# Conditional Expectation of integral of Wiener process

Let $W_t$ be a standard Wiener process. How can we calculate: $$\mathbb{E}\left[\int_0^t|W_r|^2\text{d}r \ |\ \mathcal{F}_s\right]$$ where $(\mathcal{F}_s)_{s\geq0}$ is the natural filtration?

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Is this a homework. If so, please tag it as such, using the (homework) tag. You would still be helped. –  Sasha Dec 19 '12 at 18:37
Can you compute $\mathbb E(|W_r|^2\mid\mathcal F_s)$ for some fixed $r$ and $s$? First for $r\leqslant s$, then for $r\gt s$... –  Did Dec 19 '12 at 18:42
It is not homework exactly. I am trying to solve some questions of past exams which a friend of my gave me. My department does not offer a graduate course in Stochastic Calculus and I'm trying to study on my own. Should I put the homework tag? –  Nick Papadopoulos Dec 19 '12 at 18:49
I think I can compute $\mathbb{E}\left[|W_r|^2\ |\ \mathcal{F}_s\right]$ for every $r>s$ as: $\mathbb{E}\left[|W_r|^2\ |\ \mathcal{F}_s\right]=\mathbb{E}\left[|W_r-W_s+W_s|^2\ |\ \mathcal{F}_s\right]=...$ –  Nick Papadopoulos Dec 19 '12 at 18:59
Then fill the . . . –  Did Dec 19 '12 at 19:09

Hints: (Approximately reproduced from David Williams' excellent Probability with martingales.)

Let $X$ and $Y$ denote some random variables on a probability space $(\Omega,\mathcal F,\mathbb P)$ and $\mathcal G$ denote a sub-sigma-algebra of $\mathcal F$.

• If $X$ is $\mathcal G$-measurable and integrable, then $\mathbb E(X\mid\mathcal G)=$ $______$.
• If $X$ is independent of $\mathcal G$ and integrable, then $\mathbb E(X\mid\mathcal G)=$ $______$.
• If $X$ is $\mathcal G$-measurable, $Y$ is independent of $\mathcal G$, and the function $u$ is measurable and such that $u(X,Y)$ is integrable, then $\mathbb E(u(X,Y)\mid\mathcal G)=v(X)$ where $v(x)=$ $______$.

These should allow you to compute $$\mathbb{E}\left[\int_0^t|W_r|^2\mathrm dr \,\bigg|\, \mathcal{F}_s\right]=\int_0^t\mathbb{E}[|W_r|^2 \mid \mathcal{F}_s]\mathrm dr$$

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So Fubini's Theorem is also true for conditional expectations, isn't that right? –  Nick Papadopoulos Dec 19 '12 at 19:04
Did you fill the boxes? –  Did Dec 19 '12 at 19:09

Note that

$$\mathbb{E} \left( \int_0^t |W_r|^2 \, dr |\mathcal{F}_s \right) = \int_0^t \mathbb{E}(|W_r|^2 |\mathcal{F}_s) \, dr$$

(see remark below). Let $t \geq s$. We have

\begin{align} \int_0^t \mathbb{E}(|W_r|^2 |\mathcal{F}_s) \, dr &= \int_0^s \underbrace{ \mathbb{E}(|W_r|^2 |\mathcal{F}_s)}_{|W_r|^2} \, dr + \int_{s}^t \underbrace{\mathbb{E}(|W_r|^2 |\mathcal{F}_s)}_{|W_s|^2-d \cdot s + d \cdot r} \, dr \\ &= \int_0^s |W_r|^2 \, dr + (t-s) \cdot |W_s|^2 + d \cdot \frac{(t-s)^2}{2} \end{align}

where we used that $(|W_t|^2-t \cdot d)_{t \geq 0}$ is a martingale.

(I was not sure whether $(W_t)_t$ is a 1-dimensional or $d$-dimensional Brownian Motion. In the first case put $d=1$.)

Remark Let $X = \int Z_r \, dr$ and $Y_r=\mathbb{E}(Z_r|\mathcal{F})$. Then $\int_F Y_r \, d\mathbb{P}= \int_F Z_r \, d\mathbb{P}$, thus $$\int \int_F Y_r \, d\mathbb{P} \, dr = \int \int_F Z_r \, d\mathbb{P} \, dr$$ and by Fubini's theorem $$\int_F \left(\int Y_r \, dr\right) \, d\mathbb{P} = \int_F \left(\int Z_r \, dr\right) \, d\mathbb{P}=\int_F X \, d\mathbb{P}$$ for all $F \in \mathcal{F}$. Therefore $$\mathbb{E} \left( \int Z_r \, dr |\mathcal{F} \right)=\mathbb{E}(X|\mathcal{F}) = \int Y_r \, dr = \int \mathbb{E}(Z_r|\mathcal{F}) \, dr$$ which means that we can interchange integration and conditional expectation.

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