hw5+fall+2011+solution

# hw5+fall+2011+solution - ECI 114 Fall 2011 University of...

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ECI 114 Fall 2011 University of California, Davis Department of Civil and Environmental Engineering Homework Assignment 5 (62 points) Due at 5 p.m Monday, Nov. 21, 2011 Show all work and explain your answers; no credit will be given for solutions only. Write your name on the TOP RIGHT CORNERS of the front of the first page and back of the last page of your assignment; do not fold it. A two-point penalty will be assessed for an inability to follow these instructions. A penalty of up to three points may also be assessed for lack of neatness. 1. The joint probability distribution for two random variables, X and Y, is given in the accompanying table. (2+4+2+4+2=14pts) x y 0 1 2 3 4 5 0 0 0.05 0.025 0 0.025 0 1 0.2 0.05 0 0.3 0 0 2 0.1 0 0 0 0.1 0.15 a) Verify that the properties of a joint probability distribution hold. (2 pts) b) Find the marginal probability distributions of X and Y. (4 pts) c) Determine E[X] and E[Y]. (2 pts) d) Find the conditional probability distributions p X|Y (x|y=0) and p Y|X (y|x=4). (4 pts) e) Are X and Y independent? Support your answer with the appropriate equation(s). (2 pts) Solution: a) A joint pdf needs to satisfy the two properties below: (i) All joint probabilities should be 0 ≤ p(x, y) ≤ 1, where x = 0, 1, …, 5, and y = 0, 1, 2. (ii) The sum of the joint probabilities is equal to one: ∑ ∑ = = = 5 0 x 2 0 y 1 ) y , x ( p . 1

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From the above table, we can see it satisfies both properties, so the properties of a joint probability mass function hold. b) The marginal probability distributions of X and Y are calculated by = 2 0 y ) y , x ( p and = 5 0 x ) y , x ( p , repectively. The marginal probability of X (Y) is the same as the sum of the joint probabilities in each column (row) (as shown in the table). = = = = = = = else , 0 5 x , 15 . 0 4 x , 125 . 0 3 x , 3 . 0 2 x , 025 . 0 1 x , 1 . 0 0 x , 3 . 0 ) x ( p X , = = = = else , 0 2 y , 35 . 0 1 y , 55 . 0 0 y , 1 . 0 ) y ( p Y c) 5 0 [ ] * ( ) X x E X x p x = = = 0*0.3 + 1*0.1 + 2*0.025 + 3*0.3+ 4*0.125 + 5*0.15 = 2.3. 2 0 [ ] * ( ) Y y E Y y p y = = = 0*0.1 + 1*0.55+ 2*0.35 = 1.25. d) Use the fact that ) 0 ( p ) y , x ( p ) 0 y | x ( p Y Y | X = = and ) 4 ( p ) y , x ( p ) 4 x | y ( p X X | Y = = .
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