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As you may know, if $(\Omega,\mathcal{F},\mathbb{P})$ is a probability space and $X\colon\Omega\to\mathbb{R}^k$ is a random variable, then the cumulative distribution function of $X$ is defined as $$F_X(a):=\mathbb{P}(\{\omega\in\Omega\,\colon X(\omega)\leq a\})=\mathbb{P}\Big(X^{-1}\Big(\prod_{i\in [k]}(-\infty,a_i]\Big)\Big),\,\text{ for each } a\in\mathbb{R}^k.$$

This function is always right-continuous. That is, for each $x\in\mathbb{R}^k$ we have $\lim_{a\downarrow x}F_X(a)=F_X(x)$.

My question is: Why is this property important? Is there any capital result in probability theory that depends on it?

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    $\begingroup$ In an alternative history, c.d.f.'s might have been defined as $F_X(a)=\mathbb P(\{\omega\in\Omega:X(\omega)<a\})$ with strict inequality, and then these functions would be continuous from the left rather than from the right. As far as I can see, the choice between the standard definition and this alternative one is purely a matter of convention. $\endgroup$ – Andreas Blass May 10 at 19:54
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    $\begingroup$ Possibly interesting discussion: Why aren't CDFs left-continuous? $\endgroup$ – Minus One-Twelfth May 10 at 19:56
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    $\begingroup$ Both the standard convention and @AndreasBlass's alternative history continuous-on-the-left convention suffer from this defect: if $Y=-X$, the cdf $F_X(x)$ is not simply $1-F_Y(-x)$. Some older authors (G.H. Hardy and D.V.Widder, eg) adopt a different convention: at jump discontinuities, $F(x)=(F(x-)+F(x+))/2$. $\endgroup$ – kimchi lover May 10 at 20:17
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Well, in a finite measure (by which I mean a finite $\sigma$-additive measure) space, if $\{A_i\}_{i\in\Bbb N}$ is a sequence of measurable sets such that $A_i\supseteq A_{i+1}$ for all $i$, then $\mu\left(\bigcap_{n\in\Bbb N}A_i\right)=\inf_{n\in\Bbb N} \mu(A_i)=\lim_{n\to\infty} \mu(A_i)$. In your special case where all the $A_i$-s are hyperrectangle in the form $R\left(a^{(i)}\right)=\left(-\infty,a^{(i)}_1\right]\times\cdots\times\left(-\infty, a^{(i)}_k\right]$ and $\mu=\Bbb P_X$, this translates to $$\mathbb P_X\left(R(a)\right)=\mathbb P_X\left(\bigcap_{i\in\Bbb N} R\left(a^{(i)}\right)\right)=\lim_{n\to\infty} \mathbb P_X\left(R\left(a^{(i)}\right)\right)$$ for all $a^{(i)}\searrow a$. Which is in fact continuity on the right of the CDF.

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  • $\begingroup$ I see you answered while I was typing mine. Nevertheless there is some non-overlapping content in the answers so I will keep mine also. $\endgroup$ – Michael May 10 at 20:23
  • $\begingroup$ @Michael no problem, it happens all the time. $\endgroup$ – Saucy O'Path May 10 at 20:24
  • $\begingroup$ Yep, totally agree. This is just how I proved it. Thanks $\endgroup$ – Ariel Serranoni May 10 at 21:13
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This can be proven from the "continuity of probability" result for events that shrink to a limiting event: $$A_n\searrow A \implies P[A_n]\rightarrow P[A]$$ (and this is derived from the countable additivity axiom).


One reason this is important is that it helps students to be precise when they draw pictures of CDF functions. They need to learn to be detail-oriented enough to respect this issue when points of discontinuity arise.

Another reason for importance is that it relates to this question:

Question: "What functions $F:\mathbb{R}\rightarrow\mathbb{R}$ are valid CDF functions?"

Answer: A function $F(x)$ is a valid CDF, meaning that there exists a random variable $X$ for which $P[X\leq x] = F(x)$ for all $x \in \mathbb{R}$, if and only if these four criteria are satisfied:

  • $F(x)$ is nondecreasing.
  • $\lim_{x\rightarrow-\infty} F(x) = 0$.
  • $\lim_{x\rightarrow\infty} F(x)=1$.
  • $F(x)$ is right-continuous.

So the right-continuous property has a place of prominence in this fundamental question.


This fact is useful to resolve this natural question: Let $\{X_i\}_{i=1}^{\infty}$ be i.i.d. random variables uniform over $[-1,1]$. Define $$ L_n = \frac{1}{n}\sum_{i=1}^n X_i \quad \forall n \in \{1, 2, 3, ...\}$$ Does there exist a random variable $Y$ for which the distribution of $L_n$ converges to the distribution of $Y$? The answer is "no" because: $$ \lim_{n\rightarrow\infty} P[L_n\leq x] = \left\{\begin{array}{ll} 1 & \mbox{ if $x >0$}\\ 1/2 & \mbox{ if $x=0$}\\ 0 & \mbox{ if $x<0$} \end{array}\right.$$ and, because this is not right-continuous, this is not a valid CDF function for any random variable.

Of course, the CDF of the always-zero random variable $0$ is the right-continuous unit step function, which differs from the above function only at the point of discontinuity at $x=0$. Such issues are the reason why the definition of "$Y_n\rightarrow Y$ in distribution" has the caveat that the convergence $P[Y_n\leq y] \rightarrow P[Y\leq y]$ only needs to take place at points $y$ where $P[Y\leq y]$ is continuous. With this caveat in mind, it is correct to say that $L_n\rightarrow 0$ in distribution (and of course we also know $L_n\rightarrow 0$ with probability 1 by the law of large numbers).

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  • $\begingroup$ I suppose this raises the natural question "which functions $h(x)$ can be pointwise limits of distributions $\lim_{n\rightarrow\infty}P[Y_n\leq y]$"? In particular, is it possible to construct an example that replaces the "$1/2$ if $x=0$" in the above example with "$1/3$ if $x=0$"? $\endgroup$ – Michael May 10 at 20:42
  • $\begingroup$ I suppose the central limit theorem can help answer the question in the special case when $Y_n$ is constructed from i.i.d. random variables with finite mean and variance, in that case the $1/2$ is unavoidable. $\endgroup$ – Michael May 10 at 20:49
  • $\begingroup$ I (think) the answer to the general $h(x)$ question above is "any $h(x)$ function that is nondecreasing and satisfies $0\leq h(x)\leq 1$ for all $x \in \mathbb{R}$." Now if $Y_n = \frac{1}{n}\sum_{i=1}^n X_i$ for $\{X_i\}$ iid with zero mean but infinite variance, I don't know if the $1/2$ is fundamental anymore. $\endgroup$ – Michael May 10 at 20:58
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    $\begingroup$ Thanks a lot for the example!! $\endgroup$ – Ariel Serranoni May 10 at 21:28
  • $\begingroup$ I asked a question about my $1/2$ question in the comment above, and it was answered, see here: math.stackexchange.com/questions/3221659/… $\endgroup$ – Michael May 14 at 14:02
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It doesn't "have" to be. A distribution function is defined either as $$F_X(x)=\mathbb{P}_X((-\infty,x])=\mathbb{P}(X\leq x)$$

Then it is right continuous (follows from continuity of measures from above). It could be defined as $$F_X(x)=\mathbb{P}_X((-\infty,x))=\mathbb{P}(X<x)=1-\mathbb{P}(X\geq x)$$ Then it is left continuous, which again follows from continuity of measures.

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