1
$\begingroup$

If $X(t)$ is a Brownian motion, how can we prove $X(a^2t)$ is also Brownian?

If $X(t)$ is brownian it is $N(0,\sigma^2*t)$ . But I am not able to see how I can use this for $X(a^2t)$

$\endgroup$
1
  • $\begingroup$ Look at any axiomatic characterization of Brownian motion: which aspects of it are you unable to demonstrate? $\endgroup$
    – whuber
    Nov 30, 2017 at 3:56

1 Answer 1

2
$\begingroup$

$X_a(t)=\frac{1}{a}X(a^2t)$ is a Standard Brownian Motion ($X(a^2t)$)is a general Brownian motion with mean 0). Any Gaussian stochastic process is completely specified by its expectation and covariance function. It is enough to prove that $X_a(t)$ has correct mean and covariance.

It is evident that $\frac{1}{a}X(a^2t)$ has zero mean (as expected in case of a Brownian motion).

For $s<t$;

$\mathrm{Cov}(X_a(t),X_a(s))=\frac{1}{a^2}\mathrm{Cov}(X(a^2t),X_a(a^2s))=\frac{1}{a^2}\mathrm{Min}(a^2t,a^2s)=\frac{1}{a^2}(a^2s)=s$ (as expected in case of a Brownian motion)

You can try looking-up scaling property of Brownian motion for additional reading.

$X(a^2t)$ also has a zero mean

$\mathrm{Cov}(X(a^2t),X(a^2s))=\mathrm{Cov}(X(a^2t),X_a(a^2s))=\mathrm{Min}(a^2t,a^2s)=(a^2s)$

$i.e \; X(a^2t) \sim N(0,a^2t)$, which is a general brownian motion

$\endgroup$
2
  • $\begingroup$ Can we say anything about X(a^2*t) ? I have to prove that . Thanks ! $\endgroup$
    – sww
    Nov 30, 2017 at 5:12
  • $\begingroup$ A Standard Brownian motion, by definition, has Gaussian increments ($Xt−Xs$, t>s) with mean 0 and variance $(t−s)$. $X(a^2t)$ has Gaussian increments with variance $a^2(t−s)$, It is a general Brownian with mean 0. I am editing the above answer to reflect this. $\endgroup$ Nov 30, 2017 at 6:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.