This is question is under the topic of Brownian motion.
The question is:
What is the distribution of $X(s) + X(t)$, when $s \leq t?$
Answer:
$X(s) + X(t) = 2X(s) + X(t) − X(s)$
Now $2X(s)$ is normal with mean $0$ and variance $4s$ and $X(t) − X(s)$ is normal with mean $0$ and variance $t − s$. As $X(s)$ and $X(t) − X(s)$ are independent, it follows that $X(s) + X(t)$ is normal with mean $0$ and variance $4s+t−s = 3s+t$.
My question is:
As I have bolded above, the answer shows that the variance of $2X(s)$ is $4s$. Why is it $4s$?
Thanks for all the help.