Exchange integral and conditional expectation

Question

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.

Answer 1

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, $$ \mathbb{E}\left(\chi_F\cdot \int_0^1X_t\ \mathrm{d}t\right)=\mathbb{E}\left(\int_0^1 X_t\cdot \chi_F\ \mathrm{d}t\right)=\mathbb{E}\left(\int_0^1\mathbb{E}(X_t\cdot \chi_F|\mathcal{F}\right)\ \mathrm{d}t)=\mathbb{E}\left(\chi_F\cdot \int_0^1\mathbb{E}(X_t|\mathcal{F})\ \mathrm{d}t\right).$$

Answer 2

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 $\omega$.

Answer 3

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)