# conditional expectation of stochastic integral

let $$M_t$$ be the following stochastic integral

$$M_t = \int_0^t \sigma_s dW_s$$

where $$\sigma_t$$ is a sufficiently regular deterministic function and $$W_t$$ is a standard Wiener process (that is $$W_t \sim \mathcal{N}(0,t)$$ with independent increments).

It can be shown that $$M_t$$ is martingale with distribution $$M_t \sim \mathcal{N}(0, \Sigma_t)$$ where (Ito's isometry) I have defined the variance

$$\Sigma_t = \int_0^t \sigma^2_s ds$$

Could you kindly check if the following two preliminary assertions are true (everywhere $$0 \leq s):

1. $$M_t$$ has independent increments, that is $$M_s$$ is independent from $$M_t - M_s$$.

2. Covariance: $${\mathbb E}[M_t M_s] = \Sigma_s$$.

Proof of 2.: If 1. holds, reasoning as in the Wiener case: \begin{align} {\mathbb E}[M_t M_s] &= {\mathbb E}[(M_t - M_s + M_s )M_s] \\ &= {\mathbb E}[(M_t - M_s)M_s] + {\mathbb E}[M^2_s] \\ &= {\mathbb E}[M_t - M_s] \cdot {\mathbb E}[M_s] + {\mathbb E}[M^2_s] \\ &= {\mathbb E}[M^2_s] \\ &= \Sigma_s \end{align}

Finally, my question: Conditional expectation: $${\mathbb E}[M_t|W_T] = ?$$

Edit I’m aware of the result $${\mathbb E}[W_t|W_T]=\frac{t}{T}W_T$$ using brownian bridge.

Thanks for your kind attention.

Edit2 This question has a follow-up which might be of interest as well: Regression of stochastic integral on Wiener process

What a great question! I've had a go at it below, I'd say I'm about 75% sure of the result I've got to but I'd love feedback from others.

I'm going to use the definition of the Ito integral, \begin{align} \int^t_0 \sigma_s dW_s = \lim_{n \to \infty} \sum_{i=1}^n \sigma_{t_{i-1}} \bigl( W_{t_i} - W_{t_{i-1}} \bigr) \end{align} where $$t_n = t$$.

Then, using the expression for Brownian Bridging that you've provided above (and neglecting the $$\lim_{n \to \infty}$$ below for breviety) \begin{align} {\mathbb E}\bigl[M_t | W_T\bigr] &= {\mathbb E}\bigl[ \int^t_0 \sigma_s dW_s | W_T\bigr] \\ &= {\mathbb E}\bigl[ \ \sum_{i=1}^n \sigma_{t_{i-1}} \bigl( W_{t_i} - W_{t_{i-1}} \bigr) \ | W_T\bigr] \\ &= \sum_{i=1}^n \sigma_{t_{i-1}} {\mathbb E}\bigl[ \bigl( W_{t_i} - W_{t_{i-1}} \bigr) | W_T\bigr] \\ &= \sum_{i=1}^n \sigma_{t_{i-1}} \bigl( {\mathbb E}\bigl[ W_{t_i}| W_T\bigr] - {\mathbb E}\bigl[ W_{t_{i-1}} | W_T\bigr] \bigr) \\ &= \sum_{i=1}^n \sigma_{t_{i-1}} \bigl( {\frac {t_i} T}W_T - {\frac {t_{i-1}} T}W_T \bigr)\\ &= {\frac {W_T} T} \sum_{i=1}^n \sigma_{t_{i-1}} \bigl( {t_i} - {t_{i-1}} \bigr) \\ &= {\frac {W_T} T} \int_0^t \sigma_{s} ds\\ \end{align}

As a sanity check, we can see that setting $$\sigma_s = 1$$ reproduces the brownian bridging expression.

• Good answer, StackG. I just wanted to point out that in your definition of Stochastic integral: either the sum out to run from $i=1$ to $n$, or it can run from $i=0$ to $n-1$ and then you'd need to change the indexes to $\sigma_{i+1}$ and $\left(W_{t_{i+1}}-W_{t_i}\right)$ (i.e. you cannot have the sum running from zero and then have indexes $t_{i-1}$) Dec 31, 2020 at 8:36
• Many thanks @Jan Stuller once again for the good feedback, I've now fixed this Dec 31, 2020 at 9:28
• Sorry to be picky, but the same is also true for all the sums in the Brownian Bridge :) (I am just pointing it out, because I revised Ito Integrals recently, and only once I started paying attention to the index in the sum of the definition, I started appreciating that the integrator is "forward looking": this is particularly interesting / important in the case of $$\int_0^tW_hdW_h=\lim_{n \to \infty}\sum_{i=1}^nW_{i-1}(W_i-W_{i-1})$$ Not writing out the indexes properly can be confusing to some readers! :) Dec 31, 2020 at 9:42
• Yes agreed, serves me right for trying to fix on my phone. Will fix the rest later! Dec 31, 2020 at 10:01

Just wanted to add to @StackG's great answer using a different approach. Please, double-check my solution as well because I'm not 100% sure.

Let $$\sigma_t$$ be sufficiently regular such that $$\dot{\sigma}_t \stackrel{def}{=}\frac{d \sigma}{dt}$$ is well defined. Then, Ito's lemma:

$$d(\sigma_t W_t) = \dot{\sigma}_t W_t dt + \sigma_t dW_t$$

which in integral form reads

$$\sigma_t W_t = \int^t_0 \dot{\sigma}_s W_s ds + \int^t_0 \sigma_s dW_s$$

We have then the representation

\begin{align} M_t & \stackrel{def}{=} \int^t_0 \sigma_s dW_s \\ &= \sigma_t W_t - \int^t_0 \dot{\sigma}_s W_s ds \end{align}

Therefore, using Fubini to interchange integral with expectation, recalling that $$\sigma_t$$ is deterministic, and that $${\mathbb E}[W_t|W_T] = \frac{t}{T} W_T$$ we can write the requested conditional expectation as

\begin{align} {\mathbb E}[M_t|W_T] & \stackrel{def}{=} {\mathbb E}\left[\int^t_0 \sigma_s dW_s \bigg| W_T \right] \\ &= \sigma_t {\mathbb E}[W_t|W_T] - \int^t_0 \dot{\sigma}_s {\mathbb E}[W_s|W_T] ds \\ &= \sigma_t \frac{t}{T} W_T - \frac{W_T}{T} \int^t_0 \dot{\sigma}_s s ds \\ &= \sigma_t \frac{t}{T} W_T - \frac{W_T}{T} \left[ \sigma_t t - \int^t_0 \sigma_s \cdot 1 ds\right] \\ &= \frac{W_T}{T} \int^t_0 \sigma_s ds \end{align} where integration by parts has been used in the next-to-last line.

• Nice solution @Gabriele Pompa! Jan 3, 2021 at 2:01
• Apparently one cannot accept more than one answer. I’m accepting yours, as it seems to attract more consensus. Thanks! Jan 3, 2021 at 9:54