Given the joint P.D.F of X and Y: f_{X,Y}(x,y)=6x^2y (x,y) in (0,1) times (0,1). f_{X,Y}(x,y)=0 elsewhere

perlejatyh8

perlejatyh8

Answered question

2022-11-02

Domain of joint continuous random variables - Geometrically
Given the joint P.D.F of X and Y:
f X , Y ( x , y ) = 6 x 2 y     ( x , y ) ( 0 , 1 ) × ( 0 , 1 )
f X , Y ( x , y ) = 0 elsewhere
The following double integral should be the probability of P ( Y > 2 X ):
P ( Y > 2 X ) = 0 1 0 y / 2 6 x 2 y d x d y
However what I don't understand is why X takes values from 0 to y/2.
I tried to sketch the corresponding area - upper left triangle of 1 × 1   square above Y = 2 X - and my best guess is that for each value of Y then X = Y 2
Finally how would one calculate the P ( Y < 2 X ) area where X and Y move?
Right now what I struggle with is finding the corresponding area of R 2 regardless of the problem/question

Answer & Explanation

Tasinazzokbc

Tasinazzokbc

Beginner2022-11-03Added 17 answers

Step 1
However what I don't understand is why X takes values from 0 to y/2.
Because you want to integrate over { x , y ( 0..1 ) 2 : y > 2 x } which is also { x , y ( 0..1 ) 2 : x < y / 2 }
P ( Y > 2 X ) = P ( X < Y / 2 ) = 0 1 0 y / 2 6 x 2 y d x d y
Step 2
Finally how would one calculate the P ( Y < 2 X ) area where X and Y move?
You want to integrate over { x , y : 0 < y < min { 1 , 2 x } < 2 } which may be done by parting that domain into a union of disjoint domains...
P ( Y < 2 X ) = 0 1 / 2 0 2 x 6 x 2 y d y d x + 1 / 2 1 0 1 6 x 2 y d y d x
Or of course...
P ( Y < 2 X ) = 1 P ( Y > 2 X ) + P ( Y = 2 X ) 0

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