Let $(X,Y)$ be a uniform random vector on the semicircle of radius $1$. Find $f_{X,Y}(x,y)$ and the marginals $f_X(x)$ and $F_Y(y)$.

My attempt: Since the random vector is uniform it will have a constant joint density of the form $\frac{1}{c}$. I also figured that $-1 \leqslant X\leqslant1$ and $0 \leqslant Y \leqslant \sqrt{(1-X^2)}$ as these bounds define the semicircle. When I integrate the joint density $\frac{1}{c}$ over the bounds I just mentioned and equating it to $1$, I find that the joint density must be $$f_{X,Y}(x,y)=\frac{2}{\pi}$$ But I do not think this is correct, as when I compute the marginals and then compute conditional probablities using this joint density, I get illogical results. Thanks for any help.

  • $\begingroup$ Semi-circle or semi-disk? $\endgroup$ – Did Feb 24 '15 at 17:59
  • $\begingroup$ Semi circle. There are quite a few similar examples online that involve semi-disks, but none where we only consider the boundary curve like here $\endgroup$ – Trawkley Feb 24 '15 at 18:01
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    $\begingroup$ Then $(X,Y)$ has no density since $P((X,Y)\in C)=1$ with $C$ the semi-circle and Leb$(C)=0$. $\endgroup$ – Did Feb 24 '15 at 18:01
  • $\begingroup$ I see. What have I done in my answer then? would my answer be correct if we were considering a semi-disk? $\endgroup$ – Trawkley Feb 24 '15 at 18:48
  • $\begingroup$ Provided you use the correct density, yes, that is, $$f_{X,Y}(x,y)=\frac2{\pi}\mathbf 1_{y\gt0}\mathbf 1_{x^2+y^2\lt1}.$$ $\endgroup$ – Did Feb 24 '15 at 18:50

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