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The probability generating function of the discrete non-negative integer-valued random variable Xhaving probability mass function pj,j≥0is defined by

ϕ(s)=EsX=∑j=0∞pjsj

Let Ybe a geometric random variable with parameter p=1-s, where 0<s<1. Suppose that Yis independent of X, and show that
Ï•(s)=P{X<Y}

Short Answer

Expert verified

Hence, the statement ϕ(s)=∑j=0∞pjsjis proved.

Step by step solution

01

Concept Introduction

The probability generating function of the discrete nonnegative integer values random variable Xhaving the probability mass function pj is,
ϕ(s)=∑j=0∞pjsj

02

:Explanation

The probability generating function of the discrete nonnegative integer values random variable Xhaving the probability mass function pj is,
ϕ(s)=∑j=0∞pjsj

Let Ybe a geometric random variable with parameter p=1-s.

Then pk:=(1-p)k-1p

GZ(z)=∑k=1∞(1-p)k-1pzk

=pz∑k=0∞[(1-p)z]k

=pz[1-(1-pz)]

03

:Explanation

Show that
Ï•(s)=P(X<Y)

=P(Y>X)

04

Explanation

Now,
P(Y>X)=∑jP(Y>X∣X=j)pj

=∑jP(Y>j∣X=j)pj

=∑jP(Y>j)pj

=∑jP(1-p)jpj

=∑j=0∞pjsj

05

:Final Answer

Hence, the statement ϕ(s)=∑j=0∞pjsjis proved.

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Most popular questions from this chapter

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

Let X1,X2,…,Xnbe independent random variables having an unknown continuous distribution function Fand let Y1,Y2,…,Ymbe independent random variables having an unknown continuous distribution function G. Now order those n+mvariables, and let

Ii=1 â¶Ä…â¶Ä…â¶Ä…if theith smallest of then+m â¶Ä…â¶Ä…â¶Ä…variables is from theXsample0 â¶Ä…â¶Ä…â¶Ä…otherwise

The random variable R=∑i=1n+miIiis the sum of the ranks of the Xsample and is the basis of a standard statistical procedure (called the Wilcoxon sum-of-ranks test) for testing whether Fand Gare identical distributions. This test accepts the hypothesis that F=Gwhen Ris neither too large nor too small. Assuming that the hypothesis of equality is in fact correct, compute the mean and variance of R.

Hint: Use the results of Example 3e.

Let X1,...be independent random variables with the common distribution functionF, and suppose they are independent of N, a geometric random variable with a parameter p. Let M=max(X1,...,XN).

(a) FindP{M…x}by conditioning onN.

(b) FindP{M…x|N=1}.

(c) FindP{M…x|N>1}

(d) Use (b) and (c) to rederive the probability you found in (a)

If 10 married couples are randomly seated at a round table, compute

(a) The expected number and

(b) The variance of the number of wives who are seated next to their husbands.

The best linear predictor of Ywith respect toX1and X2is equal to a+bX1+cX2, where a, b, and care chosen to minimizeEY-a+bX1+cX22 Determine a, b, and c.

See all solutions

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