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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 $(X,Y)$.

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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 $F:\mathbb{R}^2\rightarrow\mathbb{R}$: A discontinuous two dimentional CDF with no atoms whose graph can be visualized using the following diagram: An aid to the visualization of the CDF's graph

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 $c < y_0 <d$ 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.

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  • $\begingroup$ 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. $\endgroup$
    – ben
    May 10 '21 at 8:21
  • $\begingroup$ @ben: You are right. Thanks for pointing this out. This is an error in Stoyanov's book. I have corrected my answer. $\endgroup$
    – Evan Aad
    May 16 '21 at 5:32

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