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This appeared on an exam I took.

$Z \sim \text{Uniform}[0, 2\pi]$, and $X = \cos Z$ and $Y = \sin Z$. Let $F_{XY}$ denote the joint distribution function of $X$ and $Y$.

Calculate $\mathbb{P}\left[X+ Y \leq 1\right]$. So this was easy -

$$\begin{align} \mathbb{P}\left[X+Y \leq 1\right] &= \mathbb{P}\left[\sin Z+ \cos Z \leq 1\right] \\ &=\mathbb{P}\left[\sqrt{2}\sin\left(Z+\frac{\pi}{4}\right)\leq 1\right] \\ &= \mathbb{P}\left[Z \leq \arcsin\frac{1}{\sqrt{2}} - \frac{\pi}{4} \right] \\ &= \dfrac{\arcsin\frac{1}{\sqrt{2}} - \frac{\pi}{4}}{2\pi} \end{align} $$

But then, the question asked if $F_{XY}$ was absolutely continuous. I don't think so, but how would I prove it?

I thought about proceeding like this

$$ \begin{align} F_{XY}(x, y) &= \mathbb{P}\left[X \leq x, Y \leq y\right],\; x, y \in [0, 1] \\ &= \mathbb{P}\left[Z \leq \min(\arccos x, \arcsin y)\right] \end{align} $$ This is definitely continuous, but is it absolutely continuous?

Thanks!

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    Be careful of your answer for $P\{X + Y\} \leq 0.5$. Remember that $Z$ can take on any value in $[0,2\pi)$. What is the value of $\sqrt{2}\sin(Z + \pi/4)$ when $Z = \pi$, say? With regard to the other question, are $X$ and $Y$ _jointly continuous?_ That is, is there a region $A$ of nonzero area such that the random point $(X,Y)$ can be _any_ point in that region?2011-11-07
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    @Dilip, for jointly continuous $(X,Y)$, the random point $(X,Y)$ cannot be **ANY** point in any region of the plane.2011-11-07
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    @Didier I didn't say any point in any region, but rather "is there a region $A$ of nonzero area" (trying to avoid using "measurable") such that $(X,Y)$ can be _any_ point in **that** region (emphasis just added). As your answer said, since $(X,Y)$ can only be on the circle which has Lebesgue measure $0$ (or in more simplisitic terms, zero area), there is no region $A$ of nonzero area such that $(X,Y)$ can take on all possible values in that region.2011-11-07
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    @Dilip, you did not get the point (!). Jointly continuous distributions give mass zero to every point hence, by definition, $(X,Y)$ with such a density cannot be any point $(x,y)$, whatever $(x,y)$ is, in the sense that $P((X,Y)=(x,y))=0$. Hence every reasoning involving **points**, to decide whether a distribution has a density or not, is doomed.2011-11-07
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    @Didier Of course $P((X,Y)=(x,y))=0$, but obviously I have not been able to explain what I am trying to say very clearly.2011-11-07
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    Got something from an answer below?2015-04-26

2 Answers 2