Wiener process - finding the distribution of $W(s) + W(s+t)$

Question

Suppose $W(t)$ is a Wiener process ($W(0)=0$ and for $t \le s, (W(s)-W(t))$ is a normal distribution with mean $0$ and variance $(s-t)$).

For $s,t>0$, I am trying to find the distribution of the random variable $X =W(s)+W(s+t)$.

My initial idea was to write $X$ as $W(s+t)-W(s)+2W(s)$. Since $W(s+t)-W(s)$ has the same distribution as $W(t)$, can I write that $X$ follows the same distribution as $W(t)+2W(s)$?

If so, how do I proceed from this point?

Answer 1

No, $W(s+t)-W(s) \sim W(t)$ does not imply that $X$ has the same law as $W(t)+2W(s)$. To reason this way you would need that $(W(t+s)-W(s),W(s))$ has the same joint law as $(W(t),W(s))$ (which is not true).

Nevertheless, it's a good idea to write $$X = (W(t+s)-W(s))+2W(s).$$ To determine the distribution of $X$ you can use the following general statement on Gaussian random variables.

If $X_1 \sim N(\mu_1,\sigma_1^2)$ and $X_2 \sim N(\mu_2,\sigma_2^2)$ are two independent Gaussian random variables, then $X_1+X_2$ is Gaussian and $X_1+X_2 \sim N(\mu_1+\mu_2,\sigma_1^2+\sigma_2^2)$.

Check that $X_1 := W(t+s)-W(s)$ and $X_2 := 2 W(s)$ satisfy the assumptions of the result and determine their expectation value and variance. Applying the result you get the distribution of $X$.

Answer 2

All linear combinations of $(W_t)$ are normal. $W(s)+W(t+s)=[W(t+s)-W(s)]+2W(s)$ is normal with mean $0$ and variance $t+4s$. Variances of independent random variables add up .