What is $var[X(0)+X(0.5)]$, where $X(t)$ is a stationary process?

Question

So I have that $\{X(t);t ∈ R\}$, with mean $\mu_X = 0.5$ is stationary Gaussian stochastic process and its autocorrelation function is $R_X(\tau)=0.5e^{-\pi\tau^2/4} $ $\forall \tau \in R. \quad$ What I want to know is $var[X(0)+X(0.5)],$ i.e. the variance. The answer to the question is :
$V[X(0)+X(0.5)] = 2R_X(0)+2R_X(0.5)-4\mu_X^2.$
Now I know that $V[X(0)+X(0.5)] = V[X(0)]+V[X(0.5)]+2cov[X(0),X(0.5)]$ but I dont know how they got the second part. Can somebody explain the whole procedure of they got to that conclusion?

Answer 1

We will write $R$ and $\mu$ instead of $R_x$ and $\mu_X$.

We have $$K(s,t) = cov(X(s), X(t)) = EX(s) X(t) - EX(s) EX(t) = R(t-s) - \mu^2,$$ because $EX(t) = \mu$ doesn't depend on $t$.

Further $$V(X(0) + X(\frac12)) = cov(X(0) + X(\frac12), X(0) + X(\frac12)) = $$ $$=cov(X(0), X(0)) +cov(X(0), X(\frac12)) + cov(X(\frac12), X(0)) + cov(X(\frac12), X(\frac12)) =$$ $$ =cov(X(0), X(0)) + 2 cov(X(0), X(\frac12)) + cov(X(\frac12), X(\frac12)) $$ $$=K(0,0) + 2K(0, \frac12) + K(\frac12,\frac12)=$$ $$ =(R(0-0) - \mu^2) + 2(R(\frac12-0) - \mu^2) + (R(\frac12-\frac12) - \mu^2)$$ $$ = 2R(0) + 2R(\frac12) - 4 \mu^2.$$