Brownian motion proof

Question

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 $X(a^2t)$

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

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

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