# Show that the following function is not a cumulative distribution function

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?

• 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. Mar 7, 2015 at 14:36
• 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. Mar 8, 2015 at 3:18
• 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 $Y \leq 1916.$ Mar 8, 2015 at 3:25
• 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? Mar 8, 2015 at 15:56
• 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. Mar 8, 2015 at 19:51

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.
• 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.) Mar 8, 2015 at 3:52