Quantitative Finance Stack Exchange is a question and answer site for finance professionals and academics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|improve this question
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

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


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

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 $[0, +\infty)$

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.