Ito representation

Question

What is the way to find a function $f$ such that for a stochastic process, for example $e^{B_T}$ we can write: $$ e^{B_T} = E[e^{B_T}] + \int_{0}^{T}f(t)dB_t$$?

Answer 1

Use Ito's formula: for a $C^2$ function $f$,

\begin{align*} f(B_t) = f(0) + \int_0^t f'(B_s)dB_s + \frac 12 \int_0^t f''(B_s)ds \end{align*}

Answer 2

Applying Ito formula to the function $\phi : x \mapsto e^{x} $ and using the process $B$. We have : \begin{equation} e^{B_T} = 1 + \frac12\int_0^Te^{B_t}dt+ \int_0^Te^{B_t}dB_t \end{equation} Now taking the expectation on both sides, we have that: \begin{equation} \mathbb{E}[e^{B_T}] = \left(1 + \frac12\int_0^Te^{B_t}dt\right) + 0 \end{equation} Note that by definition of the stochastic integral, we have $\mathbb{E}\left[\int_0^Te^{B_t}dB_t\right]=0$. This follows from the fact that the function $f : (t,\omega) \mapsto \exp(B(t,\omega))$ belong to $H^2[0,T]$ where $H^2$ is the space of progressively measurable function such that : \begin{equation} \mathbb{E}\int_0^T f^2(t,\omega)dt < \infty \end{equation} Therefore by identification, we have: $e^{B_T} = \mathbb{E}[e^{B_T}] + \int_0^Tf(t)dB_t$