# Does a discontinuity of a joint CDF correspond to an atom?

Let $(\Omega, \mathcal{A}, P)$ be a probability space. The following result is known.

If $X:\Omega\rightarrow\mathbb{R}$ is a random variable, then the CDF $F_X$ is discontinuous at $a\in \mathbb{R}$ iff $a$ is an atom of $X$.

Does this result extend to the $\mathbb{R}^2$, i.e. is the following statement correct?

If $X,Y:\Omega\rightarrow\mathbb{R}$ are random variables, then the joint CDF $F_{X, Y}$ is discontinuous at $(a,b) \in \mathbb{R}^2$ iff $(a,b)$ is an atom of $(X,Y)$.

## 1 Answer

No, the second highlighted statement is not true. It is possible to find a two-dimensional random variable $$(X,Y)$$ with no atoms, whose CDF is discontinuous. An example can be given based on item 5.6 (The continuity property of one-dimensional distributions may fail in the multi-dimensional case) on p. 34 of Jordan M. Stoyanov's Counterexamples in Probability, 3rd Edition, Dover, 2013, ISBN-13: 978-0486499987.

Consider the following function $$F:\mathbb{R}^2\rightarrow\mathbb{R}$$: whose graph can be visualized using the following diagram:

It can be verified that:

• This is a 2-dimensional CDF. (This can be verified by checking the 3 characterizing properties of a multidimensional CDF.)
• This CDF has no atoms. (This can be verified by checking that for every $$x_0, y_0 \in \mathbb{R}^2$$ and for every $$\epsilon - \epsilon(x_0,y_0) \in (0,\infty)$$, there exists some $$\delta \in (0,\infty)$$ such that whenever $$a < x_0 < b$$, and $$c < y_0 are such that $$b-a, d-c < \delta$$, $$F(a,c)+F(b,d)-F(a,d)-F(b,c) < \epsilon$$.)
• However, every point with coordinates $$(1,y)$$, where $$y \in (\frac{1}{2},\infty)$$, is a discontinuity point of $$F$$. (This can be verified by checking that, for such a $$y$$, $$\lim_{\substack{x\rightarrow1\\x<1}}F(x,y) = \frac{1}{2} \neq y = \lim_{\substack{x\rightarrow1\\x>1}}F(x,y)$$.)

* Note that the example given above is very similar, but no identical, to Stoyanov's example, since Stoyanov's example is wrong, as pointed out by ben in a comment below, since, for instance, at $$(1,\frac{3}{4})$$ Stoyanov's $$F$$ is not right-continuous in the first parameter.

• For the example given in book, would you mind explaining how $F(x,y)$ is right continuous in $x$ and $y$? It is presumably supposed to be right continuous in $x$ and $y$, as otherwise it wouldn't count as a cdf, right? My issue is that I've calculated $\lim_{h\rightarrow 0^{+}}F(1+h,3/4)=3/4\neq F(1,3/4)=1/2$, which, if my understanding on the notion of "right continuous in $x$" is correct, would suggest that $F(x,y)$ is not a cdf.
– ben
May 10 '21 at 8:21
• @ben: You are right. Thanks for pointing this out. This is an error in Stoyanov's book. I have corrected my answer. May 16 '21 at 5:32