I'm considering the Ornstein – Uhlenbeck process
$$
V_t = e^{\lambda t} v_o + \int_0^t e^{-\lambda (t-s)} dB_s
$$
with $ \lambda > 0$, $v_0 \in \mathbb{R}$, and $B$ a brownian motion.
I want to calculate the expected value for each $V_t$ and the covariance $\mbox{Cov}(V_t, V_s)$.
For the expected value I calculate
$$
E(V_t) = E \left(e^{\lambda t} v_o + \int_0^t e^{-\lambda (t-s)} dB_s\right)
= e^{\lambda t} v_o + e^{-\lambda t} E \left( \int_0^t e^{\lambda s} dB_s \right)
$$
Then the stochastic integral involved is a local martingale because $B$ is a local martingale. I wanted it to be a martingale so I could use the property that expected values for martingales stay constant. Thus, I calculates the variation
$$
\left[ \int_0^{\cdot} e^{\lambda s} dB_s \right]_t
$$
And because its expected value was finite for each $t$ I had that the stochastic integral was really a martingale and then I could calculate its expected value as 0.
First Question:
Is there a more simple way to doing this?
Now for the variance I tried to do something similar.
I needed to calculate
$$
E \left( \left(\int_0^t e^{\lambda s} dB_s \right)\left(\int_0^s e^{\lambda s} dB_s \right)\right)
$$
Using the fact that if $M, N$ are martingales bounded in $L^2$, then $MN – [M,N]$ is a martingale, I wanted to calculate the value before by reducing my calculations to $[M,N]$ which is easier to handle. My problem is that the stochastic integrals involved don't seem to be martingales bounded in $L^2$ or I'm not able to prove it.
Second Question:
Is there any way to prove the involved stochastic integrals are martingales?
Third Question:
Is there an easier way of doing this?
I first tried using Itô Isometry but for what I know the Isometry is only true for martingales bounded in $L^2$ which is not the case for the brownian motion
$$
What is the most general context in which I can use Itô Isometry?
Thanks in advance…
$$
Best Answer
Every stochastic integral involved is a true martingale bounded in $L^2$, when restricted to some bounded interval $[0,T]$. And this is all one needs to deal with each $(t,s)$, since one can choose some finite $T\geqslant\max(t,s)$ and work in $L^2([0,T])$.
No, this is the way to go.
NARQ.