# Ito's Lemma - Integrand depends on upper limit of integration

A problem I came across while practicing using Ito's Lemma had a process with an integral whose integrand depends on the upper limit of integration (the goal is to find $dZ_{t}$):

$Z_{t}=\int_{0}^{t}e^{\frac{t-s}{2}}\sin(B_{s})dB_{s}$, where $B$ is a standard Brownian motion

In what way do I need to take this into account in my solving the problem, if at all?

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Factor out the $e^{t/2}$ part and then use the product rule. This question should be useful: quant.stackexchange.com/questions/4733/… – quasi Apr 12 '13 at 13:08

Note that the $$dX_t = b_t dt + \sigma_s dB_t$$ notation for a (local) semi-martingale $X = (X_t)_{t \in [ t_0, T]}$ is an abreviation for

$$X_t = X_{t_0} + \int _{t_0} ^t b_s~ ds + \int _{t_0} ^t \sigma_s ~dB_s$$

where $b$ and $\sigma$ can be for example of the form $b_s = b(\omega, s, X_s)$ and $\sigma_s = \sigma(\omega, s, X_s)$ under condition that they are progressivelly measurable prosses and that $$\ \int _{t_0} ^T b_s ds + \int _{t_0} ^T \sigma_ s^2 ds \ < \infty \quad \mathbb P - as$$

So, since $Z_t = e^{\frac{t}{2}} Y_t$ where $Y_t:= \int _{0} ^t e^{\frac{-s}{2}} \sin( B_s) ~dB_s$, you have by Itô's Lemma

$$d Z_t = e^{\frac{t}{2}} ~dY_t + \frac{1}{2}e^{\frac{t}{2}} Y_t ~dt+d\langle e^{\frac{t}{2}} ,Y_t\rangle_t$$

then,

$$d Z_t = e^{\frac{t}{2}} e^{\frac{-t}{2}} \sin( B_t) ~dB_t + \frac{1}{2}e^{\frac{t}{2}} \int _{0} ^t e^{\frac{-s}{2}} \sin( B_s) ~dB_s$$

for all $s\in [0, +\infty)$ (note that $Z_0 =0$)

Also note that you must verifie that $Z$ is well defined as an stochastic integral, wich is evidently true since the integrand is bounded in $[0, +\infty)$

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This isn't right. The $e^{t/2}$ term can't be treated as a constant, and there shouldn't be an $s$ and a $t$ in your final answer. – quasi Apr 12 '13 at 13:12
Of course not. It's a typo, I forgott the terms. Thank you for note that – Paul Apr 12 '13 at 13:50