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Sample $U \sim \text{Uniform}(0,\sqrt{2}-1)$. Accept $U$ with probability $1/(1+U^2)$ (else reject and sample again). Set $X = 2U/(1+U^2)$ and $Y = 1-UX = (1-U^2)/(1+U^2)$. With probability 1/2, switch the values of $X$ and $Y$. With probability 1/2, independently, change the signs of $X$ and $Y$.

Prove: The points $(X,Y)$ are uniformly distributed on the circumference of the unit circle.

I've simulated the points on R and can confirm that this method actually works. It's also clear that $X^2+Y^2 = 1$. But how does one show analytically that the points generated using this method lie on the unit circle?

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    $\begingroup$ Hint: write $U=\tan(\theta)$ and show that $\theta$ is uniformly distributed on the interval $[-\pi/8, \pi/8].$ The rest is just trigonometry. $\endgroup$ – whuber Jul 8 at 16:03
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    $\begingroup$ If $X^2+Y^2=1$ it should not be hard to prove "the points generated using this method lie on the unit circle". $\endgroup$ – Xi'an Jul 8 at 18:36
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Thanks to @whuber's hint, I was able to solve it myself. Posting a sketch of the solution here for those interested:

Let $T$ be the random variable sampled after accepting $U$. For $t \in (0,\sqrt{2}-1)$,

$$P(T \leq t) = P(U \leq t \vert V \leq 1/(1+U^2))$$ where $V \sim \text{Unif}(0,1)$ independent of $U$. This gives the following density for $T$: $$f_T(t) = \frac{1}{\arctan(\sqrt{2}-1)}\frac{1}{1+t^2} = \frac{1}{\pi/8}\frac{1}{1+t^2}, \ \ t \in (0,\sqrt{2} - 1)$$

Now, using the hint, let $T = \tan \theta$. Transform the density of $T$ to get $\theta \sim \text{Unif}(0,\pi/8)$. Write $X$ and $Y$ in terms of $\theta$ to get $X = \sin 2\theta$ and $Y = \cos 2 \theta$. Without applying the switch and sign change, $(X,Y)$ is uniformly distributed on the first quandrant of the unit circle. Apply the switch and sign change to conclude the proof.

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