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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.

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    $\begingroup$ 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? $\endgroup$
    – Glen_b
    Commented Dec 8, 2013 at 1:31
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    $\begingroup$ 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. $\endgroup$
    – Dilip Sarwate
    Commented Dec 8, 2013 at 2:15
  • $\begingroup$ @DilipSarwate Indeed, the second half of your comment was what I was getting at in my comment. $\endgroup$
    – Glen_b
    Commented Dec 8, 2013 at 2:21
  • $\begingroup$ @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 $? $\endgroup$
    – JohnK
    Commented Dec 8, 2013 at 9:24
  • $\begingroup$ @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. $\endgroup$
    – JohnK
    Commented Dec 8, 2013 at 9:47

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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 $U^2+V^2 < 1$.

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  • $\begingroup$ Sorry, visiting this post after a long time! However, still the computation of the conditional joint density doesn't seem rigorous. $\endgroup$
    – Ashok
    Commented Jan 11, 2021 at 7:50
  • $\begingroup$ @Ashok Unfortunately, I do not have time to edit my answer to make it as rigorous as you desire. However, since my answer does not meet your standards for rigor, you have the option of posting your own answer that meets your standards. If you wish, your answer can point out the shortcomings of my answer, and perhaps the OP will choose to switch his acceptance of my answer to an acceptance of your answer. $\endgroup$
    – Dilip Sarwate
    Commented Jan 12, 2021 at 15:25

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