Probability of x between two random variables Given are $Z_1, Z_2$ i.i.d. standard normal.
Find
$P[Z_1 < t < Z_2]$
I have difficulties with working out how I should split the condition.
Is $P[Z_1 < t < Z_2] = P[Z_1 < t, t < Z_2]$ or is some additional condition required?
My question
How should I approach this problem?
 A: The question seems to be a self-study question, but since there is already an answer that attempts to fully answer the question, I provide a full answer assuming that $t$ is a constant.
Indeed, $P(Z_1 < t < Z_2) = P(Z_1 < t, t < Z_2)$. By the independence of $Z_1$ and $Z_2$ the joint probability is equal to the product of the marginal probabilities. Thus,
$$
P(Z_1 < t < Z_2) = P(Z_1 < t, t < Z_2) = P(Z_1 < t) P(t < Z_2) = \Phi(t) \{1 - \Phi(t)\} ,
$$
where $\Phi(\cdot)$ denotes the distribution function of the standard normal distribution.
If $t = 0$, then $P(Z_1 < t < Z_2) = 0.25$ as one would expect since events $\{Z_1 < t\}$ and $\{Z_2 > t\}$ are independent and both have probability $0.5$.
A: As the variables are independent and standardnormal symmetric*, their joint probability equals the product of the marginals:
$P(Z_1<t<Z_2)=P(Z_1<t,Z_2>t)=\Phi_{\rho=0}(t,-t)=\Phi(t)\cdot\Phi(-t)=\Phi(t)((1-\Phi(t))$
*By the normal symmetry we have $\{Z>t\}=\{Z<-t\}$, and independence implies $\rho=0$.
