I am new to Brownian motion, and I have problem with verifying a Brownian motion. Let $B(t)$ be the standard Brownian motion. Is true that $W(t)=B(5t)-B(t)$ is also a Brownian motion (not necessarily standard). Thank you for any comment/hint.
My attempt: We know $W(0)=0$ holds. I tried to check the independent increments property: for any $s<t<r$, $W(t)-W(s)=B(5t)-B(t) - (B(5s)-B(s)) = B(5t)-B(5s)- (B(t)-B(s))$, and also $W(r)-W(t)=B(5r)-B(r)-(B(5t)-B(t))$. I am not sure how to proceed to prove/disprove the independence of $W(t)-W(s)$ and $W(r)-W(t)$.
What is a standard Brownian Motion? One of the properties is that Var(B(t)) = t.
This process $W(t) = B(5t)-B(t)$ does not satisfy this property, so it cannot be a standard Brownian Motion.
However, a general Brownian Motion has stationary increments, but the process $W(t)$ is not even stationary: $$\begin{aligned} W(t+h)-W(t) &= B(5t+5h)-B(t+h)-(B(5t)-B(t)) = \\ &= [B(5t+5h)-B(5t)] - [B(t+h)-B(t)] . \end{aligned} $$
For $h$ small enough, $5t > t+h$, then the random variables $B(5t+5h)-B(5t)$ and $B(t+h)-B(t)$ are independent since $B$ has independent increments. Furthermore, by definition of $B$, $$ B(5t+5h)-B(5t) \sim \mathcal{N}(0,5h) $$ $$ B(t+h)-B(t) \sim \mathcal{N}(0,h) .$$
If we sum (resp. subtract) two independent gaussian random variables, we obtain another gaussian random variable with mean equals to the sum (resp. the difference) of the means and variance equals to the sum of the variances. So $$ W(t+h)-W(t) = [B(5t+5h)-B(5t)] - [B(t+h)-B(t)] \sim \mathcal{N}(0,6h) ,$$ while $$ W(h) = B(5h) - B(h) \sim \mathcal{N}(0,4h). $$