Why $P\{\frac{|W(1/n^4)|}{1/n^4}>n\}=P\{\frac{|V(1/n^4)|}{1/n^4}>n\}$ for a Wiener process $W$.

Question

I stuck on an explanation to a problem in the book "Basic Stochastic Processes" by Z. Brzeniak and T. Zastawniak. Consider a Wiener process $W(t)$ and a sequence of Wiener processes $(V_n(t))$ where $V_n(t)=\frac{1}{n^2}W(n^4t)$ for each natural $n$. The explanation casually states the following equality for any $n$ $$ P\left\{\frac{|W(1/n^4)|}{1/n^4}>n\right\} =P\left\{\frac{|V_n(1/n^4)|}{1/n^4}>n\right\} $$ without any justification. I simply can't see how this equality follows trivially from the given info. I could see how we could maybe make a $\geq$ inequality, which would actually suffice for the proof. So why does this equality hold?

Disclaimer: I'm attempting a course on stochastic calculus with not a single statistics course behind my name (I did physics instead in my junior years) , and so I am not well versed at all in statistics and probability. There's a high probability (haha) that I may be missing some standard property of probability, but even so I'd appreciate knowing what it is.

Answer 1

For each $n$ and for each $t$, $V_n(t)$ has the same distribution as $W(t)$: a normal distribution with mean zero and a variance equal to $\sqrt t$.