# Integrated Brownian motion $\int_{0}^{t} W_s \, ds$ properties

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

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!

• 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 $s$. – J.G Jul 5 at 18:38
• @JasonGaitonde Thanks. However, with regards to question 1, please see the edition in the thread above, in particular showing $E[Z_T|\mathcal F_t]=Z_t+W_t(T-t)\quad\forall \,t<T$ – Galvin Ng Jul 6 at 3:56

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.

• I suppose it is a typo by the author then. I was suspicious about it. Thanks – Galvin Ng Jul 6 at 5:11

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 $$W_t$$.

• 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? – Galvin Ng Jul 6 at 3:41
• 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. – Hongjiang Qian Jul 6 at 14:27
• No worries! Thank you for your reply nevertheless – Galvin Ng Jul 6 at 14:34