Crosscorrelation of stochastic process Let 
$Z_1,Z_2 $ i.i.d. standard normal 
$$
X(t) =
\begin{cases}
0, & \text{if } t<Z_1, t<Z_2\\
1, & \text{if } Z_1\le t <Z_2 \text{ or } Z_2\le t <Z_1 \\
2, & \text{if } t\ge Z_1, t\ge Z_2
\end{cases}
$$
Asked
$E[X(t)X(t')]$
Approach
$E[X(t)X(t')] = P(X(t)X(t')=1)+2 P(X(t)X(t')=2)+ 4 P(X(t)X(t')=4)$
I work out $P(X(t)X(t')=1)$ as follows:
$P(X(t)X(t')=1)=2 P(Z_1<t<Z_2,Z_1<t'<Z_2)$  (x2 because switching Z1,Z2 gives same probability)
If $t'>t$
$P(Z_1<t<Z_2,Z_1<t'<Z_2)= 2*P(Z_1<t'<Z_2 |Z_1<t<Z_2)P(Z_1<t<Z_2)$
$=2P(t'<Z_2)P(Z_1<t<Z_2)$
$=2(1-\Phi(t')) * \Phi(t)(1-\Phi(t))$
This can be generalized to
$P(X(t)X(t')=1)=2 (1-\Phi(\max(t,t'))\Phi(\min(t,t'))(1-\Phi(\min(t,t'))$
However, according to my solutions this is wrong.
Correct answer
$P(X(t)X(t')=1)=2 (1-\Phi(\max(t,t'))\Phi(\min(t,t'))$
My Question
What am I doing wrong?
 A: I would suggest drawing diagrams.
In Cartesian coordinates representing $(Z_1, Z_2)$ (with axes oriented to have values increasing to the right for $Z_1$ and up for $Z_2$), a value $t$ determines four regions according to how $Z_1$ and $Z_2$ compare to $t$.  $X(t)$ attains a constant value in each region, which we might sketch as 
1 2
0 1

A similar diagram holds for $t^\prime$.  Overlaying the two diagrams creates nine separate regions with the following values for $X(\min(t,t^\prime))$ and $X(\max(t,t^\prime))$:
(1,1) (2,1) (2,2)
(1,0) (2,0) (2,1)
(0,0) (1,0) (1,1)

For instance, the $(2,1)$ entry along the first row occurs in the region where $\min(t,t^\prime) \le Z_1 \lt \max(t,t^\prime)$ and $\max(t,t^\prime)\le Z_2$.  In this case both $Z_1 \ge \min(t,t^\prime)$ and $Z_2 \ge \min(t,t^\prime)$, justifying the value of $2$, and $Z_1 \lt \max(t,t^\prime) \le Z_2$, justifying the value $1$.
Multiplying the components of each ordered pair show the product $X(t)X(t^\prime) = X(\min(t,t^\prime)) X(\max(t,t^\prime))$ attains the values
1 2 4
0 0 2
0 0 1

We can write this random variable in terms of sums and differences of simple random variables (that is, those which attain only one nonzero value):
1 2 4     2 2 2     0 0 2     1 0 0     0 0 0
0 0 2  =  0 0 0  +  0 0 2  -  0 0 0  -  0 0 0
0 0 1     0 0 0     0 0 2     0 0 0     0 0 1

To compute expectations the constant values appearing in each term have to be multiplied by the probabilities of the cells where they appear.  But the first two (which correspond to the events $Z_2\ge\max(t,t^\prime)$ and $Z_1\ge\max(t,t^\prime)$, respectively) are each easily seen to have probabilities $1 - \Phi(\max(t,t^\prime))$ while the independence of the $Z_i$ gives each of the last two terms the probability $\left(1 - \Phi(\max(t,t^\prime))\right)\Phi(\min(t,t^\prime)).$  The answer is immediate.
A: One mistake is I think that $P(Z_1<t'<Z_2|Z_1<t<Z_2)$ does not simplify to $\neq P(t'<Z_2)$ for $t'>t$ because the lower bound would be ignored.
I think its
$P(Z_1<t'<Z_2,Z_1<t<Z_2)=P(t'<Z_2)\cdot P(t>Z_1)$
because of independence and  $t'>t$ and then the solutions should be identical.
