# What is the distribution of $B(s) + B(t)$ , $s \leq t$? Answer already given. Just curious of a process within the answer.

This is question is under the topic of Brownian motion.

The question is:

What is the distribution of $X(s) + X(t)$, when $s \leq t?$

$$X(s) + X(t) = 2X(s) + X(t) − X(s)$$

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.

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In general for some constant $a$, $$\mathrm{Var}(aX) = E((aX)^2)-(E(aX))^2$$ $$=a^2E(X^2)-aE(X)aE(X)$$ $$= a^2(E(X^2)-(E(X))^2)$$ $$= a^2\mathrm{Var}(X)$$
We know that a Brownian Motion Process $X(t) \sim N(0,t)$, so $\mathrm{Var}(aX(t))=a^2t$.