Let $x_k$ be a Ito process defined by the equation $dx_t=-ax_tdt+\sigma dB_t$, where a is a real constant, $\sigma$ is a positive real constant and $B_t$ is a standard Brownian motion. Let $x_t=x_0$ at $t=0$.
By applying Ito's Lemma on $y_t=f(t)x_t$, where $f(t)$ is a deterministic function of $t$, show that,
i) the expectation value of $x_t$ is $x_0e^{-at}$,
ii) the variance of $x_t$ is $\frac{\sigma^2}{2a}(1-e^{-2at})$.
I tried to substitute $y_t=f(t)x_t$ into the Ito's Lemma, gives
\begin{equation}
dy=\left(-a y_t + \frac{\partial f(t)}{\partial t}x_t\right) dt + \sigma f(t) dB_t\end{equation}
What should I do next?
Best Answer
You have $$y_t=f(0)x_0+\int_0^t (f'(s)x_s-ay_s)\,ds+\sigma\int_0^t f(s)\,dB_s.$$ Take $f(t)=e^{at}$. In this way, you have $f'(s)x_s-ay_s=0$ and therefore $$e^{at}x_t=x_0+\sigma\int_0^t e^{as}\,dB_s\Rightarrow x_t=e^{-at}\left(x_0+\sigma\int_0^t e^{as}\,dB_s\right).$$ When you have a deterministic function $f\in L^2[0,t]$, one has $\int_0^t f(s)\,dB_s\sim N(0,\Vert f\Vert_{L^2[0,t]}^2)$. Hence, $$ x_t\sim N\left(e^{-at}x_0,\,e^{-2at}\sigma^2\int_0^t e^{2as}\,ds=\frac{1-e^{-2at}}{2a}\sigma^2\right).$$