CDF Convergence at Jumps and Atoms

Why CDF Convergence Has an Exception

Convergence in distribution is described through the cumulative distribution functions (CDFs) of the random variables. The key criterion is pointwise convergence of those CDFs at every continuity point of the limiting CDF. The qualification matters: at a jump of the limit, the CDF values of the sequence need not converge to the value at the jump.

A jump is not a technical nuisance. It records probability concentrated at a single point. When distributions move toward that point from one side, their CDF values at the point can remain on the other side of the jump, even while the distributions converge in distribution. We will make this precise, including bounds on what can happen at a jump.

Definition: Let $X_n$ and $X$ be real-valued random variables with CDFs $F_n$ and $F$. We say $X_n$ converges in distribution to $X$, written $X_n\xrightarrow{d}X$, if $$ F_n(x)\longrightarrow F(x) $$ for every continuity point $x$ of $F$.

The definition does not require convergence at discontinuities of $F$. It also does not say that convergence at a discontinuity is impossible: it may happen for a particular sequence. The point is that convergence in distribution alone does not guarantee it.

Jumps of a CDF Are Atoms

For a real-valued random variable $X$, the CDF is $F(x)=P(X\leq x)$. Its left limit at $x$ is $$ F(x-)=\lim_{y\uparrow x}F(y)=P(X<x). $$ For the last equality, take any sequence $y_k\uparrow x$. The events $\{X\leq y_k\}$ increase to $\{X<x\}$, so continuity of probability for increasing events gives the stated limit.

Consequently the size of the jump at $x$ is $$ F(x)-F(x-)=P(X\leq x)-P(X<x)=P(X=x). $$ A point with $P(X=x)>0$ is called an atom of the distribution. Thus the discontinuities of a CDF are exactly its atoms, and the jump size is the probability assigned to that point.

Worked Example: The CDF of a Point Mass

Let $X=0$ with probability one. Its CDF is $$ F(x)= \begin{cases} 0, & x<0,\\ 1, & x\geq 0. \end{cases} $$ For $x<0$, the event $\{X\leq x\}$ is empty; for $x\geq0$, it has probability one.

At $x=0$, the left limit is $F(0-)=0$, while $F(0)=1$. The jump is therefore $F(0)-F(0-)=1=P(X=0)$. Every $x\neq0$ is a continuity point of $F$; the atom itself is the one point where the CDF is discontinuous.

What Can Happen at a Jump

Although convergence at a jump is not required, the monotonicity of CDFs constrains the values there. The following bounds apply whenever $X_n\xrightarrow{d}X$:

Theorem: If $X_n\xrightarrow{d}X$, with CDFs $F_n$ and $F$, then for every real $x$, $$ F(x-)\ \leq\ \liminf_{n\to\infty}F_n(x) \ \leq\ \limsup_{n\to\infty}F_n(x) \ \leq\ F(x). $$ In particular, at a continuity point of $F$, these bounds force $F_n(x)\to F(x)$. At a jump, the limiting lower and upper bounds may differ.

Proof. A CDF has at most countably many discontinuities. To see this, each discontinuity is a positive jump, and the sum of the jump sizes over any finite set of points is at most 1. For each positive integer $m$, there can therefore be only finitely many jumps larger than $1/m$. Every positive jump is larger than $1/m$ for some $m$, so the set of jumps is at most countable. It follows that continuity points are dense in the real line.

Fix $x$. Choose continuity points $a_k<x<b_k$ such that $a_k\uparrow x$ and $b_k\downarrow x$. By monotonicity, $F_n(a_k)\leq F_n(x)\leq F_n(b_k)$ for every $n$. Since $a_k$ and $b_k$ are continuity points, distributional convergence gives $F_n(a_k)\to F(a_k)$ and $F_n(b_k)\to F(b_k)$. Thus, for each fixed $k$, $$ F(a_k)\leq\liminf_{n\to\infty}F_n(x) \leq\limsup_{n\to\infty}F_n(x)\leq F(b_k). $$ As $k\to\infty$, $F(a_k)\to F(x-)$ and $F(b_k)\to F(x)$, by the left-limit property and right-continuity of a CDF. Taking these limits proves the bounds. At a continuity point, $F(x-)=F(x)$, so the bounds force convergence there. $\square$

This result pinpoints the role of continuity. At a continuity point there is no gap between the lower and upper bounds, so the CDF values must converge to the limiting CDF value. At an atom, the available bounds span an interval of width $P(X=x)$; convergence at that point is not forced.

A Deterministic Sequence Can Miss the Jump

The simplest example is a sequence of constants approaching an atom from the right. A deterministic random variable is one that takes a single value with probability one. Its CDF is a step function, so the comparison at the limiting atom can be computed directly.

Worked Example: Approaching an Atom from the Right

On a probability space with one outcome, define $X_n=1/n$ and $X=0$ for every $n$. The limiting variable has CDF $F(x)=0$ for $x<0$ and $F(x)=1$ for $x\geq0$. For each $n$, $$ F_n(x)= \begin{cases} 0, & x<1/n,\\ 1, & x\geq1/n. \end{cases} $$

If $x<0$, then $x<1/n$ for every $n$, so $F_n(x)=0=F(x)$. If $x>0$, choose $N$ such that $1/N<x$. For every $n\geq N$, $1/n\leq1/N<x$, so $F_n(x)=1=F(x)$. Hence $F_n(x)\to F(x)$ at every continuity point $x\neq0$, and $X_n\xrightarrow{d}X$.

At the atom $x=0$, however, $F_n(0)=0$ for every $n$, whereas $F(0)=1$. The CDF values do not converge at the jump. The sequence still converges in distribution because $0$ is not a continuity point of the limiting CDF.

Worked Example: Approaching the Same Atom from the Left

Now take the deterministic sequence $X_n=-1/n$, again with limiting variable $X=0$. Its CDF is $$ F_n(x)= \begin{cases} 0, & x<-1/n,\\ 1, & x\geq-1/n. \end{cases} $$ For $x<0$, eventually $-1/n>x$: choose $N$ with $1/N<-x$, so for $n\geq N$, $-1/n\leq? $

We need check signs: for x<0 and n large, -1/n > x, hence x < -1/n, giving F_n(x)=0. For x>0, -1/n < x always, yielding 1. At x=0 F_n=1. Complete.

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CDF Convergence at Jumps and Atoms

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