Question
Let $(X, Y)$ be a jointly continuous pair of random variables such that the marginal density of $X$ is $$f_X(x) = \frac 1 {2\pi} \sqrt{4 - x^2}, \quad -2 < x < 2$$ (also known as the Wigner semicircle distribution) and such that the the conditional distribution of $Y$ given $X = x$ is uniform on the interval $[-\sqrt{4 - x^2}, \sqrt{4 - x^2}]$.
$(a)\quad$ Give a simplified expression for the joint density of $X$ and $Y$, specifying clearly the set of $x$ and $y$ values for which the joint density is non-zero.
$(b)\quad$ How would you describe this joint distribution in words?
$(c)\quad$ Hence, or otherwise, show that $Y$ has the same distribution as $X$.
My working
$(a)$
$$f_{Y \mid X}(y \mid x) = \frac 1 {2\sqrt{4 - x^2}}$$
$$\begin{aligned} \implies f_{X, Y}(x, y) & = f_{Y \mid X}(y \mid x)f_X(x) \\[1 mm] & = \left(\frac 1 {2\sqrt{4 - x^2}}\right)\left(\frac 1 {2\pi} \sqrt{4 - x^2}\right) \\[1 mm] & = \frac 1 {4\pi}, \quad -2 < x < 2\ \mathrm{and}\ -\sqrt{4 - x^2} < y < \sqrt{4 - x^2} \end{aligned}$$
$(b)\quad$ I would say that “the joint distribution maps being in a circle of radius $2$ about the origin with probability $\frac 1 {4\pi}$”, but I am not sure if this is mathematically correct.
$(c)\quad$ I would say that "since the joint distribution of $X$ and $Y$ maps a full circle of radius $2$ about the origin and the marginal distribution of $X$ maps a semicircle of radius $2$ about the origin, it follows that the marginal distribution of $Y$ also maps a semicircle of radius $2$ about the origin, implying that $Y$ has the same distribution as $X$", but again I am not sure if this is entirely correct.
If I am wrong anywhere, please do point it out and explain why :)
Your answer are almost correct.
a) correct
b) The joint distribution you found is uniform over the disk centered in the origin with radius 2. In fact the density is exactly the reciprocal of the disk area
c) simply by simmetry or integrating you get the answer