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I approached this problem by using the formula Var(Y) = $E(Y^2)-E(Y)^{2}$.

E(Y) = $\frac{3}{4}$

However, my difficulty was finding $E(Y^2)$. Can someone explain why this integral setup is wrong?

$E(Y^2) = \int_{0}^{1} \int_{x}^{1} y^2 dy dx $

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1 Answer 1

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It should be $\int_0^1 \int_x^1 \frac{y^2}{1-x} \, dy \, dx$, since the density of $Y$ given $X=x$ is $f_{Y \mid X=x}(y) = \frac{1}{1-x}$ on $[x, 1]$.

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