The random variable $(X,Y)$ is uniformly distributed in the circle $x^2+y^2\leq 1$.
- Find the joint density $f_{XY}(x,y)$ and the marginal density of $X$ and $Y$.
$$
\begin{split}
f_{XY}(x,y) &= \frac{1}{\pi} \\
f_X(x) &= \frac{2}{\pi}\sqrt{1-x^2} \\
f_Y(y) &= \frac{2}{\pi}\sqrt{1-y^2}
\end{split}
$$
- Say if $X \perp Y$ or not.
$\rightarrow f_X(x)f_Y(y)\neq f_{XY}(x,y)\Rightarrow X,Y$ are not independent
- Find the conditional density of $X$ given $Y=y$.
$$f_{X|Y}(x|y) := \frac{f_{XY}(x,y)}{f_Y(y)} = \frac{1}{2\sqrt{1-y^2}}$$
- Calculate $\mathbb{P}(Y<X)$.
$$
\mathbb{P}(Y<X)
= \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}
\left[\int_{-\sqrt{1-y^2}}^{x}\frac{1}{\pi}dy\right]dx
$$
Is it correct? In particular, I have a doubt as to the correctness of integration extremes at the fourth point… Should I use, for example, $x\in (-1,1)$ or not? Thanks in advance.
Best Answer
The last one is wrong. LHS is a constant but your RHS depends on $y$.
By inspection is it obvious you are asking about the proportion of the unit circle that lies below the line $y=x$, which by symmetry is $1/2$. Since the problem is symmetric in $X,Y$ that conclusion is anyhow warranted even before the geometric argument.
If you insist in doing it the hard way, I would write $$ \mathbb{P}(Y<X) = \int_{-1}^1 \int_{-\sqrt{1-x^2}}^x \frac{dydx}{\pi} $$