Let $\{ X(t), t \ge 0\}$ be standard Brownian motion.
How do I find Cov$[X(3) - 2X(2), X(4)]$?
The answer is $-1$, but I can't seem to get there no matter what I hit it with. I know that $X(3) \sim N(0,3)$, $2X(2) \sim N(0, 4)$, and $X(4) \sim N(0,4)$. Furthermore, I know that $\text{Cov}[X(s),X(t)] = \min (s,t)$; but this is only for Brownian motion, which isn't preserved (I assume) by the new normal random variable created by $X(3) - 2X(2)$. I imagine that I need to somehow transform $X(3) - 2X(2)$ in a way that preserves Brownian motion so that I can use the above property, but I'm at a loss here.
\begin{align*} \text{cov}\left[x\left(3\right)-2x\left(2\right),x\left(4\right)\right] & =\text{cov}\left[x\left(3\right),x\left(4\right)\right]-2\text{cov}\left[x\left(2\right),x\left(4\right)\right]\\ & =3-2\cdot2=-1 \end{align*}