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

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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 0 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 . 1. (b) The gain is G = 100 X + 200 Y , i.e., E [ G ] = 100 E [ X ] + 200 E [ Y ] = 300 . 2. Let X i be the Bernoulli random variable taking the value 1 if the first member of the i th couple is alive, and 0 otherwise.
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