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 $X(a^2t)$

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

1 Answer 1


$X_a(t)=\frac{1}{a}X(a^2t)$ 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


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

  • $\begingroup$ Can we say anything about X(a^2*t) ? I have to prove that . Thanks ! $\endgroup$
    – sww
    Commented 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. $\endgroup$ Commented Nov 30, 2017 at 6:44

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