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.

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