# Showing that two Brownian Motions are equal in distribution

I must show that $\{B(ct), t\geq 0\}$ is equal in distribution to $\{c^{1/2}B(t), t\geq 0\}$ where $B(t)$ is a Brownian Motion and $c$ is some constant.

So, I'll be honest. I'm at a loss. I've tried taking the Moment Generating Function, but it seems to be getting me nowhere because I might be doing it incorrectly. That is, if I can show that the joint moment generating function of $\{a_1, ..., a_n\}$ drawn from $\{c^{1/2}B(t), t\geq 0\}$ is the same as the joint moment generating function of $\{b_1, ..., b_n\}$ drawn from $\{B(ct), t\geq 0\}$, then the two are equal in distribution. Any hints or partial solutions would be extremely helpful. Brownian Motion seems to be flying far over my head.

## 1 Answer

This isn't the most rigorous approach, but we could think that from definition, $$B_t \sim N(0,t)$$ and further that $$B_{ct}\sim N(0,ct).$$ Then noticing that $c^{1/2}B_t$ is a scalar multiple of $B_t$, we check to see that $$E[c^{1/2}B_t]=c^{1/2}\cdot E[B_t]=c^{1/2}\cdot 0= 0$$ and also $$Var[c^{1/2}B_t]=c\cdot Var[B_t]=ct$$ which implies that $$c^{1/2}B_t\sim N(0,ct).$$

• You proved that the two processes have the same one-dimensional marginal distributions, yet we need the result for $n$-dimensional margins and arbitrary $n \geq 1$. Hint. use the independent increments property. – Yves Apr 17 '18 at 13:01