Homework Help is Here – Start Your Trial Now!
Math
Probability
(14-i) Let Z e {0, 1,2, 3, 4}, Y € {0, 1, 2} with the joint density f(0, 0) = =f(4, 0), f(1, 1) = = f(2, 0) = f(3, 1), and f(2, 2) = . Show that the two random variables are dependent The following reasons are proposed. (a) The greatest common divisor of 1. is positive, therefore Z, Y are dependent (b) The least common multiple of +, . is positive, therefore Z, Y are dependent (c) X, Y are in fact independent. (d) P(Z = 0, Y = 1) = 0 + P(Z = 0)P(Y = 1) (e) None of the above (a) (b) (c) (d) (e) N/ (Select One)

(14-i) Let Z e {0, 1,2, 3, 4}, Y € {0, 1, 2} with the joint density f(0, 0) = =f(4, 0), f(1, 1) = = f(2, 0) = f(3, 1), and f(2, 2) = . Show that the two random variables are dependent The following reasons are proposed. (a) The greatest common divisor of 1. is positive, therefore Z, Y are dependent (b) The least common multiple of +, . is positive, therefore Z, Y are dependent (c) X, Y are in fact independent. (d) P(Z = 0, Y = 1) = 0 + P(Z = 0)P(Y = 1) (e) None of the above (a) (b) (c) (d) (e) N/ (Select One)

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
See similar textbooks
icon
Related questions
Question
(14-i) Let \( Z \) (0, 1, 2, 3, 4), \( Y \) (0, 1, 2) with the joint density \( f(0, 0) = \frac{1}{16} = f(4, 0), f(1, 1) = \frac{1}{4} = f(2, 0) = f(3, 1), \) and \( f(2, 2) = \frac{1}{8} \). Show that the two random variables are dependent. The following reasons are proposed. (a) The greatest common divisor of \( \frac{1}{16}, \frac{1}{4}, \frac{1}{8} \) is positive, therefore \( Z, Y \) are dependent (b) The least common multiple of \( \frac{1}{16}, \frac{1}{4}, \frac{1}{8} \) is positive, therefore \( Z, Y \) are dependent (c) \( X, Y \) are in fact independent. (d) \( P(Z = 0, Y = 1) = 0 \neq P(Z = 0)P(Y = 1) \) (e) None of the above (Select One) (a) ○ (b) ○ (c) ○ (d) ○ (e) ○ N/A ● (14-ii) For the problem (14-i) above, show that the two random variables are uncorrelated. The following answers are proposed. (a) \( E(ZY) = 3/2 = E(Z)E(Y) \). So, they are uncorrelated. (b) \( E(ZY) = 3/2 \) and \( E(Z) = 1 \) and \( E(Y) = \frac{3}{2} \) making \( E(Z) + E(Y) = \frac{3}{2} \). (c) \( Z \) and \( Y \) are, in fact, positively correlated. (d) Since \( Z \) and \( Y \) are independent, therefore they are uncorrelated. (e) None of the above. (Select One) (a) ○ (b) ○ (c) ○ (d) ○ (e) ○ N/A ●
Transcribed Image Text:(14-i) Let \( Z \) (0, 1, 2, 3, 4), \( Y \) (0, 1, 2) with the joint density \( f(0, 0) = \frac{1}{16} = f(4, 0), f(1, 1) = \frac{1}{4} = f(2, 0) = f(3, 1), \) and \( f(2, 2) = \frac{1}{8} \). Show that the two random variables are dependent. The following reasons are proposed. (a) The greatest common divisor of \( \frac{1}{16}, \frac{1}{4}, \frac{1}{8} \) is positive, therefore \( Z, Y \) are dependent (b) The least common multiple of \( \frac{1}{16}, \frac{1}{4}, \frac{1}{8} \) is positive, therefore \( Z, Y \) are dependent (c) \( X, Y \) are in fact independent. (d) \( P(Z = 0, Y = 1) = 0 \neq P(Z = 0)P(Y = 1) \) (e) None of the above (Select One) (a) ○ (b) ○ (c) ○ (d) ○ (e) ○ N/A ● (14-ii) For the problem (14-i) above, show that the two random variables are uncorrelated. The following answers are proposed. (a) \( E(ZY) = 3/2 = E(Z)E(Y) \). So, they are uncorrelated. (b) \( E(ZY) = 3/2 \) and \( E(Z) = 1 \) and \( E(Y) = \frac{3}{2} \) making \( E(Z) + E(Y) = \frac{3}{2} \). (c) \( Z \) and \( Y \) are, in fact, positively correlated. (d) Since \( Z \) and \( Y \) are independent, therefore they are uncorrelated. (e) None of the above. (Select One) (a) ○ (b) ○ (c) ○ (d) ○ (e) ○ N/A ●
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer