Brownian Motion Distribution

Question

Let $(W(t))_{t\geq 0}$ be a Brownian motion and $0<s<t.$ What is the distribution of $2W(s)-W(t)$?

If $s>t$, then $2W(s)-W(t) \sim W(4s)-W(t) \sim N(0, 4s-t)$, but here we have $t>s$.

Does this calculations still hold?

Answer 1

The Brownian motion can be realized as the (continuos realization) of a Gaussian process $W_t$ such that $$ \mathbb{E}[W_t]=0 $$ $$ \operatorname{Cov}(W_t,W_s)= \min (s,t) $$

The variance of the difference is $$ \operatorname{Var}(2W(s)-W(t))= 4\mathbb{E}[W(s)^2]+\mathbb{E}[W(t)^2] -4\operatorname{Cov}(W(s),W(t))=4s+t-4 \min(s,t) $$

So we have $$ 2W(s)-W(t) \sim \mathcal{N}(0,4s+t-4 \min(s,t)) $$ or, if you prefer $$ 2W(s)-W(t) \sim W(4s+t-4 \min(s,t)) $$

Notice how, if $t>s$ you have $$ 2W(s)-W_t \sim \mathcal{N}(0,t) $$

while if $t<s$ you have $$ 2W(s)-W(t) \sim \mathcal{N}(0,4s-3t) $$

consistently with the other answers