prelim 2 solutions

C as mentioned above the conditional pdf of x given y

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(c) As mentioned above, the conditional PDF of X given Y is uniform. The mean of a uniform random variable on the interval [ a,b ] is a + b 2 . Hence for all y ( a,a ) , E [ X | Y = y ] = a −| y | +( a + | y | ) 2 = 0 . We conclude then that E [ X | Y ]=0 . (d) Define S = { ( x,y ) square: x + y a 2 } . Since X and Y are uniformly distributed over the square, we can write P ( X + Y a 2 )= Area of S Area of the square . The region S is simply the rectangle with vertices ( a, 0) , (0 , a ) , ( 3 a 4 , a 4 ) and ( a 4 , 3 a 4 ) . The area of this rectangle is then 6 4 a 2 . Which means that P ( X + Y a 2 )= 6 4 a 2 2 a 2 = 3 4 . 6. (a) Let X be a Bernoulli ( p ) random variable, then Var( X ) = p (1 p ) . The function p (1 p ) is increasing on [0 , 1 2 ] and decreasing on [ 1 2 , 1] , hence the maximum of this function is achieved at p = 1 2 . The maximum variance possible for a Bernoulli RV is then 1 4 . (b) First note that X is a uniform random variable on { 0 , 1 , 2 , 3 } , indeed P ( X = 0) = P ( X = 0 ,Y = 0)+ P ( X = 0 ,Y = 1) = 2 8 = 1 4 and similarly we have P ( X = 1) = P ( X = 2) = P ( X = 3) = 1 4 . Define the function f ( x ) as follows f ( x )= braceleftbigg 1 if x 1 0 otherwise Then f ( X ) is a Bernoulli ( 1 2 ) random variable and has the maximum possible variance. Indeed P ( f ( X ) = 1)= P ( X =0)+ P ( X =1)= 1 2 and similarly P ( f ( X )=0)= 1 2 . (c) We can let the function g ( · ) be the same as the function f ( · ) constructed in part (b), then we would have P ( f ( X ) = g ( Y )) = P ( f ( X ) = f ( Y )) = 1 . Indeed, f ( X ) = 1 if and only if X 1 , but from the joint PMF of X and Y we know that X 1 if and only if Y 1 , that is f ( X ) = 1 if and only if f ( Y ) = 1 . Similarly f ( X )=0 if and only if f ( Y )=0 , in other words P ( f ( X )= f ( Y ))=1 . 7

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(d) There are several possible choices for the functions f ( · ) and g ( · ) that will work here. We choose f ( · ) and g ( · ) as follows f ( x )= braceleftbigg 1 if x =0 , 2 0 otherwise g ( y )= braceleftbigg 1 if y =1 , 3 0 otherwise Then we can verify that both f ( X ) and g ( Y ) are Bernoulli ( 1 2 ) and hence have maximum variance. Indeed P ( f ( X ) = 1) = P ( X = 0)+ P ( X = 2) = 1 2 and P ( f ( X ) = 0) = P ( X = 1)+ P ( X = 3) = 1 2 . The verification that g ( Y ) is Bernoulli ( 1 2 ) follows along the same lines. Now it remains to show that f ( X ) and g ( Y ) are independent. We have P ( f ( X )=1 ,g ( Y )=1)= P ( X ∈ { 0 , 2 } ,Y ∈ { 1 , 3 } ) = p XY (0 , 1)+ p XY (2 , 3)= 1 4 = P ( f ( X )=1) P ( g ( Y )=1) P ( f ( X )=1 ,g ( Y )=0)= P ( X ∈ { 0 , 2 } ,Y ∈ { 0 , 2 } ) = p XY (0 , 0)+ p XY (2 , 2)= 1 4 = P ( f ( X )=1) P ( g ( Y )=0) P ( f ( X )=0 ,g ( Y )=1)= P ( X ∈ { 1 , 3 } ,Y ∈ { 1 , 3 } ) = p XY (1 , 1)+ p XY (3 , 3)= 1 4 = P ( f ( X )=0) P ( g ( Y )=1) P ( f ( X )=0 ,g ( Y )=0)= P ( X ∈ { 1 , 3 } ,Y ∈ { 0 , 2 } ) = p XY (1 , 0)+ p XY (3 , 2)= 1 4 = P ( f ( X )=0) P ( g ( Y )=0) , which shows that f ( X ) and g ( Y ) are independent. To receive full credit, you did not need to write all of this explanation. 8
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