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Chapter 7: Q. 7.25 (page 354)

Let X1,X2,…be a sequence of independent and identically distributed continuous random variables. Let N≥2be such that

X1≥X2≥⋯≥XN-1<XN

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

Hint: First find P{N≥n}.

Short Answer

Expert verified

PX1≥X2≥…≥XN-1<Xn=1-1(N-1)!N-1N=1(N-2)!N

We have proved thatE[N]=e.

Step by step solution

01

Given Information

X1,X2,…be independent and identically distributed continuous random variables.

Let N≥2be such that X1≥X2≥…≥XN-1<XN

02

Calculation

PX1≥X2≥…≥XN-1<XN

=PX1≥X2≥…≥XN-1PXN>XN-1∣X1≥X2≥…≥XN-1

We known X1≥X2≥…≥XN-1because (N-1)!

we have PX1≥X2≥…≥XN-1=1(N-1)!

PXN>XN-1∣X1≥X2≥…≥XN-1=1-1N=N-1N

PX1≥X2≥…≥XN-1<Xn=1-1(N-1)!N-1N

=1(N-2)!N

03

Final Answer

E[N]=∑N=2∞N1(N-2)!N

=∑N=2∞1(N-2)!

=∑N=0∞1N!

=e.

Therefore,E[N]=e.

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Most popular questions from this chapter

Cards from an ordinary deck of 52playing cards are turned face upon at a time. If the 1st card is an ace, or the 2nd a deuce, or the 3rd a three, or ...,or the 13th a king,or the 14an ace, and so on, we say that a match occurs. Note that we do not require that the (13n + 1) card be any particular ace for a match to occur but only that it be an ace. Compute the expected number of matches that occur.

Suppose that X1and X2 are independent random variables having a common mean μ. Suppose also that VarX1=σ12 and VarX2=σ22. The value of μ is unknown, and it is proposed that μ be estimated by a weighted average of X1 and X2. That is, λX1+(1-λ)X2 will be used as an estimate of μ for some appropriate value of λ. Which value of λ yields the estimate having the lowest possible variance? Explain why it is desirable to use this value ofλ.

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If E[X]=1and Var(X)=5find

(a)E[(2+X2)]

(b)Var(4+3X)

In the text, we noted that

E∑i=1∞Xi=∑i=1∞EXi

when the Xiare all nonnegative random variables. Since

an integral is a limit of sums, one might expect that E∫0∞X(t)dt=∫0∞E[X(t)]dt

whenever X(t),0≤t<∞,are all nonnegative random

variables; this result is indeed true. Use it to give another proof of the result that for a nonnegative random variable X,

E[X)=∫0∞P(X>t}dt

Hint: Define, for each nonnegative t, the random variable

X(t)by

role="math" localid="1647348183162" X(t)=1ift<X\\0ift≥X

Now relate4q

∫0∞X(t)dttoX
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