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Let $D$ be a unit disc $|z| \leq 1.$ Consider the following three ways of choosing a random point in $D$:
1.Chose a point $(x,y)$ where $x=\sqrt{r} \cos t,y=\sqrt{r} \sin t$,where $r =U(0,1),t = U(0,2 \pi)$
2. Chose a point $(x,y)$where the joint distribution of x and y is given by:$$ f(x,y)=\frac{1}{\pi} \text{if}\, (x,y) \in D,\text{else}\, 0 $$
3.Chose a point $(x,y)$ where X and y are iid with distribution $N(0,1/2)$
could somebody kindly explain the diffence betwenn the three?Lots of thanks for any responces

sajjad veeri
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    Does this answer your question? [Simulate a uniform distribution on a disc](https://stats.stackexchange.com/questions/120527/simulate-a-uniform-distribution-on-a-disc) That clearly deals with the first 2 methods, and the problem with the third should be clear from the problem with the first and its over-representation of values close to the origin. – EdM Jan 16 '22 at 18:33
  • I know how to simulate the points.I want to know the distinction between the above three ways – sajjad veeri Jan 16 '22 at 18:52
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    What do you mean by "the distinction" between them. Clearly you see that the methods are different. If you actually simulate them you see that the distributions are different. ... so what else do you wish to know, exactly? – Glen_b Jan 17 '22 at 00:50
  • for instance , how do we express the joint pdf in the other two cases, and how do the quantities,say,$\sqrt(x^2+y^2)$ vary for each of these – sajjad veeri Jan 17 '22 at 07:04
  • The duplicate contains answers to those questions for cases (1) and (2). For case (3), *iid* implies the joint density function is the product of the marginal densities and therefore is of the form $f(x,y)=\exp(-(x^2+y^2)/(2\sigma^2))/(2\pi\sigma^2).$ The value of $\sigma$ is determined by whatever the "$1/2$" in your notation is referring to (it could be $\sigma,$ $\sigma^2,$ or even $1/\sigma^2,$ according to three popular conventions). – whuber Jan 17 '22 at 15:49

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