# Conditional Distribution over the unit disc

How can I show that, $U$ and $V$, two independent uniform $(-1,1)$ random variables have a conditional distribution, given that $U^2 + V^2 <1$, that takes the form:

$$f_{U,V|U^2+V^2<1} (u,v|w<1) =1/{\pi}, \quad u^2+v^2<1$$

I tried using the CDF technique for $W=U^2+V^2$

$$\frac{P\left[ U^2+V^2 \leq w , U^2+V^2 <1 \right]}{P \left[ U^2+V^2<1 \right]}=P \frac{\left[ U^2+V^2 \leq w \right]}{\frac{\pi}{4}}$$

because the event $[U^2+V^2 \leq w] \subset [U^2+V^2<1]$

But after that, I am stuck as even if I was to evaluate the double integral at the numerator, it does not seem that I would get the required distribution. Any hints? Thank you.

• One approach: what is the probability of being in an element $(u,u+du)\times (v,v+dv)$ before the restriction to the circle? What is the conditional probability of being in the element, given you're in the circle? Dec 8, 2013 at 1:31
• Hint: to evaluate $P\{U^2+V^2 \leq w\}$ switch to polar coordinates. But give some thought to the uniform distribution and think of ways of computing this probability without the formality of integration. Dec 8, 2013 at 2:15
• @DilipSarwate Indeed, the second half of your comment was what I was getting at in my comment. Dec 8, 2013 at 2:21
• @Glen_b I understand your point but what about the joint probability of being in the element and in the circle at the same time? Is it just $1/4 \times 1$? Dec 8, 2013 at 9:24
• @DilipSarwate I cannot get integration to work unfortnately. Taking the bounts as $(0,\sqrt{w}) \times (0,2\pi)$ and after differentiating wrt to $w$, I am left with simply $1/4 \pi$, not the $1/4$ we were looking for. Dec 8, 2013 at 9:47

In general, the conditional pdf of $X$ given that $X \leq a$ is just $$f_{X \mid \{X \leq a\}}(x) = \begin{cases} \displaystyle \frac{f_{X}(x)}{P\{X \leq a\}}, & x \leq a,\\0, &x > a,\end{cases}$$ that is, it is just the pdf of $X$ scaled to have total area $1$ (as all pdfs must have) in the region of the conditioning event, and $0$ in the complementary event. The same applies to conditional joint pdfs. Because of the uniform distribution, the probability that the random point $(U,V)$ is in the unit disc is just the ratio of the area $\pi$ of the disc to that of the square ($4$), and the conditional joint pdf thus has value
$$\frac{f_{U,V}(u,v)}{P\{U^2+V^2 < 1\}} = \frac{1/4}{\pi/4} = \frac{1}{\pi}$$ in the interior of the unit disc and $0$ outside. Put another way, $(U,V)$ is conditionally distributed uniformly on the unit disc given that $U^2+V^2 < 1$.