/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q.7.4 Use the conditional variance for... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use the conditional variance formula to determine the variance of a geometric random variable X having parameter p.

Short Answer

Expert verified

The variance of a geometric random variable Xhaving parameter pisVar(X)=1-pp2.

Step by step solution

01

Given Information

The variance of a geometric random variable X having parameter p.

02

Explanation

Suppose a sequence of independent Bernoulli random variables Xnn≥1 with the parameter of success p. Let Xmark the index of first random variable which has yielded a success. We know that Zhas Geometric distribution with parameterp. Using the formula for conditional variance, we can condition on random variable X1. We have that

Var(X)=EVarX∣X1+VarEX∣X1

EVarX∣X1=EEX2∣X1-EX∣X12

=PX1=0·EX2∣X1=0-EX∣X1=02+ PX1=1EX2∣X1=1-EX∣X1=12

=(1-p)EX2-E(X)2+0

=(1-p)Var(X)

03

Explanation

EX∣X1=1·IX1=1+1+1pIX1=0

Applying the variance,

VarEX∣X1=p(1-p)+1+1p2p(1-p)-2p(1-p)1+1p

Var(X)=(1-p)Var(X)+1p2p(1-p)

Var(X)=1-pp2

04

Final Answer

The variance of a geometric random variable X having parameterp isVar(X)=1-pp2.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The county hospital is located at the center of a square whose sides are 3 miles wide. If an accident occurs within this square, then the hospital sends out an ambulance. The road network is rectangular, so the travel distance from the hospital, whose coordinates are (0,0), to the point(x,y) is |x|+|y|. If an accident occurs at a point that is uniformly distributed in the square, find the expected travel distance of the ambulance.

7.2. Suppose that Xis a continuous random variable with

density function f. Show that E[IX-a∣]is minimized

when ais equal to the median of F.

Hint: Write

E[IX-al]=|x-a|f(x)dx

Now break up the integral into the regions where x<a

and where x>a, and differentiate.

A total of n balls, numbered 1through n, are put into n urns, also numbered 1through nin such a way that ball iis equally likely to go into any of the urns 1,2,..i.

Find (a) the expected number of urns that are empty.

(b) the probability that none of the urns is empty.

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

Consider the following dice game, as played at a certain gambling casino: Players1and 2roll a pair of dice in turn. The bank then rolls the dice to determine the outcome according to the following rule: Playeri,i=1,2,wins if his roll is strictly greater than the banks. Fori=1,2,let

Ii=1 â¶Ä…â¶Ä…â¶Ä…ifiwins0 â¶Ä…â¶Ä…â¶Ä…otherwise

and show that I1and I2are positively correlated. Explain why this result was to be expected.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.