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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 $0 \leq s<t<T$):

  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

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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.

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  • $\begingroup$ 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 $t_{i-1}$) $\endgroup$ – Jan Stuller Dec 31 '20 at 8:36
  • $\begingroup$ Many thanks @Jan Stuller once again for the good feedback, I've now fixed this $\endgroup$ – StackG Dec 31 '20 at 9:28
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    $\begingroup$ 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! :) $\endgroup$ – Jan Stuller Dec 31 '20 at 9:42
  • $\begingroup$ Yes agreed, serves me right for trying to fix on my phone. Will fix the rest later! $\endgroup$ – StackG Dec 31 '20 at 10:01
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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.

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

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