I am pretty shaky on the theory of Gaussian processes and it has manifested itself through the following problem. Let $N$ be a standard normal random variable, mean zero and variance 1. $B_t$ is a Brownian motion, and $N$ and $B_t$ are independent. Consider the process
$$X_t = tN + (1-t)\int_0^t\frac{1}{1-s}dB_s$$
which has the SDE
$$dX_t=\left( \frac{N-X_t}{1-t}\right)dt + dB_s$$
I need to deduce that this processes is a Gaussian process and also find its expectation and covariance.
I know there are several ways to do this. I feel like I really should use the characteristic function approach, where we want to show for a discretization of time that
$$\mathbb{E}\left[ \exp \left( i \sum_{l=1}^k s_l X_{t_l} \right) \right] = \exp\left( -\frac{1}{2}\sum_{l,j}^k \sigma_{lj} s_l s_j + i \sum_{l=1}^k \mu_l s_l \right)$$
for any $s_j,\ j \in \{1,…,k\}$ and concrete values for $\sigma$ (convariance) and $\mu$ mean. If I could make this representation then I would in one pass
- prove the process is Gaussian
- find the expected value $\mu$
- find the covariance $\sigma_{lj}$.
However, I started writing it out and i am thrown off by having random elements in the equation. I am having a hard time making progress with this equation.
$$\mathbb{E}\left[ \exp \left( i \sum_{l=1}^k s_l X_{t_l} \right) \right] = \mathbb{E}\left[ \exp \left( i \sum_{l=1}^k s_l \left( t_l N + (1-t_l) \int_0^{t_l} \frac{1}{1-s}dB_s \right) \right) \right]$$
I was able to compute the expected value.
$$E[tN + (1-t)\int_0^t\frac{1}{1-s}dB_s]=0$$
This is because the expected value of a normal r.v. is zero and so is the expectation of a stochastic integral, by the martingale representation.
I was also able to compute the variance:
$$E[X_t^2]=E\left[ t^2 N^2 + (1-t)^2 \int_0^t \frac{1}{(1-s)^2}ds + 2tN(1-t)\int_0^t dB_s \right]$$
$$=E\left[ t^2 N^2 + (1-t)^2 \left( \frac{1}{1-t} – 1 \right) \right]$$
$$=t$$
This seems promising. We may be looking at a Brownian motion. However I'm not so certain about the covariance. Without loss of generality assume $s<t$, then
$$E[X_s X_t]=E\left[ st N^2 + (1-s)(1-t) \int_0^s\frac{1}{(1-u)}du + (1-s)(1-t)\int_s^t\frac{1}{(1-u)}dB_u (1-t)\int_0^t \frac{1}{(1-u)}dB_u + tN(1-t)\int_0^t dB_u + sN(1-s)\int_0^s dB_u \right]$$
$$=st + (1-s)(1-t) \left( \frac{1}{1-s} – 1 \right)$$
$$=s$$
Just showing the value of expected value, variance, covariance, is not by itself showing that the process is a Gaussian Process. So I really think I should go with the characteristic function approach, but just wanted to show some of what I did. And in this case I think if I show independent increments then we have all the properties of Brownian Motion, and BM is a Gaussian Process.
Thanks!
Best Answer
The stochastic integral $I:=\int_0^t (1-s)^{-1} dB_s$ is normal, mean $0$ and variance $\int_0^t (1-s)^{-2} ds = t/(1-t)$ (for $0\le t<1$), by the Ito isometry. Because $N$ is independent of $I$, the linear combination $tN+(1-t)I$ is also normal.