hw6_sol - ECE 3100 Fall 2009 Homework#6 Solutions 1(a...

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Unformatted text preview: ECE 3100 Fall 2009 Homework #6 Solutions 1. (a) Notice that there are 21 pairs of integers ( x,y ) that are in the set: Ω = { ( x,y ) :- 2 ≤ x ≤ 4 ,- 1 ≤ y- x ≤ 1 } . Consequently since we are considering a uniform distribution we deduce that the joint PMF of X and Y is P X,Y ( x,y ) = 1 21 for ( x,y ) ∈ Ω and otherwise. Note that for each value of x ∈ {- 2 ,..., 4 } , the random variable Y can take three values, namely x- 1 , x or x + 1 . Consequently, P X ( x ) = P X,Y ( x,x- 1)+ P X,Y ( x,x )+ P X,Y ( x,x +1) = 3 21 for all x ∈ {- 2 ,..., 4 } and P X ( x ) = 0 otherwise. The mean of X is thus equal to E [ X ] = ∑ 4 x =- 2 x P X ( x ) = 3 21 ∑ 4 x =- 2 x = 1 . Similarly for each value of y ∈ Ω , the random variable X can take at most three values, namely y- 1 , y and y + 1 , since x has to be also in {- 2 ,..., 4 } we deduce that for each value of y ∈ Ω the random variable X ∈ Ω y = { y- 1 ,y,y + 1 } ∩ {- 2 ,..., 4 } . Consequently, P Y ( y ) = ∑ x ∈ Ω y P X,Y ( x,y ) = 1 21 | Ω y | . We obtain therefore the following values for the PMF of Y , P Y ( y ) = 1 21 for y ∈ {- 3 , 5 } , P Y ( y ) = 2 21 for y ∈ {- 2 , 4 } , P Y ( y ) = 3 21 for y ∈ {- 1 ,..., 3 } and P Y ( y ) = 0 otherwise. The mean of Y is thus equal to E [ Y ] = ∑ 5 y =- 3 y P Y ( y ) = 1 ....
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hw6_sol - ECE 3100 Fall 2009 Homework#6 Solutions 1(a...

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