I want to show $e^{-\alpha t}B(e^{2\alpha t})$ is a Gaussian process and I find mean and covariance functions 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 .
 A: Maybe I'm missing something, but this question seems to be easier than it, at first, appears to be. It's not a change of variable problem (which would be messy) but simply a change in labeling with a change of scale. $t$ is an index, not a random variable.
A process is Gaussian if every random subset of variables from the index set has a multivariate Gaussian distribution. Take a collection of variables with indeces $t_1,t_2, \ldots, t_k$. They are jointly Brownian (modulo the scale change), and so jointly multivariate normal.
Every random variable from a Brownian process has mean 0, so the process defined here has mean 0 everywhere.
Now, to the covariance function. Let's look at the variance first. Call the new process $X(t)$.
The variance of the Brownian at $t$ is $t$. The variance of $X(t)$ is $e^{-2\alpha t} e^{2 \alpha t}$, which is 1. That's the point of the scaling factor, clearly.
The covariance of the Brownian at $t$ and $s$ is $\text{min}(s,t)$, so the covariance of $X(t)$ and $X(s)$ will be 
$e^{-\alpha(s+t)} \text{min}(e^{2\alpha t}, e^{2 \alpha s})$.
So if $s < t$, the covariance will be $e^{- \alpha (t-s)}$. Which is kinda cool.
A: 
The proof is very similar to how you prove a bm is a gaussian process.
What i did not say at the end of proof is that, notice, by the stationary independent increment property of a BM, the increments in the brackets i lablelled normal are INDEPENDENT normal distributions, so linear combinations of independent normal distributions are also normal, which proves the claim that $\sum a_i X_{t_i}$ is Gaussian for any choice of $a_i$ and $t_i$
