The joint probability density function of X and Y is f(x,y) = 2/3 for 0 < x < 1, 0 < y < 2, x < y, and 0 otherwise. I am trying to use the formula Var(Y|X) = E(Y^2|X)-(E(Y|X))^2. I have already calculated E(Y|X) to be (2+x)/2. But I'm unsure how to get f(Y^2) in order to find E(Y^2|X). Any tips please?
2026-04-07 11:08:33.1775560113
Find Var(Y|X=x).
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$E[Y|X] = \frac {\frac 23\int_x^2 y\ dy}{\frac 23\int_x^2 \ dy} = \frac {\frac 12(4-x^2)}{2-x} = \frac 12(2+x)$
$E[Y^2|X] = \frac {\frac 23 \int_x^2 y^2\ dy}{\frac 23\int_x^2 \ dy} = \frac {\frac 13(2^3-x^3)}{2-x} = \frac 13(4+2x+x)$
$E[Y^2|X] - (E[Y|X])^2 = \frac 13(4 + 2x + x^2) - \frac 14(4 + 4x + x^2) = \frac {1}{12} (4 -4x + x^2)$