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I am looking to take the partial derivative of an integral with respect to brownian motion. For Simplicity I will make it the same integral as in this post (don't have enough reputation to comment): Partial Derivative of an Integral

My integral is:

$X(t)=\int^t_0 f(s)dB(s)$

My question is about how the the answerers arrived at their answer exactly. I already know that by definition,

$d(X(t))=d(\int^t_0 f(s)dB(s))=f(t)dB(t)$

However I would like to understand how to arrive at this through Leibniz rule. Leibniz' Rule is defined for when the integrator, which is in this case dBs, is the same variable as the secondary variable in f, i.e. f would have to be a function f(t,Bs). But we are given a function f(t,s) (well, actually it is just a function of s), so I don't understand how to apply the rule in this case. Of course if we ignore the dBt, we end up with d(X(t))=f(t)dt, but this is of course wrong. I think this is just a simple change of variables but could someone please help me out?

Thanks, Paul

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Paul
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    There is no reason to assume that $t\mapsto X(t)$ is differentiable, and most of the time it is not. Relatedly, the accepted answer to the question you link to is incorrect, as is the answer this answer itself refers to. I feel sorry for you that you got misled by this web of incorrect answers (but, to be honest, not too sorry because plenty of other MSE pages explain things correctly). – Did Aug 05 '13 at 19:04
  • @Did, But saying that dX exists is not the same as saying it is differentiable. For example, t->B_t is non-differentiable, but dB_t is defined. And are you saying that the answer is the same as above but with the caveat "if the function is differentiable?" If not could you help me out with a link? – Paul Aug 05 '13 at 21:19
  • sorry, I don't know why I said t->B_t, I meant B_t is non-differentiable... – Paul Aug 05 '13 at 21:29
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    @Paul: Leibniz rule only works for Riemann integrals. It doesn't work for Ito Integrals. You could potentially make it work for Riemann Stieltjes integrals with some care but all of this becomes junk in the Ito integral mainly because Brownian motion isn't of bounded variation. The only analogue that exists is Ito's formula for $dX(t)$ and it doesn't have any "classical" derivation. – Alex R. Aug 05 '13 at 21:41

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