I want to find Ito representation for variable $X = W(T)$, where $W$ is a Wiener process.
Let $W_{t}$ be a Wiener process on a standard filtered probability space $(\Omega ,{\mathcal {F}},{\mathcal {F}}_{t},P)$ and let ${\mathcal {G}}_{t}$ be the augmented filtration generated by $B$. If $X$ is a square integrable random variable measurable with respect to ${\mathcal {G}}_{\infty }$, then there exists a predictable process $\Theta$ which is adapted with respect to ${\displaystyle {\mathcal {G}}_{t}}$, such that
$$X = E[X] + \int_0^T\Theta_sdWs$$
I found somewhere on internet that solution to this problem is $W[T] = \int_0^TdWs$, however I'm not sure exactly what it means. Does it mean that this expression equals to $\Theta_s$ ? Could you please explain to me why this formula is proper for Ito representation ?
Best Answer
This is simply the statement that when you compute the Ito integral $\int_0^T \Theta_s\,dW_s$ with $\Theta_s$ being the constant process $1$ (not depending on either $s$ or $\omega$), you get $\int_0^T 1\,dW_s = W_T$. (This is an easy exercise that can be done directly from the definition of the Ito integral.)
Combining this with the fact that $E[W_T] = 0$ (from the definition of Brownian motion as a mean-zero Gaussian process), we see that $$W_T = E[W_T] + \int_0^T 1\,dW_s$$ and so this is the Ito representation of $W_T$. The notation $\int_0^T \,dW_s$ is just shorthand for $\int_0^T 1\,dW_s$.