Distribution-Function Convergence in Discrete and Continuous Examples

Distribution Functions as a Weak-Convergence Test

For probability measures on the real line, a distribution function turns a measure into a nondecreasing step-shaped or continuous function. Comparing these functions can make weak convergence concrete, especially when the laws have atoms that move with the index.

The important qualification is that the limiting distribution function may jump. Weak convergence requires convergence at its continuity points; it does not require convergence at every real number. In the examples below, locating those exceptional points is part of the calculation, not an afterthought.

Definition: If $\mu$ is a probability measure on $\mathbb{R}$, its distribution function is the function $F_\mu:\mathbb{R}\to[0,1]$ defined by $$ F_\mu(x)=\mu((-\infty,x]). $$ A sequence $(\mu_n)$ converges weakly to $\mu$, written $\mu_n\Rightarrow\mu$, if $\int f\,d\mu_n\to\int f\,d\mu$ for every bounded continuous function $f:\mathbb{R}\to\mathbb{R}$.

We use the standard distribution-function criterion: $\mu_n\Rightarrow\mu$ if and only if $F_{\mu_n}(x)\to F_\mu(x)$ at every continuity point $x$ of $F_\mu$. This criterion is a convenient form of the earlier material on weak convergence and continuity sets. Its emphasis on continuity points is crucial: an atom of $\mu$ produces a jump, and convergence at that jump need not hold.

For a point mass $\delta_a$, the distribution function is

F_{\delta_a}(x)= \begin{cases} 0, & x<a,\\ 1, & x\ge a. \end{cases}

It is a step function with a jump at $a$. More generally, the distribution function of a finite atomic law is a cumulative sum: as $x$ passes an atom, the function increases by that atom’s probability.

A Moving Point Mass

Worked Example: A Point Mass Moving Toward Zero

Let $\mu_n=\delta_{1/n}$ and propose $\mu=\delta_0$. For a fixed $x$, the distribution function of $\mu_n$ is

F_{\mu_n}(x)= \begin{cases} 0, & x<1/n,\\ 1, & x\ge 1/n. \end{cases}

The proposed limit has $F_\mu(x)=0$ for $x<0$ and $F_\mu(x)=1$ for $x\ge0$. If $x<0$, then $x<1/n$ for every $n$, so $F_{\mu_n}(x)=0=F_\mu(x)$. If $x>0$, choose $N$ with $1/N<x$. For every $n\ge N$, $1/n\le1/N<x$, so $F_{\mu_n}(x)=1=F_\mu(x)$. Thus convergence holds at every $x\ne0$, precisely the continuity points of $F_\mu$.

At the jump, however, $F_{\mu_n}(0)=0$ for every $n$, while $F_\mu(0)=1$. The failure at zero does not prevent weak convergence: the criterion does not require convergence at a discontinuity of the limiting distribution function. Therefore $\delta_{1/n}\Rightarrow\delta_0$.

Proposition: If $a_n\to a$ in $\mathbb{R}$, then $\delta_{a_n}\Rightarrow\delta_a$.

Proof. Let $f:\mathbb{R}\to\mathbb{R}$ be bounded and continuous. Integration against a point mass evaluates the function at its location, so $$ \int f\,d\delta_{a_n}=f(a_n) \qquad\text{and}\qquad \int f\,d\delta_a=f(a). $$ Since $a_n\to a$ and $f$ is continuous at $a$, $f(a_n)\to f(a)$. This holds for every bounded continuous $f$, which is exactly $\delta_{a_n}\Rightarrow\delta_a$. $\square$

The proposition gives the same conclusion as the first example without requiring a separate case calculation for each $x$. The distribution-function calculation remains useful, though: it makes the limiting jump and the possible failure at its location explicit.

Several Moving Atoms

Worked Example: Two Atoms with Different Locations

For $n\in\mathbb{N}$, set $$ \mu_n=\frac{2}{5}\delta_{1/n}+\frac{3}{5}\delta_{2+1/n}, \qquad \mu=\frac{2}{5}\delta_0+\frac{3}{5}\delta_2. $$ The proposed limiting distribution function has jumps at $0$ and $2$:

F_\mu(x)= \begin{cases} 0, & x<0,\\ 2/5, & 0\le x<2,\\ 1, & x\ge2. \end{cases}

For each $n$, $F_{\mu_n}(x)$ is $0$ below $1/n$, $2/5$ from $1/n$ up to but not including $2+1/n$, and $1$ at and above $2+1/n$. Fix $x$ at a continuity point of $F_\mu$, so $x\ne0,2$. If $x<0$, it lies below both atoms, giving $F_{\mu_n}(x)=0$. If $0<x<2$, then for all sufficiently large $n$, $1/n\le x<2+1/n$, giving $F_{\mu_n}(x)=2/5$. If $x>2$, then for all sufficiently large $n$, both atom locations are at most $x$, giving $F_{\mu_n}(x)=1$. These are the corresponding values of $F_\mu(x)$, so the criterion gives $\mu_n\Rightarrow\mu$.

At $x=0$, the sequence of distribution functions is eventually zero, whereas $F_\mu(0)=2/5$. At $x=2$, it is eventually $2/5$, whereas $F_\mu(2)=1$. Both mismatches occur exactly at jumps of the limit.

This example illustrates a reliable procedure for finite atomic laws: write down the ordered atom locations and their cumulative weights, then check fixed points on either side of each limiting jump. A location that moves toward a limiting atom can change which side of the jump contains a given test point, but for a fixed continuity point that side eventually stabilizes.

Mixtures and Their Distribution Functions

A mixture chooses between probability laws according to specified weights. If $\mu$ and $\nu$ are probability measures and $p\in[0,1]$, then $p\mu+(1-p)\nu$ assigns to each measurable set $A$ the probability $p\mu(A)+(1-p)\nu(A)$. Its distribution function is the same weighted average of the component distribution functions:

F_{p\mu+(1-p)\nu}(x) = pF_\mu(x)+(1-p)F_\nu(x).

Theorem: Suppose $\mu_n\Rightarrow\mu$, $\nu_n\Rightarrow\nu$, and $p_n\to p$ with $p_n,p\in[0,1]$. Then $$ p_n\mu_n+(1-p_n)\nu_n \Rightarrow p\mu+(1-p)\nu. $$

Proof. Let $f:\mathbb{R}\to\mathbb{R}$ be bounded and continuous. By the definition of a mixture and linearity of integration, $$ \int f\,d\bigl(p_n\mu_n+(1-p_n)\nu_n\bigr) = p_n\int f\,d\mu_n+(1-p_n)\int f\,d\nu_n. $$ Weak convergence gives $\int f\,d\mu_n\to\int f\,d\mu$ and $\int f\,d\nu_n\to\int f\,d\nu$. Since $p_n\to p$, and each integral is bounded in absolute value by $\sup_x|f(x)|$, taking limits in the displayed identity gives $$ \lim_{n\to\infty}\int f\,d\bigl(p_n\mu_n+(1-p_n)\nu_n\bigr) = p\int f\,d\mu+(1-p)\int f\,d\nu. $$ The right side is $\int f\,d(p\mu+(1-p)\nu)$. The equality holds for every bounded continuous $f$, proving the stated weak convergence. This argument also covers $p=0$ and $p=1$. $\square$

Worked Example: A Moving Atom Mixed with a Uniform Law

Let $U$ have the uniform distribution on $[0,1]$, and define $$ \mu_n=\frac{1}{3}\delta_{1/2+1/n}+\frac{2}{3}\mathcal{L}(U), \qquad \mu=\frac{1}{3}\delta_{1/2}+\frac{2}{3}\mathcal{L}(U), $$ where $\mathcal{L}(U)$ denotes the law of $U$. Write $G$ for the uniform distribution function:

G(x)= \begin{cases} 0, & x<0,\\ x, & 0\le x\le1,\\ 1, & x>1. \end{cases}

The mixture formula gives $$ F_{\mu_n}(x)=\frac{2}{3}G(x)+\frac{1}{3}\mathbf{1}_{\{x\ge1/2+1/n\}}, \qquad F_\mu(x)=\frac{2}{3}G(x)+\frac{1}{3}\mathbf{1}_{\{x\ge1/2\}}. $$ The uniform component is continuous everywhere, while the atom makes $F_\mu$ discontinuous only at $x=1/2$. For any fixed $x<1/2$, the indicator terms are both zero. For any fixed $x>1/2$, $1/2+1/n\le x$ eventually, so both indicators are eventually one. Consequently $F_{\mu_n}(x)\to F_\mu(x)$ at every required continuity point.

At $x=1/2$, $F_{\mu_n}(1/2)=\frac{2}{3}G(1/2)=1/3$ for every $n$, while $F_\mu(1/2)=1/3+1/6=1/2$. This is again a permitted discrepancy at the atom. The mixture theorem also gives the convergence directly, since the uniform component is fixed and the moving point masses converge.

A Continuous Limit: Every Point Must Be Checked

Worked Example: Uniform Distributions Shifted by a Vanishing Amount

Let $\mu_n$ be the law of $U+1/n$, where $U$ is uniform on $[0,1]$, and let $\mu$ be the law of $U$. For a fixed real $x$, $$ F_{\mu_n}(x)=\mathbb{P}(U+1/n\le x)=G(x-1/n), \qquad F_\mu(x)=G(x). $$ The function $G$ is continuous at every real number, including $0$ and $1$. Since $x-1/n\to x$, continuity gives $G(x-1/n)\to G(x)$ for every fixed $x\in\mathbb{R}$. Thus the distribution functions converge at all points, and in particular at all continuity points of the limit; hence $\mu_n\Rightarrow\mu$.

For instance, if $x<0$, both functions are eventually zero; if $x>1$, both are eventually one. For $0<x<1$, the values are eventually $x-1/n$ and $x$, respectively. At the endpoints, $G(-1/n)=0\to G(0)=0$, and $G(1-1/n)=1-1/n\to G(1)=1$. This also verifies explicitly why the shifting boundary causes no jump in the limiting law.

Why Continuity Points Matter

The distribution-function criterion is not a rule to compare values indiscriminately. First identify the limiting law; next find the discontinuities of its distribution function; then check convergence at every other point. For a purely continuous limiting law, every real number is a continuity point, so no points can be omitted. For a law with atoms, its jump locations are exactly where the criterion does not impose pointwise convergence.

A common mistake is to see a mismatch at an atom and conclude that weak convergence fails. The moving-point-mass example rules out that conclusion. The opposite mistake is to check only a few convenient points and overlook another continuity point where convergence fails. The test requires all continuity points, not merely a selected collection.

In mixtures, the jump structure can be read from the components: an atom contributes a jump of size equal to its mixture weight times its mass, while a continuous component contributes no jump. The displayed examples show why the location and size of a jump should be calculated from the proposed limiting law before comparing the sequence.

Check Your Understanding

Use the distribution-function criterion, and identify the continuity points of each proposed limiting distribution before drawing a conclusion.

For $\delta_{-2/n}$, what is the proposed weak limit, and at which real number can convergence of the distribution functions fail without violating the criterion?
Let $\mu_n=\frac{1}{4}\delta_{1/n}+\frac{3}{4}\delta_3$. Write the proposed limiting law and list its discontinuity points.
For a mixture of an atom at $a$ and a uniform law on $[0,1]$, when does the mixture distribution function have a jump?
Why is checking a finite number of continuity points not enough to establish convergence by the distribution-function criterion?
If $\mu_n\Rightarrow\mu$ and $\nu_n\Rightarrow\nu$, what additional condition on the mixing weights ensures convergence of the corresponding mixtures?