Suppose RVs X Y are uniform on x2 y2 less than equal to 1 F

Suppose RV\'s X, Y are uniform on x^2 + y^2 less than equal to 1. Find 1. P(Y 0) 2. marginal pdf fx(x) 3. condition pdf fX(x| y) 5. P(X

Solution

X,Y r.v\'s are uniform on x^2 + y^2 <=1
f(x,y) = 1 / pi when x^2 + y^2 <=1
   = 0 otherwise
x^2 + y^2 <=1 or, y^2 <= 1-x^2 .....so, the range of y is (-1 , sqrt(1- x^2 ) )
2) marginal pdf of f(x) = integration from -1 to sqrt(1 - x^ 2 ) ( dy / pi ) = ( 1+ sqrt ( 1 - x^2) ) / pi ......
3) conditional pdf f (x | y) = f ( x ,y ) / f (x ) = 1 / ( 1+ sqrt ( 1 - x^2) )
4) E(x | y ) = integration from -1 to sqrt ( 1 - y ^2 ) ( xdx /  ( 1+ sqrt ( 1 - x^2) ) )
= ln( 1 + sqrt ( 1 - x^2) ) - sqrt ( 1 - x^2) [without putting the limits ] = ln ( |y| +1) - |y|

5)
Here the condition x^2 + y ^ 2 <=1 becomes x ^ 2 < 3 / 4....so, - sqrt (3) /2 < x < sqrt(3) / 2
again, x < 1/2....so the range of x becomes ( - sqrt (3) /2 , 1/2 )
p ( x< 1 / 2 | y=1/2 ) = integration from -sqrt(3 / 2) to 1/2 ( f (x , y) )dx / f ( y = 1/2 )
= ( ( sqrt(3) + 1 ) / 2*pi ) / (( 1 + sqrt( 1 - ( 0.5)^2 ) ) /pi )
=( 1 + sqrt( 3) ) / ( 2 + sqrt (3) )

(1 ) p( y < x and x >0) =double int. from ( -1 to x ) and ( 0 to 1 ) [ dx.dy / pi ]

= int from 0 to 1 [ (x +1)dx / pi = 3 / ( 2*pi)