# Rationale behind defining distribution function with strict inequality

A friend from Italy pointed me to a nice Probability Theory book written in Italian by G. Dall'Aglio, a well-known Italian probabilist; Original title: Calcolo delle Probabilità, terza edizione, Zanichelli, ISBN 978-88-08-17676-9.

I discovered that the author defines the cumulative distribution function of a random variable $$X$$ as

$$F(x) = P(X i.e. using strict inequality instead of the usual $$\leq$$.

In this unusual definition (I am not aware of other scholars defining the distribution function this way), the $$F(x)$$ turns out to be left-continuous, though the other properties are the same as those of the usual distribution function. The book, as far as I can tell, fails to motivate such an unusual choice.

Is there a reason to prefer $$\eqref a$$ to the usual definition or, is the choice between the two definitions a matter of taste?

• You are free to work with $-X$ instead of $X.$
– whuber
Nov 21, 2022 at 21:20
• $\Pr(-X \lt x) = \Pr(X \gt -x) = 1 - \Pr(X \le -x) = 1-F_X(-x)$ interchanges the two conventions.
– whuber
Nov 21, 2022 at 21:40
• Nov 22, 2022 at 6:47
• @BCLC when I click it brings me to the top of the page where the first row is the (*) equation, so I guess it works. Nov 23, 2022 at 9:47
• :-O what a prompt response @User1865345 Nov 23, 2022 at 12:42

In Section "1.1 Monotone Functions" of Chung's classic probability textbook A Course in Probability Theory, it is stated:

How can $$f_1$$ and $$f_2$$ differ at all? This can happen only when $$f_1(x)$$ and $$f_2(x)$$ assume different values in the interval $$(f_1(x-), f_1(x+))= (f_2(x-), f_2(x+))$$. It will turn out in Chapter 2 (see in particular Exercise 21 of Sec. 2.2) that the precise value of $$f$$ at a jump is quite unessential for our purposes and may be modified, subject to (3), to suit our convenience. More precisely, given the function $$f$$, we can define a new function $$\tilde{f}$$ in several different ways, such as \begin{align} \tilde{f}(x) = f(x-), \tilde{f}(x) = f(x+), \tilde{f}(x) = \frac{f(x-) + f(x+)}{2}, \end{align} and use one of these instead of the original one. The third modification is found to be convenient in Fourier analysis, but either one of the first two is more suitable for probability theory. We have a free choice between them and we shall choose the second, namely, right continuity.

You can interpret $$f_1(x), f_2(x)$$ in the above quote as $$P(X < x), P(X \leq x)$$ respectively (in the original text, $$f_1$$ and $$f_2$$ are two nondecreasing functions agreed on a dense set $$D \subset (-\infty, +\infty)$$, which are discussed for introducing the probability distribution function aftermath). And as the highlighted text shows, in probability theory, there is no particular reason of preferring $$f_2(x)$$ to $$f_1(x)$$ (or vice versa). The cited "Exercise 21 of Sec 2.2" basically stated that, with some obvious minor modifications on the relationship between $$f_2$$ and $$\mu$$, the fundamental theorem

Each distribution function $$f_2$$ determines a probability measure $$\mu$$ on $$\mathscr{B}^1$$.

still holds for $$f_1$$ and $$\mu$$, supporting the highlighted statement above.

To conclude, my short answer to your question is "yes, it is just a matter of taste" -- I remember that my college probability professor, who learned probability theory from Soviet Union's (where the father of modern probability theory A. N. Kolmogorov is from) literature, told us that in Russian probability texts (e.g., the reference mentioned by @User1865345 in the comment), $$F(x)$$ is defined as $$P(X < x)$$. On the other hand, in most classical English probability texts that I have seen, such as

• A Course in Probability Theory by Kai Lai Chung
• Probability and Measure by Patrick Billingsley
• Probability: Theory and Examples by Rick Durrett
• Probability with Martingales by David Williams

$$F(x)$$ is defined as $$P(X \leq x)$$. One exception is Probability Theory: Independence, Interchangeability, Martingales by Yuan Shih Chow and Henry Teicher, where $$F(x)$$ is defined as $$P[X < x]$$.

• (+1) Indeed, I could not find any good reason for such a choice as the implied properties are quite the same. But, since my knowledge of measure theory is only basic, I suspected I must be missing something. Nov 21, 2022 at 21:21
• The usage indeed prevailed for quite some time in the Soviet literatures. The most prominent that I can remember right now is Theory of Probability by B. V. Gnedenko. Nov 21, 2022 at 21:33
• @User1865345 that's right! Gnedenko seems to define $F$ with strict inequality. Thanks for the edit by the way! Nov 21, 2022 at 21:42
• I am seeing Marek Fisz in his book on Probability Theory and Mathematical Statistics also used this formalism @utobi. Dec 29, 2023 at 14:26

Like others have mentioned the use of $$<$$ instead of $$\leq$$ is just a convention. Here I want to post a remark which is, I believe, interesting enoughy to be placed as an answer.

The convention $$<$$ was used by Kolmogorov in 'Foundations of the Theory of Probability' (see page 15 of this 1956 second edition of the English translation).

The function $$F^{(x)}(a) = P^{(x)}(-\infty,a) = P \{ x where $$-\infty$$ and $$\infty$$ are allowable values of $$a$$ is called the distribution function of the randome variable $$x$$.

It is remarkable that he uses the interval $$(-\infty,a)$$ instead of $$[-\infty,a)$$. Maybe Kolmogorov used the $$<$$ convention, instead of the $$\leq$$ convention, because he didn't want to use an interval with infinity inclusive.

If $$a = \infty$$ then $$F^{(x)}(a) = P^{(x)}(-\infty,\infty)$$ and using $$P^{(x)}(-\infty,\infty]$$ instead would be problematic.

I believe this might be a reason for Kolmogorov's convention, although I would not understand exactly why there would be a problem with including infinity in the interval. Maybe because it is not an element in the sample space?

• That is an interesting take. Hm. Unfortunately it would be hard to trace any explicit reason as to why he stuck to this for $\leq$ wasn't uncommon back then either. Nov 22, 2022 at 7:25
• @User1865345 the reasoning here is explicit. He was defining the distribution in terms of sets $F^{(x)}(a) = P^{(x)}(-\infty,a)$ and the expression $F^{(x)}(a) = P \{ x<a \}$ is derived from that. The use of $<$ instead of $\leq$ stems from the use of $(-\infty,a)$ instead of $[-\infty,a]$. Nov 22, 2022 at 19:48
• That is what I am wondering as to why he chose $[a, b)$ over $(a, b].$ Was there any precedent? Or was it simply just a choice? That is not explicit. Nov 22, 2022 at 23:01
• @User1865345 It is not just a 'choice' on the right side endpoint bracket. $$(-\infty,a) \text{ versus } (-\infty,a]$$ it is also the left side where the bracket was chosen to be non-inclusive. That left side has nothing to do with the sign $<$ or $\leq$. The reasons may have to do with $-\infty$ and $\infty$ not being part of the sample space. The value $F^{(x)}(-\infty) = P^{(x)}(-\infty,-\infty) = 0$, but what about $F^{(x)}(-\infty) = P^{(x)}[-\infty,-\infty)$ or $F^{(x)}(-\infty) = P^{(x)}[-\infty,-\infty]$ ? Nov 22, 2022 at 23:18
• He didn't choose [𝑎,𝑏) over (𝑎,𝑏], he chose (a,b). Nov 22, 2022 at 23:22

A "conceptual consistency criterion" to choose between them per case, is when a value $$x^*$$ in the support is a "threshold", and then it matters whether consequences kick in if one reaches this threshold or if ones exceeds it.

• In the first case (consequences kick in when reaching the threshold), using $$F_X(x) \equiv \Pr(X appears to be the appropriate choice to reflect the conceptual separation of the two "states of the world," since $$\Pr(X describes the probability of "no consequences".
• In the second case (consequences kick in when exceeding the threshold), one would use $$F_X(x) \equiv\Pr(X\leq x)$$, which is the "no-consequences" situation in this case.

This is just conceptual for a continuous r.v., but becomes important when the random variable has jumps or is discrete.