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Suppose $Y$ is uniformly distributed on $(0,1)$ and that thet conditional distribution of $X$ given that $Y=y$ is uniform on $(0,y)$ Find $E(X)$ and $Var(X)$

My attempt : We have $Y \sim$ Unif$(0,1) $ and $f_{X|Y}(x,y) = \frac{1}{y}$

Thus $$E(X)=E(E(X|Y))$$ $$= E(\int_{0}^{\infty}xf_{X|Y}(x,y) dx) $$ $$= E(\int_{0}^{\infty}\frac{x}{y}dx)$$

I am sure I am making mistake here. Can you help ?

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You have written your conditional density incorrectly. It should be $f_{X\mid Y}(x,y) = \frac1y \mathbb 1_{[0,y]}$, i.e. it is $\frac1y$ on $[0,y]$ and $0$ outside. Then,

$$\begin{align} E[X] &= E[E[X\mid Y]]\\ &=E\left[\int_0^\infty xf_{X\mid Y}(x,y) dx\right]\\ &=\int_0^1\left[\int_0^y \frac{x}{y} dx\right]dy\\ &=\int_0^1\frac y2dy\\ &=\frac14 \end{align} $$

You can find the variance similarly.

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