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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?

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2  
Is $t$ a constant? I'd interpret the notation here so, but @NeilG's comment at emcor's answer considers a random $t$. –  Juho Kokkala Jun 23 at 10:19

2 Answers 2

up vote 5 down vote accepted

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 $0.5$.

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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 $\rho=0$.

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Even though $Z_1$ is independent of $Z_2$, that doesn't mean that $Z_1<t$ is independent of $Z_2>t$. Consider $Z_1$ and $Z_2$ as low-variance normals at -1, 1, and t is -10, 0, or 10 with equal probability. According to your formula, the probability is around 4/9; in fact it should be around 1/3. –  Neil G Jun 23 at 9:28
    
@NeilG Thanks for your comment, indeed I corrected it. –  emcor Jun 23 at 9:56
    
This can't be right since the result should be a probability and $1/\{2\Phi(t)\} > 1$ for $t < 0$. –  QuantIbex Jun 23 at 10:05
    
Yes, I reedited it.. –  emcor Jun 23 at 10:06
2  
Please take a while to ponder and double check your answer. This is about the 4th different expression you provide as the answer. –  QuantIbex Jun 23 at 10:08

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