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 .

  • 6
    $\begingroup$ What do you know about $B(t)$ and Gaussian processes that might be relevant (e.g., what definitions are you using)? Have you developed any intuition for what this procedure is doing (stretching time by $e^{2\alpha t}$ and rescaling the values by $e^{-\alpha t}$)? If so, what approach would that suggest? (Could you create a simulation of this new process or draw a picture of it?) $\endgroup$
    – whuber
    Nov 29 '13 at 20:22

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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 $t_i$

  • $\begingroup$ What is the covariance function for this process? $\endgroup$
    – whuber
    Jan 3 '14 at 23:19
  • 1
    $\begingroup$ @whuber i didn't bother doing this as the other answer already answered this, i have a tendency of no posting something already been said by someone else. In this particular case, admittedly, i also forgot. I am using an ipad so i want to minimise latex writing. The other answer answers this correctly. $\endgroup$
    – Lost1
    Jan 3 '14 at 23:26
  • $\begingroup$ I approve of avoiding duplication ;-). Some things to consider in future posts would be (a) acknowledging the things you find to be well-stated in other answers and (b) pointing out which aspects of your answer you think are unique or better. That would help readers understand what has motivated your contribution and help focus their attention on the important parts. $\endgroup$
    – whuber
    Jan 3 '14 at 23:40

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.


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