/*! 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 15E Suppose that X1, . . . ,Xn are... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose thatX1, . . . ,Xnare random variables suchthat thevariance of each variable is 1 and the correlationbetween each pair of different variables is 1/4. DetermineVar(X1+. . .+Xn).

Short Answer

Expert verified

\(Var\left( {{X_1} + {X_2} + ... + {X_n}} \right) = n + \frac{{n\left( {n - 1} \right)}}{4}\)

Step by step solution

01

Given information

The variance of each variable is 1.

The correlation between each pair of different variables is \(\frac{1}{4}\).

02

Compute the variance 

\(\begin{align}Var\left( {{X_1} + {X_2} + ... + {X_n}} \right) &= \sum\limits_i {Var\left( {{X_i}} \right)} + 2\sum {\sum\limits_{i < j} {Cov\left( {{X_i},{X_j}} \right)} } \\ &= n\left( 1 \right) + 2\left( {\frac{{n\left( {n - 1} \right)}}{2}} \right)\left( {\frac{1}{4}} \right)\\ &= n + \frac{{n\left( {n - 1} \right)}}{4}\end{align}\)

Hence,\(Var\left( {{X_1} + {X_2} + ... + {X_n}} \right) = n + \frac{{n\left( {n - 1} \right)}}{4}\).

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

Suppose that one word is to be selected at random from the sentence the girl put on her beautiful red hat. If X denotes the number of letters in the word that is selected, what is the value of E(X)?

Suppose that a point\({{\bf{X}}_{\bf{1}}}\)is chosen from the uniform distribution on the interval\(\left( {{\bf{0,1}}} \right)\)and that after the value\({{\bf{X}}_{\bf{1}}}{\bf{ = }}{{\bf{x}}_{\bf{1}}}\)is observed, a point\({{\bf{X}}_{\bf{2}}}\)is chosen from a uniform distribution on the interval\(\left( {{{\bf{x}}_{\bf{1}}}{\bf{,1}}} \right)\). Suppose further that additional variables\({{\bf{X}}_{\bf{3}}}{\bf{,}}{{\bf{X}}_{\bf{4}}}{\bf{,}}...\)are generated in the same way. Generally,\({\bf{j = 1,2,}}...{\bf{,}}\)after the value\({{\bf{X}}_{\bf{j}}}{\bf{ = }}{{\bf{x}}_{\bf{j}}}\)has been observed,\({{\bf{X}}_{{\bf{j + 1}}}}\)is chosen from a uniform distribution on the interval\(\left( {{{\bf{x}}_{\bf{j}}}{\bf{,1}}} \right)\). Find the value of\({\bf{E}}\left( {{{\bf{X}}_{\bf{n}}}} \right)\).

LetYbe a discrete random variable whose p.f. is the

functionfin Example 4.1.4. LetX= |Y|. Prove that the

distribution ofXhas the p.d.f. in Example 4.1.5

\({\bf{f}}\left( {\bf{x}} \right){\bf{ = }}\left\{ \begin{array}{l}\frac{{\bf{1}}}{{{\bf{2}}\left| {\bf{x}} \right|\left( {\left| {\bf{x}} \right|{\bf{ + 1}}} \right)}}{\bf{,x = \pm 1, \pm 2 \ldots ,}}\\{\bf{0,Otherwise}}\end{array} \right.\)

\({\bf{f}}\left( {\bf{x}} \right){\bf{ = }}\left\{ \begin{array}{l}\frac{{\bf{1}}}{{{\bf{x}}\left( {{\bf{x + 1}}} \right)}}{\bf{,x = 1,2,3 \ldots ,}}\\{\bf{0,Otherwise}}\end{array} \right.\)

Suppose that an observed value of X is equally likely to come from a continuous distribution for which the pdf is for from one for which the pdf is g. Suppose that \(f\left( x \right) > 0\) for \(0 < x < 1\) and \(f\left( x \right) = 0\) otherwise, and suppose also that \(g\left( x \right) > 0\) for \(2 < x < 4\) and \(g\left( x \right) = 0\) otherwise. Determine:

  1. the mean and
  2. the median of the distribution of X.

Let X be a random variable. Suppose that there exists a number m such that \(\Pr \left( {X < m} \right) < \frac{1}{2}\)and\(\Pr \left( {X > m} \right) < \frac{1}{2}\).Prove that m is the unique median of the distribution of X.

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.