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$\Theta$ is a uniform random variable on the interval $[0, 2\pi)$, and we have $X = \cos(\Theta)$ and $Y = \sin(\Theta)$.

Suppose $(x,y)$ is a point inside the unit circle. Then, the arc length that is shown in white in Fig. 3.2.a on page 56 of An Introduction to Copulas (Nelson, 2006) is $2\pi H(x,y)$, where $H(x,y)$ is a joint distribution function of RVs $X$ and $Y$.

Could you please help me to understand why the arc length is $2\pi H(x,y)$?

[EDIT] Captured from the book

DEVA
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    By definition, $H(x,y)$ is the chance that a random point $\Theta$ on the unit circle has coordinates $X$ and $Y$ not exceeding either $x$ or $y$, respectively. The white arc shown in the figure depicts that event. Because the distribution of $\Theta$ is uniform, the chance is proportional to the arc length. The normalization factor is the total arc length, equal to $2\pi$. Because both the question and this explanation will be unintelligible to anyone without access to the book, we have to close the question pending further explanation. – whuber Dec 22 '16 at 15:07
  • @whuber thanks! I just put the figure. Yes, I should've seen it if in the form of $l/2\pi = H(x,y)$, where $l$ is the arc length. – DEVA Dec 22 '16 at 16:35
  • You should also add the [self-study] tag & read its wiki. – gung - Reinstate Monica Dec 22 '16 at 17:46
  • How is $(x,y)$ in your second picture considered as being inside the unit circle? – Dilip Sarwate Dec 23 '16 at 01:04
  • @Dilip Sarwate sorry. It is 3.2.a. I edited my question – DEVA Dec 23 '16 at 04:05

1 Answers1

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$H$ refers to a bivariate distribution function. In general this means

$$H(x,y) = \Pr(X \le x,\ Y \le y).$$

In the $(x,y)$ plane, such a region is a rectangle bounded at the top by $y$, at the right by $x$, and infinite to the left and down, as shown by the gray regions in both figures.

These particular random variables are confined to the unit circle because $X^2 + Y^2 = \cos^2(\Theta) + \sin^2(\Theta) = 1$. Moreover, they are uniformly distributed on that circle because $\Theta$ has a uniform distribution on the interval $[0, 2\pi)$. Uniform means the chance that $(X,Y)$ lies along any particular (measurable) part of the circle is directly proportional to the length of that part. Since the total length is the perimeter $2\pi$, the constant of proportionality must be $1/(2\pi)$.

Consequently, $H(x,y)$ must be proportional to the total length of any parts of the circle lying within the semi-infinite rectangle determined by $(x,y)$. The figures show two examples, one where $(x,y)$ is in the disk bounded by the circle and another where $(x,y)$ is outside that disk. In either case, though, $H(x,y)$ must equal $1/(2\pi)$ times this total arc length. Multiplying this equality yields the statement in the question: $2\pi H(x,y)$ is the total length of arcs to the left of and below the point $(x,y)$.

whuber
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