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Let $\xi$ be uniformly distributed on $\left[-\pi,\,\pi\right]$, $X = \cos \xi$, $Y = \sin \xi$. Is it true that $\Pr \left( X=1\mid Y=0 \right) = 0.5$?

It is obvious this problem cannot be solved in term of events as $\Pr \left( Y=0 \right) = 0$. Therefore I am to compute conditional pdf $p \left( x \mid y \right)$. But joint pdf $p \left( x, y \right)$ is distributed on the zero-measured set. So, I'm a bit confused with this.

EDIT: The key problem here is that the distribution on the unit circumference is singular in $\mathbb{R}^2$. However I still don't know if this equality is correct in any sense.

Denis Korzhenkov
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1 Answers1

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$X$ and $Y$ are individually continuous random variables but they are not jointly continuous random variables, and they do not enjoy a joint pdf; the random point $(X,Y)$ perforce lies on the unit circle which is a set of measure $0$ in the plane. Nonetheless, in this case, conditional distributions do exist (though conditional densities do not), and in this particular case, conditioned on $Y$ having value $0$ (an event of probability $0$ but not the impossible event $\emptyset$), the conditional distribution of $X$ is a discrete distribution, that is, $X$ is conditionally a discrete random variable that takes on values $+1$ and $-1$ with equal probability $\frac 12$. Thus, it is perfectly correct to write $$P\{X = 1\mid Y = 0\} = \frac 12.$$

Dilip Sarwate
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    How would you prove this using Kolmogorov probability theory? B.t.w. it's not clear enough that $\Pr (X=1 | Y=0) = \Pr (X=-1 | Y=0) $ – Denis Korzhenkov Nov 01 '17 at 17:35
  • The absolute value of the slope of the circle is the same in both points in which it crosses the line $y=0$ (namely, $dx/dy$ is zero in both points). This fact is important to derive $P(X=1|Y=0)=1/2$. Where have you used it? – fonini Nov 06 '17 at 13:35