TITLE: Correlation between stochastic processes
QUESTION [6 upvotes]: Question: if $W(t)$ is a standard Brownian motion with $W(0)=0$, what is the linear coefficient between the stochastic processes $W(t)$ and $I(t)=\int_0^t W(s)ds$?
I argued as follows: what we want is the coefficient of the differential product $dW(t)dI(t)$. Then, write this as $dW(t)\cdot W(t)dt$. Now the presence of a product of the form $dW(t)dt$ which is equal to zero forces the prduct $dW(t)dI(t)=0$, which seems to imply that the correlation is zero. Is it this an acceptable argument? Or did I make a mistake??

REPLY [9 votes]: We can compute the covariance between $W_t$ and $I_t$ as follows:
\begin{align*}
covar(W_t, I_t) &=E\left(W_t \int_0^t W_s ds \right)\\
&= E\left( \int_0^t W_t W_s ds \right)\\
&= \int_0^t E\left(W_t W_s \right) ds\\
&= \int_0^t \min(t, s)\, ds\\
&=\frac{t^2}{2}.
\end{align*}
Alternaticely, note that 
\begin{align*}
d(tW_t) = W_t dt + t dW_t.
\end{align*}
Then,
\begin{align*}
I_t = \int_0^t W_s ds = tW_t - \int_0^t sdW_s.
\end{align*}
Therefore, 
\begin{align*}
covar(W_t, I_t) &= E(W_t I_t) \\
&=tE(W_t^2) - E\left(\int_0^t dW_s\int_0^t sdW_s \right)\\
&=t^2 - \int_0^t sds\\
&=\frac{t^2}{2}.
\end{align*}
The correlation can be computed similarly.