Question

Let$U1,U2,...$be a sequence of independent uniform$(0,1)$random variables. In Example $5i$, we showed that for $0\le x\le 1,E\left[N\right(x\left)\right]=e^x$, where

$N\left(x\right)=min\left\{n:\sum _{i=1}^n U_i>x\right\}$

This problem gives another approach to establishing that result.

(a) Show by induction on n that for $0<x\le 1$0 and all $n\ge 0$

$P\left\{N\right(x)\ge n+1\}=\frac{x^n}{n!}$

Hint: First condition on$U1$and then use the induction hypothesis.

use part (a) to conclude that

$E\left[N\right(x\left)\right]=e^x$

Short answer

We concluded that$E\left[N\right(x\left)\right]=e^x$by using part (a) answers.

Step-by-step solution

Step 1:Given Information(part a)

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

Step 2:Explanation(part a)

For $n=0$,

$P\left\{N\right(x)\ge n\}=\frac{x^n}{(n-1)!}$

for $n>0$,

$P\left\{N\right(x)\ge n+1\}={\int }_0^1 P\left\{N\left(x\right)\ge n+1\mid U_1=y\right\}dy$

$={\int }_0^x P\left\{N\right(x-y)\ge n\}dy$

$={\int }_0^x P\left\{N\right(u)\ge n\}du$

$={\int }_0^x \frac{u^{n-1}}{(n-1)!}du$

$=\frac{x^n}{n!}$

Step 3:Final Answer(part a)

$P\left\{N\right(x)\ge n+1\}=\frac{x^n}{n!}$

Step 4:Conclusion

$E\left(N\right(x\left)\right)=\sum _{n=0}^{\mathrm{\infty }} P\left\{N\right(x)>n\}$

$=\sum _{n=0}^{\mathrm{\infty }} P\left\{N\right(x)>n+1\}$

$=\sum _{n=0}^{\mathrm{\infty }} \frac{x^n}{n!}$

$=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\dots$

$=1+\frac{x^1}{1!}+\frac{x^2}{2!}+\dots$

$=e^x$