What CDF Convergence Requires
A sequence of random variables can converge in distribution even when its values do not settle down along individual sample paths. The definition compares their probability laws: at locations where the limiting cumulative distribution function is continuous, the probabilities of being at or below that location must converge. The qualification about continuity points is essential, especially when the limit has atoms.
Let \(X_n\), for \(n\ge 1\), be real-valued random variables, and let \(X\) be a real-valued random variable. They may be defined on a common probability space \((\Omega,\mathcal{F},\mathbb{P})\); the definition only uses their distributions, so a common space is not necessary. Write \(F_n(x)=\mathbb{P}(X_n\le x)\) and \(F(x)=\mathbb{P}(X\le x)\) for their cumulative distribution functions (CDFs).
The inequality in the CDF definition is \(\le\), not \(<\). This convention includes any probability mass located exactly at \(x\). At a continuity point, however, there is no mass at \(x\), so the distinction does not affect the limiting CDF's jump behavior.
Proof. For each positive integer \(m\), the events \(\{X\le x-1/m\}\) increase as \(m\) increases, and their union is \(\{X<x\}\). Indeed, if \(X(\omega)<x\), then \(1/m\le x-X(\omega)\) for some sufficiently large \(m\); if \(X(\omega)\le x-1/m\), then \(X(\omega)<x\). Continuity of probability for increasing events gives \(F(x-)=\lim_{m\to\infty}F(x-1/m)=\mathbb{P}(X<x)\). Also \(F(x)=\mathbb{P}(X\le x)\), and \(\{X\le x\}\) is the disjoint union of \(\{X<x\}\) and \(\{X=x\}\). Subtracting proves the jump formula. A CDF is right-continuous, since the events \(\{X\le x+1/m\}\) decrease to \(\{X\le x\}\), so continuity at \(x\) is equivalent to having no left jump there. The jump formula gives the stated equivalence. \(\square\)
This proposition tells us exactly where to test convergence: every point without an atom in the limiting distribution is a continuity point, while points carrying positive limiting mass are excluded from the definition. Values at those excluded points can still converge, but they do not have to.
Three CDF Calculations
Worked Example: A Discrete Limiting Distribution
Let \(U\) be uniform on \([0,1]\), with Lebesgue probability measure, and define \(p_n=\tfrac13+\tfrac{1}{6n}\). Set \(X_n=0\) when \(U\le p_n\) and \(X_n=2\) otherwise. Define \(X=0\) when \(U\le\tfrac13\) and \(X=2\) otherwise. First, \(p_n\) is a valid probability: for \(n\ge1\), \(\tfrac13<p_n\le\tfrac12<1\), so both possible outcomes have nonnegative probability.
The CDFs, including their endpoint conventions, are $$ F_n(x)= \begin{cases} 0,&x<0,\\ p_n,&0\le x<2,\\ 1,&x\ge2, \end{cases} \qquad F(x)= \begin{cases} 0,&x<0,\\ \tfrac13,&0\le x<2,\\ 1,&x\ge2. \end{cases} $$ The limiting CDF has jumps at \(0\) and \(2\), so its continuity points are all \(x<0\), all \(0<x<2\), and all \(x>2\). At \(x<0\), \(F_n(x)=F(x)=0\); at \(0<x<2\), \(F_n(x)=\tfrac13+\tfrac{1}{6n}\to\tfrac13=F(x)\); and at \(x>2\), both CDFs equal \(1\). Thus \(X_n\xrightarrow{d}X\).
Check: \(F_n(1)=p_n=\tfrac13+\tfrac{1}{6n}\), and \(1\) is a continuity point of \(F\). For \(n=2\), \(F_2(1)=\tfrac13+\tfrac{1}{12}=\tfrac5{12}\); in general, \(F_n(1)=\tfrac13+\tfrac{1}{6n}\to\tfrac13\) as \(n\to\infty\). The limiting CDF assigns probability \(\tfrac13\) to \(0\) and \(\tfrac23\) to \(2\), which sum to \(1\).
Worked Example: Discrete Approximations to a Continuous Limit
Let \(U\) be uniform on \((0,1)\) with Lebesgue probability measure, and set \(X_n=\lceil nU\rceil/n\), where \(\lceil y\rceil\) is the smallest integer greater than or equal to \(y\). Each \(X_n\) takes values in \(\{1/n,2/n,\ldots,1\}\). We show that \(X_n\) converges in distribution to \(U\).
For \(0\le x<1\), the event \(X_n\le x\) is equivalent to \(\lceil nU\rceil\le nx\). Since the left side is an integer, this is equivalent to \(\lceil nU\rceil\le\lfloor nx\rfloor\), which, up to endpoints of probability zero, means \(U\le\lfloor nx\rfloor/n\). Hence $$ F_n(x)= \begin{cases} 0,&x<0,\\ \lfloor nx\rfloor/n,&0\le x<1,\\ 1,&x\ge1. \end{cases} $$ The limiting CDF is \(F(x)=0\) for \(x<0\), \(F(x)=x\) for \(0\le x\le1\), and \(F(x)=1\) for \(x>1\). This is continuous at every real \(x\). For \(0\le x<1\), the defining property of the floor gives \(0\le x-\lfloor nx\rfloor/n<1/n\). Therefore \(F_n(x)\to x=F(x)\); for \(x<0\) and \(x\ge1\), the CDFs agree exactly. This proves \(X_n\xrightarrow{d}U\).
Check: At \(n=4\) and \(x=\tfrac38\), \(nx=\tfrac32\), so \(\lfloor nx\rfloor=1\) and \(F_4(\tfrac38)=\tfrac14\). The limiting value is \(F(\tfrac38)=\tfrac38\), and the difference is \(\tfrac38-\tfrac14=\tfrac18<\tfrac14=1/n\), as the bound predicts.
Worked Example: Convergence to a Degenerate Distribution
Fix \(c\in\mathbb{R}\), take a probability space consisting of a single point of probability \(1\), and let \(X_n=c+1/n\) and \(X=c\) on that space. The limiting distribution is degenerate: it places probability \(1\) at \(c\), so its CDF is \(F(x)=0\) for \(x<c\) and \(F(x)=1\) for \(x\ge c\). It is discontinuous at \(c\) and continuous everywhere else.
For any fixed \(x<c\), \(X_n=c+1/n>x\) for every \(n\), so \(F_n(x)=0=F(x)\). For any fixed \(x>c\), choose an integer \(N\) with \(1/N\le x-c\). For \(n\ge N\), \(c+1/n\le x\), and therefore \(F_n(x)=1=F(x)\). These are all the continuity points of \(F\), so \(X_n\xrightarrow{d}X\).
Check at the excluded point: At \(x=c\), \(F_n(c)=\mathbb{P}(c+1/n\le c)=0\) for every \(n\), whereas \(F(c)=1\). This does not contradict convergence in distribution because \(c\) is a discontinuity point of the limiting CDF. The calculation also illustrates why one must identify the continuity points before checking CDF values.
A Useful Consequence for Interval Probabilities
Proof. For any real-valued random variable \(Y\), the event \(\{a<Y\le b\}\) is the difference of the nested events \(\{Y\le b\}\) and \(\{Y\le a\}\). Thus \(\mathbb{P}(a<X_n\le b)=F_n(b)-F_n(a)\). Since \(a\) and \(b\) are continuity points, the definition gives \(F_n(b)\to F(b)\) and \(F_n(a)\to F(a)\). Subtracting these convergent real sequences yields \(F_n(b)-F_n(a)\to F(b)-F(a)\). Applying the same event identity to \(X\) gives \(F(b)-F(a)=\mathbb{P}(a<X\le b)\), as required. \(\square\)
The endpoints matter. This proposition uses the interval \((a,b]\), matching the CDF convention \(\mathbb{P}(Y\le x)\). If an endpoint is an atom of the limit, it is not a continuity point, and this argument does not establish convergence for an interval using that endpoint. In applications, choose endpoints where the limiting CDF is continuous whenever possible.
Common Pitfalls and Why the Definition Matters
Common mistake: demanding convergence at every real number. In the degenerate example, a reader might calculate \(F_n(c)=0\) and \(F(c)=1\), then conclude that \(X_n\) does not converge in distribution to \(X\). The calculation is correct, but the conclusion is not: \(F\) jumps at \(c\), so the definition makes no demand there. The correct check is to verify convergence at each continuity point, as done for \(x<c\) and \(x>c\).
Another common mistake: confusing a discrete approximation with a discrete limit. In the uniform example, every \(X_n\) is discrete, but its limiting CDF is continuous. The floor calculation verifies convergence at every real number, not only at the support points of \(X_n\). Conversely, for a discrete limit, its jump locations must be recognized before applying the definition. The first example has no requirement at \(0\) or \(2\), even though convergence happens to hold at both.
Distributional convergence is useful when the target is a law rather than a pointwise value: discrete approximations may approach a continuous law, and random variables may approach a point mass in distribution. The definition gives a precise way to make those statements without asserting that sample paths converge or that moments converge. Each of those is a different claim and requires its own conditions.
Check Your Understanding
For each question, identify the limiting CDF's continuity points before testing convergence.
- A sequence has CDF \(F_n(x)=0\) for \(x<0\), \(F_n(x)=\tfrac12+\tfrac{1}{4n}\) for \(0\le x<1\), and \(F_n(x)=1\) for \(x\ge1\). What is the limiting CDF, and at which points must convergence be checked?
- Why does the CDF of a point mass at \(c\) fail to be continuous at \(c\)? State the jump size.
- In the uniform approximation example, calculate \(F_5(\tfrac{7}{10})\) and compare it with the limiting CDF value.
- If \(a\) and \(b\) are continuity points of \(F\), write the CDF expression for \(\mathbb{P}(a<X_n\le b)\) and state its limit.
- For \(X_n=c+1/n\) and \(X=c\), calculate \(F_n(c)\) and \(F(c)\). Explain why their difference does not invalidate convergence in distribution.