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

brownian motionmartingalesprobabilityprobability theorystochastic-processes

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!

Best Answer

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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