Assume the random variable $(X,Y)$ is uniformly distributed on the disc $B=\{(x,y)\in\mathbb{R}^2:x^2+y^2\leqslant 1\}$ and on $[-1,1]^2$, respectively.
In both cases determine the the conditional distribution of $Y$ given $X=x$.
My attempt:
$f(x,y)=\frac{1}{\text{area of B}}=\frac{1}{2\pi}$ if $(x,y)\in B$
$P(A)=\int\int_{A}f(x,y)dxdy=\int\int\frac{1}{2\pi}=\frac{\text{area of A}}{2\pi}$
I need to compute:
$P(Y\in dy|X=x)=f_{Y|X}(x,y)=\frac{f(x,y)}{f_{X}(x)}$
where $f_{X}(x)=\int_B f(x,y)dy$
However it is precisely this last integral that is difficult for me to define, since I am integrating on a disc. On which domain shall I integrate in what concerns $Y$?
Question:
How should I solve this problem? Am I right insofar?
Thanks in advance!
Area is $\pi$ not $2\pi$. $f_X(x)=\int f(x,y)dy=\int_{-\sqrt {1-x^{2}}}^{\sqrt {1-x^{2}}} \frac 1 {\pi} dy=\frac {2\sqrt {1-x^{2}}} {\pi}$ for $|x| \leq 1$. Also $\int \frac 1 {\sqrt {1-x^{2}}}dx=\arcsin \, x$. Can you take it from here?
For the square you have to integrate w.r.t $y$ from $-1$ to $1$.