I want to show $e^{-\alpha t}B(e^{2\alpha t})$ is a Gaussian process.

Question

Let $B(t)$ be Brownian motion. Show that $e^{-\alpha t}B(e^{2\alpha t})$ is a Gaussian process. Find its mean and covariance functions.

thanks .

Answer 1

We have to prove that if $(t_1,\dots,t_d)$ are non-negative numbers and $(a_1,\dots,a_d)$ real numbers, then $\sum_{j=1}^da_jX_{t_j}$, with $X_{t}=e^{-\alpha t}B(e^{2\alpha t})$, is Gaussian.

For $d=2$, we write $$a_1X_{t_1}+a_2X_{t_2}=a_1e^{-\alpha t_1}B(e^{\alpha t_1})+a_2e^{-\alpha t_2}\left(B(e^{2\alpha t_2})-B(e^{2\alpha t_1})\right)+a_2e^{-\alpha t_1}B(e^{2\alpha t_1}).$$ We then rearrange the terms and use the independence of increments of Brownian motion. From this computation, we can get a formula for the characteristic function of $(X_{t_1},X_{t_2})$.