How does Variance become an Autocorrelation Function? "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? 
 A: 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
$$E[Y^2(t)]=2(R(0)-R(0.5))=2(1-0.5)=1$$
A: 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 $0$.
