Why does a C.D.F need to be right-continuous?

Question

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?

Answer 1

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.

Answer 2

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).

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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.

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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).

Answer 3

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.