Answer to Question #249598 in Statistics and Probability for aditi

Question #249598

Let X and Y have the joint pdf f(x, y) = 3x, 0 < y < x < 1, zero elsewhere. Are X and Y independent? If not, find Var(Y|x).


1
Expert's answer
2021-10-12T06:06:51-0400

"f(x,y)=3x," "0\\lt y \\lt x\\lt1"

"0,\\space elsewhere"

Given the "p. d. f" above, we shall determine whether "X" and "Y" are independent using the property for testing independence given as,

"E(XY)=E(X)*E(Y)"

Now,

"E(XY)=\\int^1_0\\int^x_0 (xy)*f(x,y)dydx"


"=\\int^1_0\\int^x_0(xy)*3xdydx"


"=\\int^1_0\\int^x_03yx^2dydx"


"=" "\\int^1_03\/2*(y^2x^2|^x_0)dx"


"=\\int^1_0(3\/2)*x^4dx"


"=(3\/10*x^5)|^1_0=3\/10"


"E(X) =\\int^1_0\\int^x_0xf(x,y)dydx"


"=\\int^1_0\\int^x_03x^2dydx"


"=\\int^1_0(3x^2y)|^x_0dx"


"=\\int^1_03x^3dx"

"=(3\/4x^4)|^1_0"


"=3\/4"

"E(Y)=\\int^1_0\\int^x_0y*f(x,y)dydx"


"=\\int^1_0\\int^x_03xydydx"


"=\\int^1_03\/2(xy^2)|^x_0dx"


"=\\int^1_03\/2(x^3)dx"


"=3\/8(x^4)|^1_0"


"=3\/8"

Let us check if the condition stated above is satisfied,

"3\/10\\not=3\/4*3\/8"

Therefore, random variables "X" and "Y" are not independent.

Since they are not independent, we move ahead to determine "V(Y|x)".

To determine this conditional variance, we need to find the conditional distribution of "Y|x" given as "f(Y|x)" as follows,

"f(Y|x)=f(x,y)\/g(x)" where "g(x)" is the marginal distribution of "X"

We already have "f(x,y)", let's find "g(x)"

Now,

"g(x)=\\int^x_0f(x,y)dy"


"=\\int^x_03xdy"


"=(3xy)|^x_0"


"=3x^2"

Therefore, the marginal distribution of "X" is given as,

"g(x)=3x^2,\\space 0\\lt x\\lt 1"

"0,\\space elsewhere"

Thus, "f(Y|x)=3x\/3x^2=1\/x,\\space 0\\lt y\\lt x\\lt1"

"0,\\space elsewhere"

Conditional variance, "V(Y|x)=E(Y^2|x)-E(Y|x)".

We determine the two conditional expectations as follows,

"E(Y|x)=\\int^x_0y*f(Y|x)dy"


"=\\int^x_0(y\/x)dy"


"=(y^2\/2x)|^x_0"


"=x^2\/2x=x\/2"


"E(Y^2|x)=\\int^x_0y^2*f(Y|x)dy"


"=\\int^x_0(y^2\/x)dy"


"=(y^3\/3x)|^x_0"


"=x^3\/3x=x^2\/3"

Now,

"V(Y|x)=x^2\/3-(x\/2)^2=x^2\/12,\\space 0\\lt x\\lt 1"

"0,\\space elsewhere"


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