Why should the hyperplane define an at most countable set of discontinuity points?

Question

In Durrett's book, 4th edition, on page 173, we have the following sentence:

Let $H^i_c = \{x : x_i = c\}$ be the hyperplane where the $i$th coordinate is $c$. For each $i$, the $H^i_c$ are disjoint so $D^i = \{c : P(X \in H^i_c) > 0\}$ is at most countable. It is easy to see that if $x$ has $x_i \notin D^i$ for all $i$ then $F$ is continuous at $x$. This gives us more than enough points to reconstruct $F$.

(Original image here.)

I don't get why $D^i$ is at most countable and supposedly has the discontinuity points on the $i$th-axis for the distribution function $F(x)=P(X\leq x)$, where $x \in \mathbb{R}^d$.

Edit: For example what's the problem with $P(X\in H^i_c)=1/(b-a)$ and zero otherwise. $P(X\in R^d)\geq \int_{[b,a]} 1/(b-a) \ d c = 1$

You must provide the context and relevant definitions, to makes some sense of such a question. — uniquesolution, Aug 20 '18 at 16:44
@uniquesolution honestly I don't know which context to provide. It's probability theory, so the definitions of e.g. distribution function, etc. should be of general knowledge. Do you have something specific in mind? — An old man in the sea., Aug 20 '18 at 16:47
F is the distribution function of a random vector $X$, i.e. $F(x)=P(X\leq x)$ — An old man in the sea., Aug 20 '18 at 16:51

saulspatz · Accepted Answer · 2018-08-20T19:17:58.890

2

The author claims that at most countably many of the hyperplanes have positive probability. Suppose, by way of contradiction, that there are uncountably many such. For $n=1,2,3,\dots$ let $S_n^i = \{H_c^i|P(H_c^i)>1/n\}.$ Each of the uncountably many $H_c^i$ with positive probability belongs to at least one of the $S_n^i.$ It cannot be the case that all the $S_n^i$ are finite, because there are countably many $S_n^i$ and the countable union of finite sets is countable. But then we have infinitely many disjoint sets with probability greater than $1/n$ for some $n,$ and this is absurd, so there are at most countably many $H_c^i$ with positive probability.

The set of such hyperplanes is in one-to-one correspondence with $D^i,$ so $D_i$ is at most countable.

edited Aug 20 '18 at 19:17

answered Aug 20 '18 at 17:40

saulspatz

53,131

Saul, I'm not really sure I follow you, could you please add some more details? – An old man in the sea. Aug 20 '18 at 17:48
You would have to add some details to the question first. All I understand is that you have a set of pairwise disjoint measurable sets of positive probability, and the author claims there can be at most countably many such. I have explained why this is so. Please tell me at what point I haven't been clear. – saulspatz Aug 20 '18 at 17:52
I've edited the question. We're not sure if the hyperplanes have positive probability,they may have... – An old man in the sea. Aug 20 '18 at 18:22
I've tried to be more explicit. Does it help? – saulspatz Aug 20 '18 at 18:32
Saul, Thanks for the help. However, I still have some doubts. Is the definition of the set $S^i_n$ correct? Having a set $S^i_n$ with infinite number of elements, means only that we may have a continuous $D^i$, like an interval. How do you conclude «we have infinitely many disjoint sets with probability greater than 1/n for some n»? – An old man in the sea. Aug 20 '18 at 19:03
My fault, I got the notation messed up. I was thinking that the $D_i$ were the hyperplanes themselves. I'll edit my answer. It may be a few minutes, while I figure out what I'm trying to say. – saulspatz Aug 20 '18 at 19:13
@Anoldmaninthesea. I think I've got it right now. Take a look. Sorry for the confusion. – saulspatz Aug 20 '18 at 19:18
I think I get what you're trying to say. I would just define $S^i_n={c: P(H^i_n)>1/n }$. Also, what about the 2nd part of my question, why can we say that the set $D^i$ contains all discontinuities? – An old man in the sea. Aug 20 '18 at 19:44
Can you do it for a function of one variable? – saulspatz Aug 20 '18 at 19:52
+1 for all the help ;) – An old man in the sea. Aug 20 '18 at 21:01
Let us continue this discussion in chat. – An old man in the sea. Aug 20 '18 at 21:01

score 1 · Answer 2 · answered Aug 20 '18 at 19:55

There's an entire "genre" of results about probability distributions which state that probability distributions can only have countably many of certain kinds of "discontinuities". All of them ultimately come from a general (and simple) mathematical principle which every student should have seen at least once: "an uncountable sum always diverges". I've posted a simple explanation and proof of this principle here.

From this we get the classic fact that a probability distribution has at most countably many "atoms" (singletons with positive probability), since otherwise it would have finite sets of arbitrarily large probability - not possible when the probability of the whole set is supposed to be $1$.

You can use the principle to prove your statement directly. If there were uncountably many of these disjoint hyperplanes of positive probability, we would be able to find a finite family of them whose union had arbitrarily large probability (notice we use the disjointness to be able to say that the probability of the union is the sum of the probabilities). In fact, more generally, a probability distribution cannot have an uncountable family of disjoint sets of positive probability.

In this case, however, since you're dealing with particularly simple sets (hyperplanes), you can also use another proof. Consider the random variable $X_i$, the $i$-th coordinate of $X$. The set $H^i_c$ is exactly the set $\{X_i = c\}$. So $P(X\in H^i_c)=P(X_i=c)$. Since the distribution of $X_i$ can only have countably many singletons of positive probability, only countably many of the $H_i$ can have positive probability.

The problem with your attempted counterexample is that when you write $P(X\in H^i_c)=\frac1{b-a}$, what you really mean is that the probability density function of $X_i$ is $\frac1{b-a}$ at $c$. We would still have $P(X\in H^i_c)=0$. Your idea is obviously based on the uniform distribution, and notice that even for a regular old uniform random variable $U$, we don't have $P(U=c)=\frac1{b-a}$, we have $P(U=c)=0$.

Jack, thanks for your answer and link. I still have some doubts though. We cannot have $P(X\in H^i_c)=\frac1{b-a}$ in an interval because then $P(X\in H^i_c)=P(X_i = c)=1/(b-a)$ and we could not allow $X_i$ to be a continuous r.v. ? — An old man in the sea., Aug 20 '18 at 21:00
@Anoldmaninthesea. I'm not sure what you mean by "we could not allow $X_i$ to be a continuous rv", but it's impossible to have a random variabel $X_i$ with $P(X_i=c)=\frac1{b-a}$ for all $c$ in an interval. — Jack M, Aug 21 '18 at 06:32

Why should the hyperplane define an at most countable set of discontinuity points?

2 Answers2