Question

Let $X_1,X_2,\dots$be a sequence of independent and identically distributed continuous random variables. Let $N\ge 2$be such that

$X_1\ge X_2\ge \cdots \ge X_{N-1}<X_N$

That is, $N$is the point at which the sequence stops decreasing. Show that $E\left[N\right]=e$.

Hint: First find $P\{N\ge n\}$.

Short answer

$P\left(X_1\ge X_2\ge \dots \ge X_{N-1}<X_n\right)=1-\frac{1}{(N-1)!}\frac{N-1}{N}$$=\frac{1}{(N-2)!}N$

We have proved that$E\left[N\right]=e$.

Step-by-step solution

Given Information

$X_1,X_2,\dots$be independent and identically distributed continuous random variables.

Let $\mathrm{N}\ge 2$be such that $X_1\ge X_2\ge \dots \ge X_N-1<X_N$

Calculation

$P\left(X_1\ge X_2\ge \dots \ge X_{N-1}<X_N\right)$

$=P\left(X_1\ge X_2\ge \dots \ge X_{N-1}\right)P\left(X_N>X_{N-1}\mid X_1\ge X_2\ge \dots \ge X_{N-1}\right)$

We known $X_1\ge X_2\ge \dots \ge X_{N-1}$because $(N-1)$!

we have $P\left(X_1\ge X_2\ge \dots \ge X_{N-1}\right)=\frac{1}{(N-1)!}$

$P\left(X_N>X_{N-1}\mid X_1\ge X_2\ge \dots \ge X_{N-1}\right)=1-\frac{1}{N}=\frac{N-1}{N}$

$P\left(X_1\ge X_2\ge \dots \ge X_{N-1}<X_n\right)=1-\frac{1}{(N-1)!}\frac{N-1}{N}$

$=\frac{1}{(N-2)!}N$

Final Answer

$E\left[N\right]=\sum _{N=2}^{\infty }N\frac{1}{(N-2)!}N$

$=\sum _{N=2}^{\infty }\frac{1}{(N-2)!}$

$=\sum _{N=0}^{\infty }\frac{1}{N!}$

$=e.$

Therefore,$E\left[N\right]=e$.