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I am reading a paper on modeling the dependencies in discrete distribution functions, and am having a hard time understanding the following. Let us define: $$r \leq min(p,q)$$ $$ B(u,v) = \begin{cases} 0 & \text{if }u=0 \ \text{or } v=0 \\ r & \text{if }(u,v) \in (0,p] \times (0,q] \\ p & \text{if }(u,v) \in (p,1] \times (0,q] \\ q & \text{if }(u,v) \in (0,p] \times (q,1] \\ 1 & \text{if }(u,v) \in (p,1] \times (q,1] \\ \end{cases} $$ $$ C(u,v) = \begin{cases} r & \text{if }(u,v) \in [0,p) \times [0,q) \\ p & \text{if }(u,v) \in [p,1] \times [0,q) \\ q & \text{if }(u,v) \in [0,p) \times [q,1] \\ 1 & \text{if }(u,v) \in [p,1] \times [q,1] \\ \end{cases} $$

The paper then says that $C(u,v)$ is a distribution function, while $B(u,v)$ is NOT a distribution function.

Can anyone explain why this is the case? I do notice that C(u,v) is defined in a right continuous manner (which is typically how distribution functions are defined?), is that why? Or is there something else that I am missing. I can provide more context if necessary, but it seems to me that this is a standalone thing.

Thanks!

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If $B$ and $C$ are supposed to be joint cumulative probability distribution functions (joint CDFs), then you need to be aware that a joint CDF $F(\cdot,\cdot)$ must satisfy the "rectangle constraint" (a name that I just made up and so don't bother Googling for it). This rectangle constraint says that for all real numbers $a,b,c,d$ such that $a<c, b<d$, $F(c,d)-F(a,d)-F(c,b)+F(a,b)$, which equals $P\{a < X \leq c, b < Y \leq d\}$ must be nonnegative: $$F(c,d)-F(a,d)-F(c,b)+F(a,b) = P\{a < X \leq c, b < Y \leq d\} \geq 0.$$ For discrete random variables in particular, if $F(u,d)$ has constant value $p$ for $a \leq u <c$ and so does $F(c,v)$ have fixed value $q$ for all $b \leq v < d$, then $F(c,d)$ must be larger than $\max(p,q)$.

As an example, consider the function $$B(u,v) = \begin{cases} 0, & u<0, v<0,\\ 0, & 0 \leq u < 1, 0 \leq v < 1,\\ 1, & \text{otherwise} \end{cases}$$ which is right-continuous in both variables and has all the properties that one might desire of a joint CDF except that $$B(1,1)-B(0,1)-B(1,0) + B(0,0) = -1$$ and so is disbarred from the ranks on joint CDFs by the rectangle constraint.

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  • $\begingroup$ +1 Your careful use of strict inequalities $\lt$ and inequalities $\le$ clearly exposes the key idea. $\endgroup$
    – whuber
    Jun 16 '15 at 20:38
  • $\begingroup$ @whuber Thanks for the upvote. I think, though, that the issue in the OP's question is one of left-continuity versus right-continuity which, as mark Stone points out, is more a matter of convention (if I recall correctly, Loeve's GTM tome also defines the CDF as $P\{X < x\}$). $\endgroup$ Jun 16 '15 at 20:48
  • $\begingroup$ Yes--and that's precisely why I think careful use of the inequalities, as well as a clear quotation of the definition, are critical in this case. $\endgroup$
    – whuber
    Jun 16 '15 at 20:52
  • $\begingroup$ @Dilip Sarwate, right you are, or should I say, left you are regarding Loeve's book . See p. 177 of books.google.com/… . Then again, he was at Berkeley, so maybe that explains his leftist leanings. So Breiman and Loeve are the only two "modern": (although they are both now deceased) non-Russian authors I know of using the left continuity convention. $\endgroup$ Jun 16 '15 at 22:33
  • $\begingroup$ Hi all, thanks again for the detailed answers! I'm still trying to sort through what is stated in the answer above. One comment is, in Nelsen's Introduction to Copula's book, the Rectangle formula you provided is called the H-Volume, and in the book he does mention that the H-volume of a joint CDF must be $\geq$ 0. See Definition 2.1 of w4.stern.nyu.edu/ioms/docs/sg/seminars/nelsen.pdf $\endgroup$
    – Kiran K.
    Jun 17 '15 at 23:19
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Probability distribution functions are defined so as to be right continuous ... unless you're in Russia, in which case they are defined so as to be left continuous.

Leo Breiman, best known to youngsters as the inventor of random forests and the person who coined the term "bagging", was perhaps a Commie, based on, for example, his 1973 book "STATISTICS: With a View Toward Applications", in which he uses the Russian, i.e., left continuous, convention, and never bothers informing the reader that he is departing from the standard convention outside of Russia. Or look at Definition 2.21 at the bottom of p. 25 in his 1968 book "Probability".

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    $\begingroup$ Let's not use pejorative terms like Commie and just call Breiman a leftist. Much more useful in getting the point across, wouldn't you say? $\endgroup$ Jun 16 '15 at 19:45
  • $\begingroup$ Just trying to interject a little levity. ,,, And I did say "perhaps". $\endgroup$ Jun 16 '15 at 19:49

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