1
$\begingroup$

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?

$\endgroup$
1
  • 2
    $\begingroup$ Is $t$ a constant? I'd interpret the notation here so, but @NeilG's comment at emcor's answer considers a random $t$. $\endgroup$ – Juho Kokkala Jun 23 '14 at 10:19
5
$\begingroup$

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 $0.5$.

$\endgroup$
2
$\begingroup$

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 $\rho=0$.

$\endgroup$
11
  • $\begingroup$ 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. $\endgroup$ – Neil G Jun 23 '14 at 9:28
  • $\begingroup$ This can't be right since the result should be a probability and $1/\{2\Phi(t)\} > 1$ for $t < 0$. $\endgroup$ – QuantIbex Jun 23 '14 at 10:05
  • $\begingroup$ Yes, I reedited it.. $\endgroup$ – emcor Jun 23 '14 at 10:06
  • 2
    $\begingroup$ Please take a while to ponder and double check your answer. This is about the 4th different expression you provide as the answer. $\endgroup$ – QuantIbex Jun 23 '14 at 10:08
  • $\begingroup$ This can't be right either since $P(Z_1 < t < Z_2) \neq P(Z_1 < t \mid Z_2 > t)$. $\endgroup$ – QuantIbex Jun 23 '14 at 10:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.