I'm given process $X(t) = \int_{0}^{t} a(s)dB(s) $ where a(s) is a square-integrable deterministic function
I need to show, that $E[e^{imX(t)}] = e^{-\frac{m^2}{2}\int_{0}^{t}a^2(s)ds}$
My attempt:
I write $e^{imX(t)}$ as an Ito process:
$f(x,t) = e^{imx}$, $f_{x} = ime^{imx}, f_{t} = 0, f_{xx} = -m^2e^{imx}, dx = a(t)dB(t), [dx*dx] = a^2(t)dt$
So $d(f(x,t)) = d(e^{imx}) = i*m*f*a(t)dB(t)-\frac{1}{2}f*m^2*a^2(t)dt$
So $e^{imx} = 1 + \int_{0}^{t}i*m*e^{imx(s)}*a(s)dB(s)-\int_{0}^{t}\frac{1}{2}e^{imx(s)}*m^2*a^2(s)ds$
And now I'm lost. When I take expectation of both sides even though I get rid of the stochastic integral, I still have something on the right which I don't know how to deal with ... I'd appreciate if anyone could give a helping hand.
Since the stochastic integral
$$M_t := i \int_0^t m e^{im X_s} a_s \, dB_s$$
is a martingale, it has, in particular, constant expectation and so $\mathbb{E}(M_t) = \mathbb{E}(M_0)=0$. Taking the expectation in the last equation in your question, we therefore get
$$\mathbb{E}(e^{im X_t}) =1+ 0 - \frac{m^2}{2} \int_0^t \mathbb{E}(e^{im X_s}) a(s)^2 \, ds.$$
This shows that the mapping
$$\varphi(t) := \mathbb{E}(e^{im X_t})$$
satisfies the ordinary differential equation
$$\frac{d}{dt} \varphi(t) = - \frac{m^2}{2} a(t)^2 \varphi(t), \qquad \varphi(0)=1.$$
Find the solution to this equation in order to prove the assertion...