=1}^{\infty}\mathcal{F}_{n}:=\sigma\left(\cup_{n=1}^{\infty}\mathcal{F}_{n}\right)$. $\endgroup$ – Danny Pak-Keung Chan Mar 1 '19 at 22:33
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    $\begingroup$ @DannyPak-KeungChan : Yes you are right it should be $\sigma(\cup_{n=1}^{\infty} F_n)$. $\endgroup$ – Michael Mar 1 '19 at 23:37
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You should state all the premises and define your notations. For example, let $(\Omega,\mathcal{F},P)$ be a probability space and let $\{\mathcal{F}_{t}\mid t\geq0\}$ be a filtration. Let $X:\Omega\rightarrow\mathbb{R}$ be an integrable random variable.

Define $\mathcal{F}_{\infty}=\sigma\left(\cup_{t}\mathcal{F}_{t}\right)$. For each $t\geq0$, note that $E\left[X\mid\mathcal{F}_{t}\right]$ is only determined a.e. What is the sense of convergence $\lim_{t\rightarrow\infty}E\left[X\mid\mathcal{F}_{t}\right]$ ? Pointwisely a.e. ?

If this is what you want, you need to be careful: For the case of sequence, if we want to talk about pointwise a.e. convergence $X_{n}\rightarrow X$, for each $n$, we may modify $X_{n}$ on a $P$-null set and it would not affect the conclusion. However, for limit process involving uncountably terms, like $X_{t}\rightarrow X$, we are not allow to "For each $t$, modify $X_{t}$ on a $P$-null set". Now, we immediately encounter a problem: What is $E[X\mid\mathcal{F}_{t}]$? It is not a concrete random variable, but it is only determined a.e.. It is true that $\{E[X\mid\mathcal{F}_{t}]\mid t\geq0\}$ is always a martingale. However, its sample paths are out of control. Note that, if the filtration is standard, we can always choose a cadlag modification for $\{E[X\mid\mathcal{F}_{t}]\mid t\geq0\}$ (a deep result due to Doob) then invoke Martingale Convergence Theorem (uniformly integrable version) and conclude that $\lim_{t\rightarrow\infty}E[X\mid\mathcal{F}_{t}]=E[X\mid\mathcal{F}_{\infty}]$ pointwisely a.e.

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