Question

it is really shame on me that I cannot find the derivative of the following integral :(

$$X(t)=e^{-bt}X(0)+\sigma\int_{0}^{b}e^{-b(t-s)}dW(t)$$

Find $dX(t)$ ? where $0 < b, \sigma\in\mathbb{R} $, $X(0)$ is initial distribution of $X(t)$, independent of the Brownian motion $W(t)$. I want so show that $dX(t)= -bX(t)dt+\sigma dW(t)$, but I am getting stuck on computing the derivative of $$\sigma\int_{0}^{b}e^{-b(t-s)}dW(t)$$

could some one please give me some ideas ? thanks so much for your time

PS. the above equation is one of type of the Langevin's equation, more detail could be found here http://en.wikipedia.org/wiki/Langevin_equation

Answer 1

If you interpret the stochastic integral in the Ito-sense (often used in finance) you'll have to use Ito's lemma to evaluate it:
See e.g. here: Ito's lemma

Alternatively you could interpret it in the Stratonovich-sense (often used in physics):
See e.g. here: Stratonovich integral

A good introduction to solving these kinds of stochastic differential equations (sde) without the use of measure theory and with lots of intuition is e.g. Wiersema: Brownian motion calculus

Answer 2

Ok here is the trick

Use Itô's lemma mentioned by vonjd to the function $f(t,X_t)=e^{bt}.X(t)$ and after some algebra you'll get what you want.

Regards