Let X1,X2,… be a sequence of independent and identically distributed continuous random variables. Let N≥2 be such thatX1≥X2≥⋯≥XN-1<XNThat is, N is the point at which the sequence stops decreasing. Show that E[N]=e.Hint: First find P{N≥n}.

PX1≥X2≥…≥XN-1<Xn=1-1(N-1)!N-1N =1(N-2)!NWe have proved that E[N]=e.

Q. 7.25 - A First Course in Probability

Let $X_{1}, X_{2}, \dots$ be a sequence of independent and identically distributed continuous random variables. Let $N \geq 2$ be such that

$X_{1} \geq X_{2} \geq \dots \geq X_{N - 1} < X_{N}$

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

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

Verified

Answer

$P (X_{1} \geq X_{2} \geq \dots \geq X_{N - 1} < X_{n}) = 1 - \frac{1}{(N - 1)!} \frac{N - 1}{N}$ $= \frac{1}{(N - 2)!} N$

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

1Step 1: Given Information

$X_{1}, X_{2}, \dots$ be independent and identically distributed continuous random variables.

Let $N \geq 2$ be such that $X_{1} \geq X_{2} \geq \dots \geq X_{N} - 1 < X_{N}$

2Step 2: Calculation

$P (X_{1} \geq X_{2} \geq \dots \geq X_{N - 1} < X_{N})$

$= P (X_{1} \geq X_{2} \geq \dots \geq X_{N - 1}) P (X_{N} > X_{N - 1} ∣ X_{1} \geq X_{2} \geq \dots \geq X_{N - 1})$

We known $X_{1} \geq X_{2} \geq \dots \geq X_{N - 1}$ because $(N - 1)$ !

we have $P (X_{1} \geq X_{2} \geq \dots \geq X_{N - 1}) = \frac{1}{(N - 1)!}$

$P (X_{N} > X_{N - 1} ∣ X_{1} \geq X_{2} \geq \dots \geq X_{N - 1}) = 1 - \frac{1}{N} = \frac{N - 1}{N}$

$P (X_{1} \geq X_{2} \geq \dots \geq X_{N - 1} < X_{n}) = 1 - \frac{1}{(N - 1)!} \frac{N - 1}{N}$

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

3Step 3: Final Answer

$E [N] = \sum_{N = 2}^{\infty} N \frac{1}{(N - 2)!} N$

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

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

$= e .$

Therefore, $E [N] = e$ .