Question

Suppose X1, X2,..., X10 are independent normal random variables with mean O and variance 1. Let max{X1, X2, ..., X10} What is the largest value of t so that P(M <t) < 0.90? That is, find the 90th percentile of M.

Answer 1

We are to find t here such that:

P( M <= t) <= 0.9

The CDF for M is obtained here as:

P(M <= m) = Probability that the maximum value is less than or equal to m

P(M <= m) = 1 - P( M > m)

P(M <= m) = 1 - (1 - Probability that all the X_i are less than m)

We are given here for each i = 1, 2, 3, .. 10 that:

$X_i \sim N(0,1)$

Therefore, the CDF for M here is obtained as:

$P(M \leq m) = 1 - (1 - \Phi(m)*\Phi(m)*\Phi(m)..... \ 10 \ times )$

$P(M \leq m) = 1 - (1 - [\Phi(m)]^{10} )$

$P(M \leq m) = [\Phi(m)]^{10}$

Therefore, we are to find t here such that:

$[\Phi(t)]^{10} \leq 0.9$

$[\Phi(t)] \leq \sqrt[10]{0.9}$

$[\Phi(t)] \leq 0.9895$

From standard normal tables, we have here:
P(Z < 2.309) = 0.9895

Therefore t = 2.309 is the required 90th percentile value of M here.