Exchange integral and conditional expectation

I know that if we have $E[\int_0^1 |X_t|dt] < \infty$ we may apply Fubini's theorem and compute $E[\int_0^1 X_tdt] = \int_0^1 E[X_t]dt$. Is there a similar version that allows the exchange of integral and conditional expectation? I want to say $E[\int_0^1 X_tdt|F_s] = \int_0^1 E[X_t|F_s]dt$ but I don't know why it should be true.

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Take $F\in\mathcal F_s$ and compute $E[\chi_F\int_0^1X_tdt]$ with Fubini's theorem. – Davide Giraudo Feb 16 '13 at 19:52

Let $(\Omega,\cal A,\mu)$ be a probability space, and assume that $X\colon [0,1]\times\Omega\to \Bbb R$ is measurable for the product $\sigma$-algebra and is integrable with respect to the product measure. In particular, we don't need the assumption of path-continuity.

We can approximate $X$ in $L^1(\lambda \otimes\mu)$ and almost everywhere by simple function, i.e. finite linear combinations of elements of the form $\chi_A$, where $A\in \mathcal{A}\otimes \mathcal{B}([0,1])$. As such an element can be approximated in $L^1(\lambda\otimes\mu)$ by a finite linear combination of elements of the form $\chi_A\chi_B$, $A\in\cal A$, $B\in\mathcal B[0,1]$, we can consider linear combinations $\sum_kc_k\chi_{A_k}\chi_{B_k}$. We will denote $\{X_n\}$ this approximating sequence, which can be chosen in order to make $|X_n|\leqslant |X|$ almost everywhere.

Using dominated convergence theorem for conditional expectation, we just have see prove the equality for $X_n$ instead of $X$. We can indeed take $E(|X(t,\cdot)| \mid\mathcal F)$ as a dominating function, since $|X_n|\leqslant X$.

Note that by dominated convergence, $$\int_0^1E[X(t,\cdot)\mid\mathcal F]dt=\lim_{n\to +\infty}\int_0^1E[X_n(t,\cdot)\mid\mathcal F]dt,$$ and $\int_0^1E[X_n(t,\cdot)\mid\mathcal F]dt$ is $\mathcal F$-measurable.

By linearity of conditional expectation, we will be done once we show that for any $A\in\cal A,$ $B\in\mathcal B[0,1]$, we have $$E\left[\int_0^1\chi_A(\cdot)\chi_B(t)dt\mid \mathcal F\right]=\int_0^1E\left[\chi_A(\cdot)\chi_B(t)\mid \mathcal F\right]dt.$$ It's the case since $\chi_B(t)$ doesn't depend on $\omega$.

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When using Dominated Convergence Theorem, it's important to know the dominating function. In your proof, can you show that with probability one, – Coiacy Feb 22 '13 at 4:27
When using Dominated Convergence Theorem, it's important to know the dominating function. In your proof, can you show that with probability one, $|\mathbb{E}(X_n|\mathcal{F})|\le |\mathbb{E}(X|\mathcal{F})|$? And can you give more details on the $\mathcal{F}$-measurability of $\int_0^1\mathbb{E}(X_n|\mathcal{F})\mathrm{d}t$? Say why $\int_{[t_k,t_{k+1}]}\mathbb{E}(\chi_A |\mathcal{F})$ is $\mathcal{F}$-measurable? Thanks! – Coiacy Feb 22 '13 at 4:41
We have $|E(X_n\mid\mathcal F)|\leqslant E(|X_n|\mid\mathcal F)\leqslant E(|X|\mid \mathcal F)$ as we took $|X_n|\leqslant X$ almost surely. – Davide Giraudo Feb 22 '13 at 10:34
Another thing is that If $\{X_n\}$ is a sequence of $r.v.'s$ and $X_n \to X\ a.s.$, then $\mathbb{E}(X_n|\mathcal{F})\ may\ not\ converge\ to\ \mathbb{E}(X|\mathcal{F})$. There exists such counter examples. Well, this is just for random variables, and here you approximate a process, so I wonder whether it works. Maybe it's better to approximate $\mathbb{E}(X_n(t,\omega)|\mathcal{F})$ instead. – Coiacy Feb 23 '13 at 23:24
@Coiacy The fact follows because there is convergence in L^1$(and also almost everywhere). I agree that almost everywhere convergence alone wouldn't give the result. – Davide Giraudo Feb 24 '13 at 10:17 So you are considering a continuous-time stochastic process$\{X_t: t\ge 0\}$.$\mathcal{F}_s$looks like an element in a filtration$\{\mathcal{F}_t: t\ge 0\}$, and I suggest you put a general$\sigma$-field$\mathcal{F}$instead. The following are some points we need pay attention to: 1. First we need make sure$X_t$is intergrable on$[\ 0,1\ ]$, say the path of$X$is continuous a.s. 2. Both$\mathbb{E}\left[\,\int_0^1 X_t\;\mathrm{d}t\,|\mathcal{F}\,\right]$and$\int_0^1 \mathbb{E}[\,X_t|\mathcal{F}\,]\mathrm{d}t$are random variables, and$\mathbb{E}[\,\int_0^1 X_t\;\mathrm{d}t\,|\mathcal{F}\,]$is$\mathcal{F}-$measurable, can you show that$\int_0^1 \mathbb{E}[\,X_t|\mathcal{F}\,]\mathrm{d}t$is also$\mathcal{F}-$measurable? (I am not sure) 3. Of course we can check by definition of conditional expectation. I think it's true $$\forall F\in \mathcal{F},\quad \mathbb{E}\left(\chi_F\cdot \int_0^1X_t\ \mathrm{d}t\right)=\mathbb{E}\left(\chi_F\cdot \int_0^1 \mathbb{E}(X_t|\mathcal{F})\ \mathrm{d}t\right)$$ In fact, $$\mathbb{E}\left(\chi_F\cdot \int_0^1X_t\ \mathrm{d}t\right)=\mathbb{E}\left(\int_0^1 X_t\cdot \chi_F\ \mathrm{d}t\right)=\mathbb{E}\left(\int_0^1\mathbb{E}(X_t\cdot \chi_F|\mathcal{F}\right)\ \mathrm{d}t)=\mathbb{E}\left(\chi_F\cdot \int_0^1\mathbb{E}(X_t|\mathcal{F})\ \mathrm{d}t\right).$$ - I'm stuck trying to show$\int_0^1 E[X_t|\mathcal{F}]$is$\mathcal{F}$-measurable. – nullUser Feb 18 '13 at 2:51 Here is a way to show$\int_0^1 \mathbb{E}[X_t|\mathcal{F}]\,\mathrm{d}t$is$\mathcal{F}$-measurable: define$I_n(\omega)=\sum_{k=1}^n \mathbb{E}[X_{\frac{k}{n}}|\mathcal{F}](\omega)\cdot \frac{1}{n}$, then$I_n$(sum of finitely many$\mathcal{F}$-measurable$r.v.$') is$\mathcal{F}$-measurable. If we know$\mathbb{E}[X_t|\mathcal{F}]$is integrable on$[0,1]$, for example,$X_t$is continuous$a.s.$such as Brownian motion, then it's Riemann integrable and $$\int_0^1 \mathbb{E}[X_t|\mathcal{F}]\,\mathrm{d}t=:I=\lim_{n \rightarrow \infty}I_n \quad a.s.$$ Note this is a pathwise property. Now since$I_n\in \mathcal{F}$, we obtain$I=\lim_{n \rightarrow \infty}I_n \in \mathcal{F}$.(because the limit of$\mathcal{F}$-measurable functions is still$\mathcal{F}$-measurable) - You are right, here I just assume it's continuous out of convenience. I think you know the fact every continuous-time martingale or submartingale$X_t$(which we usually take interest in) has a version that is right continuous with left limits(then its Lebesgue integral equals Riemann integral). But it still remains to be seen whether$\mathbb{E}(X_t|\mathcal{F})$is continuous when$X_t\$ is continuous. – Coiacy Feb 22 '13 at 4:15