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).

“f(x,y)=3x,” “0lt y lt xlt1”

“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_0int^x_0 (xy)*f(x,y)dydx”

“=int^1_0int^x_0(xy)*3xdydx”

“=int^1_0int^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_0int^x_0xf(x,y)dydx”

“=int^1_0int^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_0int^x_0y*f(x,y)dydx”

“=int^1_0int^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,

“3/10not=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 0lt xlt 1”

“0,space elsewhere”

Thus, “f(Y|x)=3x/3x^2=1/x,space 0lt ylt xlt1”

“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 0lt xlt 1”

“0,space elsewhere”

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