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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?


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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? – Dilip Sarwate Nov 7 '11 at 1:05
@Dilip, for jointly continuous $(X,Y)$, the random point $(X,Y)$ cannot be ANY point in any region of the plane. – Did Nov 7 '11 at 1:51
@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. – Dilip Sarwate Nov 7 '11 at 2:14
@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. – Did Nov 7 '11 at 2:19
@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. – Dilip Sarwate Nov 7 '11 at 3:06

The absolute continuity meant here is not the absolute continuity of the function $F_{X,Y}(x,y)$, but rather the absolute continuity of the measure $\mathrm{d}F_{X,Y}(x,y)$ with respect to Lebesgue measure on $\mathbb{R}^2$.

According to Wikipedia the measure is absolutely continuous is for every $\epsilon > 0$, there exists $\delta > 0$, such that $\int_A \mathrm{d}F_{X,Y}(x,y) < \epsilon$ for all subsets $A$ such that $\int_A \mathrm{d}x \mathrm{d}y < \delta$.

It is clear that if you choose $A$ to be an anulus with inner radius $1-\delta/4$ and outer radius $1+\delta/4$, its Lebesgue measure if going to be $ \pi \delta$, while $\int_A \mathrm{d}F_{X,Y}(x,y) = 1$, because the annulus contains the entire support of $(X,Y)$ random vector, which is the unit circle.

By the way, I am afraid you incorrectly determined the $F_{X,Y}(x,y)$. Remember $x$ and $y$ both range from $-1$ to $1$.

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Sasha, thanks for your reply. I was wondering if you could point me to more examples of measures that are not absolutely continuous with respect to the Lebesgue measure. Clearly, my intuition could do with some of those. :) – tchakravarty Nov 7 '11 at 17:07
@FgNu See wikipedia page, it states that absolutely continuous measures are those that have the finite probability density functions, i.e. $\partial_{x,y} F_{X,Y}(x,y)$ exists. Thus measures that correspond to discrete random variables are not absolutely continuous as well. – Sasha Nov 7 '11 at 17:11

Absolutely continuous distributions are measures with a density with respect to the Lebesgue measure. The distribution of $(X,Y)$ is concentrated on the unit circle, which has Lebesgue measure zero, hence this distribution is not absolutely continuous.

Regarding the computation of $\mathrm P(X+Y\leqslant1)$, note that $\arcsin(1/\sqrt2)=\pi/4$ hence your formula would yield $\mathrm P(X+Y\leqslant1)=0$ (a quite unlikely result). Rather, drawing a picture of the unit circle and of the line of equation $x+y=1$ shows that $x+y\leqslant1$ corresponds to the three quarters of the circle from the point $(0,1)$ to the point $(1,0)$ through the points $(-1,0)$ and $(0,-1)$, hence $\mathrm P(X+Y\leqslant1)=3/4$.

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Thanks for your reply. So you are saying that if a distribution has support on a set of Lebesgue measure zero, it can never be absolutely continuous? So, I guess there are two ways in which absolute continuity can fail - by having support on a set of Lebesgue measure zero and by actually being discontinuous in its arguments on sets of positive Lebesgue measure. I would really appreciate pointers to some textbook examples of that illustrate these two cases. – tchakravarty Nov 7 '11 at 17:06
You might wish to read Lebesgue's decomposition theorem and to keep in mind Cantor's staircase function. – Did Nov 7 '11 at 17:41

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