Suppose $F(x,y)=0$ if $x+y<2016$ and $F(x,y)=1$ if $x+y>2016$. Prove that $F$ is not a bivariate cumulative distribution function.

My attempt:

$F(1008+\varepsilon,1008+\varepsilon)+F(1008,1008)-F(1008,1008+\varepsilon)-F(1008+\varepsilon,1008)=1+F(1008,1008)-1-1=F(1008,1008)-1$ which is actually $P(X\in[1008,1008+\varepsilon],Y\in[1008,1008+\varepsilon])$ if we suppose that $F$ is indeed a d.f.

But this probability being non-negative, it follows that $F(1008,1008)=1$. But this means that $P(X\in[1008,1008+\varepsilon],Y\in[1008,1008+\varepsilon])=0$ for ANY $\varepsilon>0$.

This indeed means that $X\leq1008$,or $Y\leq1008$. But $F(1008,1008)=1$ also implies that both $X\leq1008,Y\leq1008$. But notice that we never really used the numbers $1008$ and $1008$ anywhere, we only used that $1008+1008=2016$. In fact, the whole thing could have been done analogously with $a,b$ instead of $1008,1008$ under the constraint that $a+b=2016$. So, similar arguments would have implied that $X\leq a,Y\leq b$ for all $a,b$ such that $a+b=2016$.

So consider $a_n=-n$ and $b_n=2016+n$. Arguments above show that $X\leq -n,Y\leq 2016+n$ for all $n$, so taking limit as $n\to\infty$ we have, $X\leq-\infty$. Oppositely taking $a_n=2016+n,b_n=-n$ we would have $Y\leq -n$ for all $n$ and hence $Y\leq-\infty$. Both of these are true and hence $X,Y$ cannot be random variables, contradicting the existence of $F$ as a d.f.

Is it right?

  • $\begingroup$ The last paragraph seems a little unclear to me. It seems you have taken the fact that $X\leq a$ OR $Y\leq b$ and have reinterpreted it as $X\leq a$ AND $Y\leq b$. In fact the joint distribution $X=Y=1008$ satisfies $X\leq a$ or $Y\leq b$ for all $a+b=2016$, although (for other reasons) it does not have the given $F$ as its cdf. $\endgroup$
    – David K
    Mar 7, 2015 at 14:36
  • $\begingroup$ You are correct. So as i have found that for a certain sequence X is less than any real number , maybe i can immediately conclude that X cannot be a real number and hence the bivariate cdf does not make sense. $\endgroup$ Mar 8, 2015 at 3:18
  • $\begingroup$ Except that you never showed that $X$ is less than any real number. You showed that $X$ cannot be greater than $2016-Y$. That is a very different thing, because if this were a legitimate cdf, then $X$ could be (for example) $100$, provided it is possible that $Y \leq 1916.$ $\endgroup$
    – David K
    Mar 8, 2015 at 3:25
  • $\begingroup$ I do not quite get you. Pardon my stupidity please. But I did show that $X$ is less than any real number; considering the sequence $a_n=-n$. I have shown that whenever $a+b=2016$ we must have $X\leq a$ and $Y\leq b$. So take $a_n=-n$ which gives $X\leq-\infty$. Where is the flaw, according to you? $\endgroup$ Mar 8, 2015 at 15:56
  • $\begingroup$ Why take $a_n = -n$? Why not $a_n = n$? Conversely, consider the joint distribution $W\sim N(0,1)$ and $Z\sim N(0,1)$ where $W = -Z$. This also obeys the rule that you can always write $a + b = 2016$ with $W \leq a$ and $Z \leq b$. If you set $a_n = -n$ you can "prove" that $W \leq -\infty$ with logic just as valid as your "proof" that $X \leq -\infty$; the only difference is that in the case of $X$ there are other reasons that coincidentally show that $X$ cannot have a random distribution. $\endgroup$
    – David K
    Mar 8, 2015 at 19:51

1 Answer 1


Here is a simpler approach:

Note that $P ( (x_1,x_2] \times (y_1, y_2]) = F(x_2,y_2)-F(x_2,y_1)-F(x_1,y_2)+F(x_1,y_1)$

If we let $A=(0,2000]^2$ and $B= A + \{(-5000,5000)\})$, then $A,B$ are disjoint but we have $P A =1, PB = 1$, which is impossible.

  • $\begingroup$ I am not really familiar with the notation. Could you kindly explain B? $\endgroup$ Mar 8, 2015 at 3:23
  • $\begingroup$ Just shift the box up and to the left, $B=(-5000,-3000]\times (5000,7000]$. (Easier to visualise than to compute, basically a disjoint box so that only the top right hand corner is in the $x+y > 2016$ space.) $\endgroup$
    – copper.hat
    Mar 8, 2015 at 3:52

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