How does Variance become an Autocorrelation Function?

Question

"For a Gaussian stochastic process $X=\{X(t)|-\infty<t<\infty\}$ with mean function $\mu(t)=0$ for all $t$, its autocorrelation function is $$ E(X(t)\cdot X(s))=R(h)=\max(0,1-|h|), h=t-s. $$ Compute the characteristic function of $$ Y=X(t)-X(t-0.5)." $$ Since I know it is Normal distributed (Gaussian), I only have to find the mean and variance of $Y$ and then find the corresponding characteristic function. I have found the mean by $$ E(Y)=E(X(t)-X(t-0.5))=E(X(t))-E(X(t-0.5))=\mu(t)-\mu(t-0.5)=0-0=0. $$ However I am stuck on computing the variance, and only get that: $$ Var(Y)=Var(X(t)-X(t-0.5))=Var(X(t))+Var(X(t-0.5))-2Cov(X(t),X(t-0.5)). $$ I know that $$ Var(X(t))=E(X(t)^2)-E(X(t))^2=E[X(t)\cdot X(t)]-\mu(t)^2=E[X(t)\cdot X(t)]-0^2 $$ But the answer states that $$ Var(Y)=R(0)+R(0)-2R(0.5)=1+1-2\cdot 0.5=1 $$ How do I get there and does this imply that I am supposed to see that $t=s$ (so that $E(X(t)^2)=R(h)$)? Any help is much appreciated.

Edit: Also, how do I see that $h=t-s=0$ and $h=0.5$ in $R(0)$ and $R(0.5)$ respectively?

Answer 1

Note that due to $\mu(t)=0$, the variance of $Y(t)=X(t)-X(t-0.5)$ is

$$E[Y^2(t)]=E[(X(t)-X(t-0.5))^2]=\\E[X^2(t)]+E[X^2(t-0.5)]-2E[X(t)X(t-0.5)]\tag{1}$$

From the definition of the autocorrelation function, the variance of $X(t)$ equals

$$E[X^2(t)]=R(0)$$

Since the autocorrelation function only depends on the difference $h=t-s$, and not on absolute time, the variance of $X(t-0.5)$ also equals $R(0)$. Finally, the third term in $(1)$ can also be expressed in terms of $R(h)$:

$$E[X(t)X(t-0.5)]=R(0.5)$$

which results in

$$E[Y^2(t)]=2(R(0)-R(0.5))=2(1-0.5)=1$$

Answer 2

Actually from the very beginning of the definition, you can already compute $E[X(t)^2]$ by putting $s = t$, and you immediately see that $$Var[X(t)] = E[X(t)^2] = R(0) = 1$$ and that's why he can call this as the auto-correlation function - because by definition this is just the auto-covariance, and they are equal when the process has a constant variance of $1$. Of course an important fact should not be missed is that $$ Var[X(t)] = Cov[X(t), X(t)]$$ so we are used to compute the variance by evaluating the auto-covariance function at $0$.