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Chapter 7: Q.7.62 (page 357)

LetU1,U2,...be a sequence of independent uniform(0,1)random variables. In Example 5i, we showed that for 0≤x≤1,E[N(x)]=ex, where

N(x)=minn:∑i=1n Ui>x

This problem gives another approach to establishing that result.

(a) Show by induction on n that for 0<x≤10 and all n≥0

P{N(x)≥n+1}=xnn!

Hint: First condition onU1and then use the induction hypothesis.

use part (a) to conclude that

E[N(x)]=ex

Short Answer

Expert verified

We concluded thatE[N(x)]=exby using part (a) answers.

Step by step solution

01

Step 1:Given Information(part a)

Given that U1,U2,... be a sequence of independent uniform(0,1) random variables.

02

Step 2:Explanation(part a)

For n=0,

P{N(x)≥n}=xn(n−1)!

for n>0,

P{N(x)≥n+1}=∫01 PN(x)≥n+1∣U1=ydy

=∫0x P{N(x−y)≥n}dy

=∫0x P{N(u)≥n}du

=∫0x un−1(n−1)!du

=xnn!

03

Step 3:Final Answer(part a)

P{N(x)≥n+1}=xnn!

04

Step 4:Conclusion

E(N(x))=∑n=0∞ P{N(x)>n}

=∑n=0∞ P{N(x)>n+1}

=∑n=0∞ xnn!

=x00!+x11!+x22!+…

=1+x11!+x22!+…

=ex

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