Why Compare CDFs?
A sequence of random variables can change its distribution as its index grows. To describe what those distributions approach, we need a notion of convergence that compares their laws rather than requiring every random variable to be observed on the same sample path. Convergence in distribution does this through cumulative distribution functions (CDFs).
This distinction matters in statistical theory because limiting distributions describe the behavior of statistics across repeated samples. A sequence can have a clear limiting law even when its random variables are defined on different probability spaces. The definition therefore asks a precise question: at which values of \(x\) must the CDFs approach the proposed limit?
Definition Through CDFs
A continuity point here is a real number \(x\) at which the limiting CDF \(F\) is continuous. The condition is imposed on the proposed limit, not on the individual \(F_n\). In particular, it does not require convergence at every real number: values where \(F\) has a jump are excluded from the definition.
The limiting random variable \(X\) need not be defined on the same probability space as the \(X_n\). Only its distribution matters. If two random variables have the same CDF, they represent the same candidate limiting law for this definition, regardless of how either is constructed.
Worked Example: A Discrete Grid Approaches a Uniform Law
For each positive integer \(n\), let \(X_n\) assign probability \(1/n\) to each of the points \(1/n,2/n,\ldots,n/n\). Verify that \(X_n\) converges in distribution to a random variable \(U\) that is uniform on \([0,1]\).
For \(x<0\), none of the possible values of \(X_n\) are at most \(x\), so \(F_n(x)=0\). For \(0\leq x<1\), the number of grid points at most \(x\) is \(\lfloor nx\rfloor\), where \(\lfloor t\rfloor\) denotes the greatest integer not exceeding \(t\). For \(x\geq 1\), all \(n\) grid points are at most \(x\). Thus
The CDF of \(U\) is \(F(x)=0\) for \(x<0\), \(F(x)=x\) for \(0\leq x\leq 1\), and \(F(x)=1\) for \(x>1\). This CDF is continuous at every real number, including 0 and 1. If \(0\leq x<1\), the defining property of the floor function gives
As \(n\) increases, the right-hand bound tends to zero, so \(F_n(x)\to x=F(x)\) for each \(x\) in \([0,1)\). For \(x<0\), both CDFs equal zero; for \(x\geq1\), both equal one. Hence \(F_n(x)\to F(x)\) at every continuity point of \(F\), and \(X_n\xrightarrow{d}U\).
Two Further Checks of the Definition
Worked Example: Normal Distributions with Changing Variance
Let \(X_n\sim N(0,1+1/n)\), and let \(Z\sim N(0,1)\). Check convergence in distribution by writing the CDFs explicitly.
Let \(\Phi\) denote the CDF of \(N(0,1)\). Standardizing \(X_n\) gives, for every real \(x\),
For each fixed \(x\), \(x/\sqrt{1+1/n}\to x\). The function \(\Phi\) is continuous, so
The limiting CDF is continuous everywhere. The required convergence therefore holds at every real \(x\), and \(X_n\xrightarrow{d}Z\). This calculation compares the laws directly; it makes no claim that the variables \(X_n\) and \(Z\) move closer on any particular sample path.
Worked Example: Alternating Point Masses Have No Distributional Limit
Suppose \(X_n=-1\) with probability one when \(n\) is even, and \(X_n=1\) with probability one when \(n\) is odd. Show that this sequence does not converge in distribution to any real-valued random variable.
Assume, to obtain a contradiction, that it converges in distribution to a random variable with CDF \(G\). A CDF has at most countably many discontinuities. Consequently, the interval \((-1,1)\) contains a continuity point \(c\) of \(G\). At that point, the CDF of an even-indexed \(X_n\) equals one, because \(-1\leq c\); the CDF of an odd-indexed \(X_n\) equals zero, because \(1>c\). Thus \(F_n(c)\) alternates between one and zero and has no limit. This contradicts the required convergence at the continuity point \(c\), so no such limiting distribution exists.
Why the Limiting Distribution Is Unique
The definition refers to a proposed limiting CDF. It is useful to know that two different CDFs cannot both satisfy it for the same sequence. The argument uses the fact that a CDF has at most countably many discontinuities, so there are enough points at which both proposed limits are continuous.
To justify the fact about discontinuities used here, recall that a CDF is nondecreasing and right-continuous. Its discontinuity at \(t\), if any, is a jump of size \(F(t)-F(t-)>0\), where \(F(t-)\) is the left-hand limit. For any finite set of distinct jump points, the sum of their jump sizes is at most one: the jumps correspond to disjoint probability masses, and total probability is one. Hence there can be only finitely many jumps of size at least \(1/m\), for each positive integer \(m\). Every positive jump is at least \(1/m\) for some \(m\). The discontinuities are therefore a countable union of finite sets and are at most countable.
The union of the discontinuity sets of \(F\) and \(G\) is countable. Its complement \(D\) is dense in the real line, so every open interval contains points where both CDFs are continuous.
For any \(y\in D\), both assumed convergence statements apply. Since the sequence \(F_n(y)\) has a limit, its limit must equal both \(F(y)\) and \(G(y)\). Thus \(F(y)=G(y)\) for every \(y\in D\).
Fix \(x\). Density of \(D\) allows a sequence \(y_k\in D\) with \(x<y_k<x+1/k\), so \(y_k\downarrow x\) in the sense that \(y_k\to x\) from the right. Right-continuity gives \(F(y_k)\to F(x)\) and \(G(y_k)\to G(x)\). Since \(F(y_k)=G(y_k)\) for every \(k\), it follows that \(F(x)=G(x)\).
Because \(x\) was arbitrary, \(F=G\) everywhere. Therefore, although the definition tests convergence only at continuity points of a proposed limit, it cannot yield two distinct limiting distributions. \(\square\)
Common Pitfall and Practical Interpretation
Wrong reasoning: “The CDFs do not converge at every real number, so the sequence cannot converge in distribution.” This uses a stronger condition than the definition asks for. The definition requires convergence only at continuity points of the limiting CDF. A failure at a point where the proposed limit is discontinuous does not, by itself, settle whether convergence in distribution holds.
Correction: First identify the proposed limiting CDF and its continuity points. Then compare \(F_n(x)\) with \(F(x)\) at each such point. Do not substitute pointwise convergence everywhere for the stated condition, and do not confuse a statement about CDFs with a statement about sample paths. The two normal models above, for example, are compared entirely through their CDFs.
This definition is a practical way to describe an asymptotic law: for each fixed threshold \(x\) where the limiting CDF is continuous, the probability of falling at or below that threshold approaches the corresponding limiting probability. It gives a precise distributional target without requiring the sequence to be realized as an increasingly close set of outcomes on one probability space.
Check Your Understanding
Use the CDF definition to answer the following questions.
- In the definition of convergence in distribution, which CDF determines the points at which convergence must be checked?
- For the discrete grid in the worked example, how many possible values of \(X_n\) are at most a fixed \(x\in[0,1)\)?
- Why does pointwise convergence of the normal CDFs at every real \(x\) establish convergence in distribution in that example?
- For the alternating point masses, why can a proposed limiting CDF not avoid the contradiction by being discontinuous at every point of \((-1,1)\)?
- Why does uniqueness of the limiting CDF imply uniqueness of the limiting distribution?