Thanks It kXy is linear then Huts in Example 446 the EXy is

Thanks

It k(X|y) is linear, then Huts, in Example 4.4-6, the E(X|y) is p^2 = 1/4. it f Because the ratio of those Let f(x,y) = (3/16)xy^2, 0

Solution

= (3/16)x(₀² (y²)dy)

= (3/16)x[[y^3]/3]₀²

= (½) x

            f_y(Y) = ₀² ((3/16)xy²)dx

= (3/16)y²(₀² (x)dx)

= (3/16)y^2[[x^2]/2]₀²

= (3/8) y^2

= ₀²( ₀^y (3/16) X Y^2 dx)dy

= ₀²(3/16)Y^2 (X^2/2)₀^y dy

= ₀²(3/16)Y^2 (Y^2/2) dy

= ₀²(3/32)Y^4 dy

= (3/32)(Y^5/5)₀²

= 3/5