Two random variables, X and Y, have the joint density function: f <mo stretchy="false">(

gaitaprepeted05u 2022-04-30 Answered
Two random variables, X and Y, have the joint density function:
f ( x , y ) = { 2 0 < x y < 1 0 i o c
Calculate the correlation coefficient between X and Y.
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Answers (2)

haillarip0c9
Answered 2022-05-01 Author has 23 answers
Step 1
Let X and Y random variables with joint density function given by
f ( x , y ) = { 2 , if 0 < x y < 1 , 0 , if otherwise .
The coefficient correlation of X and Y is given by,
ρ X Y = C o v ( X , Y ) σ X σ Y = σ X Y σ X σ Y
where C o v ( X , Y ) = E [ X Y ] E [ X ] E [ Y ] is the covariance of X and Y and σ X and σ Y standard deviations.
Now,
f X ( x ) = x 1 f ( x , y ) d y = 2 ( 1 x ) .
f Y ( y ) = 0 y f ( x , y ) d x = 2 y .
E [ X ] = + x f X ( x ) d x = 0 1 x ( 2 2 x ) d x = 1 3
E [ X 2 ] = + x 2 f X ( x ) d x = 0 1 x 2 ( 2 2 x ) d x = 1 6
E [ Y ] = + y f Y ( y ) d y = 0 1 y ( 2 y ) d y = 2 3
E [ Y 2 ] = + y 2 f Y ( y ) d y = 0 1 y 2 ( 2 y ) d y = 1 2
E [ X Y ] = + + x y f ( x , y ) d x d y = 0 1 0 y x y ( 2 ) d x d y = 0 1 y 3 d y = 1 4
C o v ( X , Y ) = E [ X Y ] E [ X ] E [ Y ] = 1 4 1 3 × 2 3 = 1 36
σ X = V a r ( X ) = E [ X 2 ] ( E [ X ] ) 2 = 1 6 ( 1 3 ) 2 = 2 6
σ Y = V a r ( Y ) = E [ Y 2 ] ( E [ Y ] ) 2 = 1 2 ( 2 3 ) 2 = 2 6
Therefore,
ρ X Y = C o v ( X , Y ) σ X σ Y = 1 / 36 ( 2 / 6 ) 2 = 1 2 > 0.
Since ρ X Y > 0 then X and Y they are positively, linearly correlated, but not perfectly so.
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Norah Small
Answered 2022-05-02 Author has 12 answers
Step 1
The correlation coefficient between X and Y is defined as follows:
ρ X , Y = E [ ( X μ X ) ( Y μ Y ) ] σ X σ Y
However, ρ can be expressed in terms of uncentered moments:
ρ X , Y = E [ X Y ] E [ X ] E [ Y ] E [ X 2 ] ( E [ X ] ) 2   E [ Y 2 ] ( E [ Y ] ) 2 .
Step 2
It seems that you are struggling with the orders of integration. It helps to recall the Law of Total Expectation, which states that
E [ X ] = E [ E [ X | Y ] ] and E [ Y ] = E [ E [ Y | X ] ]
Step 3
Then, the integrals you need to compute are:
E [ X ] = E [ E [ X | Y ] ] = 0 1 0 y x f ( x , y ) d x d y
E [ X 2 ] = E [ E [ X 2 | Y ] ] = 0 1 0 y x 2 f ( x , y ) d x d y
E [ Y ] = E [ E [ Y | X ] ] = 0 1 x 1 y f ( x , y ) d y d x
E [ Y 2 ] = E [ E [ Y 2 | X ] ] = 0 1 x 1 y 2 f ( x , y ) d y d x
E [ X Y ] = 0 1 x 1 x y f ( x , y ) d y d x
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