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I want to compute $E\left[W_t \int_0^t s \, dW_s\right]$ where $W_t$ is a Brownian motion. My attempt below is based on some very shaky mathematics; in particular I have no justification of the 4th equality but it leads me to the right answer. Can anyone show me the correct way to compute this expectation?

$$ E\left[W_t \int_0^t s \, dW_s\right] = E\left[\int_0^t dW_s \int_0^t s \, dW_s\right] = E\left[\int_0^t \int_0^t s \, dW_s \, dW_s\right] = E\left[\int_0^t s \, dt\right]=\frac{t^2}{2} $$

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  • $\begingroup$ The equality between the second and third terms is also bogus because you've conflated the two dummy variables. $\endgroup$ Aug 20, 2013 at 23:58

1 Answer 1

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Hint: The expression $E\left[\int_0^t dW_s \int_0^t s\,dW_s\right]$ looks helpful. Now recall the Itô isometry.

Hint 2: If you only know the Itô isometry in the form $E\left[\left(\int_0^t A_s\,dB_s\right)^2\right] = E\left[\int_0^t A_s^2\,ds\right]$, then think about the simple identity $ab = \frac{(a+b)^2-(a-b)^2}{4}$. (Or the magic word "polarization".) Or, stated in a more sophisticated way, a linear map between Hilbert spaces that preserves the (squared) norm also preserves the inner product.

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  • $\begingroup$ If the integrand $s$ was not in the second integral, I would be able to apply Itô's isometry. But I'm not sure what to do as things stand $\endgroup$
    – AUK1939
    Aug 21, 2013 at 0:14
  • $\begingroup$ @aukie: Ok, I added another hint. $\endgroup$ Aug 21, 2013 at 0:21
  • $\begingroup$ Thanks, the second hint did the trick. Was not aware of other forms of Itô's isometry $\endgroup$
    – AUK1939
    Aug 21, 2013 at 1:18

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