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I'm thinking about the problem of deriving the stochastic differential of an integral with both time and state part of the integrand but not in a way that you can easily factor it out - for example I want to derive the partial derivative with respect to $t$ of

$\int_0^t e^{tX_s}X_s dB_s$

or similar. In the case the integrand $f(t, X_t)$ can be represented as a product of state and time terms I can use integration by parts sometimes but is there a general way to work with these types of problems?

Edit: Before the suggestion comes, I also want to close out the case where I could substitute into a function which I could then tackle wit the product rule of differentiation.

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  • $\begingroup$ no, the whole point of Ito integrals is that they allow for dealing with non-continuous functions. the notion of classic integration (eg, by parts, product rule) doesn't apply in the same way. have you gone through Shreve (or a similar text) $\endgroup$ – Chris Feb 6 '20 at 9:37
  • $\begingroup$ I did. But your answer confuses me, neither is the integrand nor the integrator non-continuous a.s. (we can assume that $X=g(t, B_t)$ continuous for simplicity) nor are non-continuous functions any problem for classical integrals (there is no problem in integrating the sign function for example with the Riemann integral). In case you mean differentiability, this is by the example above also not a problem, hence not "the whole point". EDIT: I also want to add that I refer to the stochastic product rule here: $d(XY) = XdY + YdX + d<X,Y>$. $\endgroup$ – not_sure95 Feb 6 '20 at 9:42
  • $\begingroup$ There is a stochastic Leibniz rule which you can apply to this type of integrals, see here or here. Is that what you are after? $\endgroup$ – Daneel Olivaw Mar 30 '20 at 16:26
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There is a simple way to transform $dB_t$ to $dt$ in the case your $X_t$ is function of the brownian motion. let's compute for example: $$\int^T_0e^{tBt}dB_t$$ Here is my way to do it:

  1. You replace Bt by x in the function inside the integral, so the function becomes: $e^{tx}$.
  2. You compute the primitive of the function with respect to x: $\int e^{tx}dx=\frac{e^{tx}}{t}$.
  3. You replace $x$ by $B_t$ in the primitive then you use ito lamma and its done: $$d\frac{e^{tB_t}}{t}=\frac{B_te^{tB_t}-e^{tB_t}}{t^2}dt+e^{tB_t}dB_t+\frac{1}{2}te^{tB_t}dt$$ $$e^{tB_t}dB_t=d\frac{e^{tB_t}}{t}-(\frac{B_te^{tB_t}-e^{tB_t}}{t^2}+\frac{1}{2}te^{tB_t})dt$$
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  • $\begingroup$ I don't understand what you're doing. It's also not the class of problems that I'm concerned with as this is a classical integral where the time derivative is trivially given by the fundamental theorem of calculus and I'm looking at Ito integrals here. I'm also not sure how and why the thing you do there should be correct. $\endgroup$ – not_sure95 Feb 6 '20 at 18:07
  • $\begingroup$ I edited the answer. If you still don't understand what i wrote you should review your comprehension of what ito lemma is! $\endgroup$ – Valometrics.com Feb 7 '20 at 16:12
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I think here you can just differentiate. The variable $t$ appears twice in the expression: as the limit of integration and inside the integrand. You get one term in the partial derivative from each of these occurances. That is, $$ \frac{d}{dt} \int_0^t e^{t X_s} X_s dB_s = e^{t X_t} X_t dB_t + \int_0^t e^{t X_s} X_s^2 dB_s . $$

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  • $\begingroup$ Can you review your formula!! I see the sum of $dB_t$ with an integral!!!! $\endgroup$ – Valometrics.com Feb 7 '20 at 16:15
  • $\begingroup$ His notation is wrong. It should be $$ d \int_0^t e^{t X_s} X_s dB_s = e^{t X_t} X_t dB_t + \bigg( \int_0^t e^{t X_s} X_s^2 dB_s \bigg)dt . $$ $\endgroup$ – Gordon Feb 7 '20 at 17:38

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