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 '16 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 '16 at 11:58
  • $\begingroup$ Alright I will post an answer later. $\endgroup$
    – Dark
    Jun 14 '16 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 '16 at 12:12
  • 1
    $\begingroup$ @Gordon has already answered :) The final result should be $\frac{\sqrt{3}}{2}$ $\endgroup$
    – Dark
    Jun 14 '16 at 17:29

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.