Let the equation :

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

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

Consider the following equation for $X(t)$:

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

where $0 < b, \sigma\in\mathbb{R} $, $X(0)$ is the 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 someone please give me some ideas? Thanks so much for your tim.

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

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

Let the equation :

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

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

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

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