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I was helping a student with a question I couldn't solve. We have the following process: X is sampled from a $U(0,1)$ distribution. Then Y is sampled from a $U(-x,x)$ distribution. Therefore I have $Y|X$ is a Uniform. How do I find the distribution for $Y$ and $X|Y$? I was unable to integrate the joint distribution for x because the integration limits depend on x, if I can solve this, then it's just a Bayes theorem application.

$$f_{X,Y}(x,y) = f_{Y|X=x}(y)f_{X}(x) = \frac{1}{2x}\mathbb{1}_{(-x,x)}(y) \mathbb{1}_{(0,1)}(x) $$ $$ f_Y(y) = \int f_{X,Y}(x,y) dx = \int \frac{1}{2x}\mathbb{1}_{(-x,x)}(y) \mathbb{1}_{(0,1)}(x) dx $$

$$f_{X|Y} = \frac{f_{X,Y}}{ f_{Y}} $$

Is there something I missed? Is it possible to avoid this integral?

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    $\begingroup$ You don't use the same symbol for the dummy in the integral as the 'real' variable in the limit. I suggest drawing a picture of the relationship between the bounds on Y and the value of X, which should make it easier to see what's going on. $\endgroup$
    – Glen_b
    Apr 16, 2015 at 1:15

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Hint: $(X,Y)$ must always be inside the triangle with vertices $(0,0), (1,1), (1,-1)$ with $f_{X,Y}(x_0,y)$ having the same value (call it $g(x_0)$) for all $y, -x_0 < y < x_0$ (else the conditional pdf of $Y$ given $X=x_0$ would not be $U(-x_0,x_0)$). Note that value of $g(x_0)$ does change with choice of $x_0$. Indeed, since the area at $x_0$ of the cross-section of the solid defined by the joint pdf is $f_X(x_0)$ which we know has value $1$ (by uniformity of $X$ on $(0,1)$), it must be that $2x\cdot g(x_0) = 1$, that is, $g(x) = \frac{1}{2x}, x\in(0,1)$.

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