Calculating probability that a linear combination of stochastic variable is greater than $2$ A gaussian process $X$ has expectation zero and covariance function 
$$r_X(\tau) = \frac{2 + \tau^2}{1+\tau^2}.$$
I'm asked to determine the probability 
$$P\left( X(2) > \frac{X(1) + X(3)}{2} + 1\right) \iff *$$
My attempt is as follows: 
$$P \left(2X(2) - X(1) - X(3) > 2  \right).$$
Let $Z = 2X(2) - X(1) - X(3).$ Z is also gaussian, therefore
$$P(Z > 2) = 1 - P(Z \leq 2) = 1-\Phi \left(\frac{E[Z] - 2}{\sqrt{V[Z]}} \right)$$
I calculate the variance to $2.4$ and $E[Z]$ is obviously $0$.
Variance calculation
However the answer is supposed to be 
$$\Phi \left(-\frac{1}{\sqrt{0.9}} \right).$$
What is wrong? 
 A: Your calculations are correct, and here is an independent verification of that.
Let $Z=\sum_i a_iX_i$ with the sum ranging over some finite set ($\{1,2,3\}$ in the problem). Then
$$
\mathbb E[Z^2]=\sum_{i,j}a_ia_j \mathbb E[X_iX_j]=\sum_{i,j}a_ia_j \frac{2+|i-j|^2}{1+|i-j|^2}=\Bigl(\sum_i a_i\Bigr)^2+\sum_{i,j} \frac{a_ia_j}{1+|i-j|^2}.
$$
When $\sum_i a_i=0$ (as is the case in this problem) we only need to focus on the second term,
$$
\sum_{i,j} \frac{a_ia_j}{1+|i-j|^2}=\sum_i a_i^2+\sum_{n=1}^{i^+-i^-}\frac{\sum_i a_i(a_{i+n}+a_{i-n})}{1+n^2},
$$
where $i^{\pm}$ denote the max and min of the index set and the inner sum includes only those terms for which $i\pm n$ belong to the index set.
In this case, $i^+=3$ and $i^-=1$ and $(a_1,a_2,a_3)=(-1,2,-1)$.
Evaluating the sum yields
$$
\mathbb E[Z^2]=(2^2+1^2+1^2) + 4\cdot \frac{-1\cdot 2}{2}+\frac{1^2+1^2}{5}=2+\frac{2}{5}=2.4,
$$
matching your calculation. Thus, if $W$ is a random variable satisfying $12W^2=5Z^2$, then $\mathbb E[W^2]=1$ and $W$ is a standard normal random variable. Thus
$$
\mathbb P(Z>2)=\mathbb P\Bigl(W>\sqrt{\frac53}\Bigr)=\mathbb P\Bigl(W<-\sqrt{\frac53}\Bigr)=\Phi\Bigl(-\sqrt{\frac53}\Bigr).
$$
