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I have 2 questions on integrated Brownian motion and would appreciate any guidance on them.

Question 1

Let $\mathcal F_t = \sigma(W_t)$, the $\sigma$-field generated by $W_t$, is $$Z_t = \int_{0}^{t} W_s \,ds$$ $\mathcal F_t$-measurable? Why so?

I asked the above because I am trying to prove:

For $\mathcal F_t = \sigma(W_t)$ and $Z_t = \int^{t}_{0}\,e^{W_u}\,du$ show $$E[Z_T|\mathcal F_t]=Z_t+W_t(T-t)\quad\forall \,t<T$$

The solution provided from the book I am using https://www.worldscientific.com/worldscibooks/10.1142/9620 is \begin{align} E[Z_T|\mathcal F_t] &= E\left[\int^{t}_{0}W_u\,du|\mathcal F_t\right] + E\left[\int^{T}_{t}W_u\,du|\mathcal F_t\right] \\ &= Z_t + E\left[\int^{T}_{t}W_u - W_t+W_t\,du|\mathcal F_t\right] \tag1 \\ &= \cdots \end{align}

Wouldn't $(1)$ imply that $Z_t$ is $\mathcal F_t$-measurable? But I am not sure why.

Question 2

In an attempt to prove $cov(Z_t, W_t) = \frac{t^2}{2}$, consider

\begin{align} cov(Z_t, W_t) &= E[Z_tW_t] - E[Z_t]E[W_t] \\ &= E[Z_tW_t] \\ &= E\left[W_t\int_{0}^{t}W_s\,ds\right]\tag1 \\ &=E\left[\int_{0}^{t}W_tW_s\,ds\right] \tag2 \\ &= \cdots \end{align}

What is the reason for $(1)$ to $(2)$? Does it have to do with Fubini's theorem? Gordan's answer in Correlation between stochastic processes may be helpful for context.

Thanks!

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    $\begingroup$ Question 1: Why would it be, unless you mean $\mathcal{F}_t=\sigma(\cup_{s\leq t} W_s)$?. Question 2: You can always push in constants into an integral. Here $W_t$ is a constant as it does not depend on the integration variable $s$. $\endgroup$ – J.G Jul 5 at 18:38
  • $\begingroup$ @JasonGaitonde Thanks. However, with regards to question 1, please see the edition in the thread above, in particular showing $E[Z_T|\mathcal F_t]=Z_t+W_t(T-t)\quad\forall \,t<T$ $\endgroup$ – Galvin Ng Jul 6 at 3:56
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To expand on my comment for Question 1: $Z_t$ is not measurable with respect to $\mathcal{F}_t$ as you have currently written it, as it depends on values of $W_s$ for all $s\leq t$, not just $W_t$. It is, however, measurable with respect to $\sigma(\cup_{s\leq t} W_s)$, that is, the sigma-algebra generated by all previous values of $W_s$. To understand why, note that the map $s\mapsto W_s$ is a.s. continuous, so the integral $Z_t$ exists and moreover is almost surely equal to the limit of Riemann sums. The Riemann sums for this integral are just linear combinations of $W_s$ for $s\leq t$, so each Riemann sum is measurable with respect to $\sigma(\cup_{s\leq t} W_s)$, and therefore the same is true of the limit.

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  • $\begingroup$ I suppose it is a typo by the author then. I was suspicious about it. Thanks $\endgroup$ – Galvin Ng Jul 6 at 5:11
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Question 1: Yes. According to Ito product rule $d(tW)=W dt+tdW$, then we could get $$ \int_{0}^t W(s)ds=tW(t)-\int_{0}^{t}s dW(s) $$ For the right-hand side of first term, it is $\mathcal{F}_t$-measurable;for the second term, it is also $\mathcal{F}_t$-measurable. For how to determine whether a random variable is $\mathcal{F}_t$-measurable, there is an sentence from Oksendal in his book named " Stochastic Differential Equations" which may be helpful. It says,

Intuitively, that $h$ is $\mathcal{F}_t$-measurable means that the value of $h(w)$ can be decided from the values of $B_s(w)$ for $s\leq t$. For example, $h_1(w)=B_{t/2}(w)$ is $\mathcal{F}_t$-measurable, while $h_2(w)=B_{2t}(w)$ is not.

Question 2: since the integral $\int_{0}^t W_s ds$ is for variable $s$, therefore you could recognize $W_t$ as a constant. The integral has no influence on $W_t$.

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  • $\begingroup$ Thank you for your reply. What I understand from $\mathcal F_t$-measurability is a random variable X is $\mathcal F_t$-measurable if $\{X \in (a,b)\} \in \mathcal F_t \, \forall a,b \in \mathbb R$. But I still do not see why by that definition, $Z_t$ is $\mathcal F_t$-measurable. Also, I have not learn Ito's product rule. Would you have another explanation then? $\endgroup$ – Galvin Ng Jul 6 at 3:41
  • $\begingroup$ sorry about the answers for question1. I did not notice that part. I thought you forgot to write it. But anyway, you got your answer. $\endgroup$ – Hongjiang Qian Jul 6 at 14:27
  • $\begingroup$ No worries! Thank you for your reply nevertheless $\endgroup$ – Galvin Ng Jul 6 at 14:34

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