Differentiating Time-Dependent Integrals with Brownian Integrators

# Partial derivative of Ito integral without product rule


# Partial derivative of Ito integral without product rule












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.

## Answer by Valometrics.com (score 1)

https://quant.stackexchange.com/a/51034

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:

- You replace Bt by x in the function inside the integral, so the function becomes: $e^{tx}$.

- You compute the primitive of the function with respect to x: $\int e^{tx}dx=\frac{e^{tx}}{t}$.

- You replace $x$ by $B_t$ in the primitive then you use ito lamma and its done: $$d\frac{e^{tB_t}}{t}=\frac{B_te^{tB_t}-e^{tB_t}}{t^2}dt+e^{tB_t}dB_t+\frac{1}{2}te^{tB_t}dt$$ $$e^{tB_t}dB_t=d\frac{e^{tB_t}}{t}-(\frac{B_te^{tB_t}-e^{tB_t}}{t^2}+\frac{1}{2}te^{tB_t})dt$$

## Answer by q.t.f. (score 0)

https://quant.stackexchange.com/a/51044

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, $$ \frac{d}{dt} \int_0^t e^{t X_s} X_s dB_s = e^{t X_t} X_t dB_t + \int_0^t e^{t X_s} X_s^2 dB_s . $$