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??

  • $\begingroup$ The covariation (what you have computed) is 0 but the correlation is not. $\endgroup$
    – Dark
    Jun 14, 2016 at 11:56
  • $\begingroup$ OK, it is clear that I am missing something here. Could you pease elaborate in an answer? Thank you. $\endgroup$
    – RandomGuy
    Jun 14, 2016 at 11:58
  • $\begingroup$ Alright I will post an answer later. $\endgroup$
    – Dark
    Jun 14, 2016 at 12:05
  • $\begingroup$ Excellent, thank you. I thought this was the correlation computation in fact so I think I am missing something. Looking forward for you answer $\endgroup$
    – RandomGuy
    Jun 14, 2016 at 12:12
  • 1
    $\begingroup$ @Gordon has already answered :) The final result should be $\frac{\sqrt{3}}{2}$ $\endgroup$
    – Dark
    Jun 14, 2016 at 17:29

1 Answer 1


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.


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