Answer in Statistics and Probability for aditi #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).
“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”
