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I have a doubt : I know that if $x_{t}=\int_{0}^{t}\gamma(s)dW_{s}$ (with $W_{s}$ a brownian motion), we have : $dx_{t}=\gamma(t)dW_{t}$ What about if $x_{t}=\int_{0}^{t}\gamma(s,t)dW_{s}$. Do I have to apply a kind of Lieibniz rule to get $dx_{t}$ ? If so what is the result ? Tx !

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We assume that $\gamma(s, t)$ is differentiable with respect to $t$. Then, \begin{align*} dx_t = \left(\int_0^t \frac{\partial\gamma(s, t)}{\partial t} dW_s \right)dt + \gamma(t, t) dW_t. \end{align*}

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  • $\begingroup$ I see in the meantime a similar answer on Willmott: wilmott.com/messageview.cfm?catid=4&threadid=51005 But it's $...dW_{t})dt$ or $...dW_{s})dt$ ? And what about the quadratic variation : do we have : $d<x>_{t}=\gamma(t,t)^2dt$ ? Seems like I missing the 1st term though ... $\endgroup$ – glork May 19 '16 at 13:20
  • $\begingroup$ The first is a $dt$ term, which disappears for quadratic variation. $\endgroup$ – Gordon May 19 '16 at 13:25
  • $\begingroup$ ok thank you. And what about my 1st question : is it : $...dW_{t})dt$ or $...dW_{s})dt$ Many thanks $\endgroup$ – glork May 19 '16 at 13:29
  • $\begingroup$ sorry, that was typo, which I have corrected. $\endgroup$ – Gordon May 19 '16 at 13:32

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