Convergence in Distribution Through Cumulative Distribution Functions

Why Compare Cumulative Distribution Functions?

In asymptotic work, the distribution of a statistic may change with sample size. A central question is whether those changing distributions approach a fixed probability law as the sample size grows. Convergence in distribution gives a precise answer by comparing cumulative distribution functions (CDFs). It is useful because the limiting law can describe the statistic's approximate behavior even when the statistic and its limit are not equal for any finite sample size.

The definition is carefully restricted to points where the limiting CDF is continuous. That qualification matters: a discrete distribution has jumps, and convergence in distribution does not require the CDFs in the sequence to converge at those jumps. The examples below show how to apply the definition for both continuous and discrete limits.

Definition: The cumulative distribution function of a random variable $X$ is $F(x)=P(X\leq x)$. A sequence of random variables $X_n$ converges in distribution to a random variable $X$ if $$ F_n(x)\longrightarrow F(x) $$ for every continuity point $x$ of $F$, where $F_n(x)=P(X_n\leq x)$ and $F(x)=P(X\leq x)$. The notation is $X_n\xrightarrow{d}X$.

A continuity point is a value $x$ at which the CDF has no jump: the probability that $X$ equals $x$ is zero. At a jump, $P(X=x)>0$, so the CDF's value just before $x$ differs from its value at $x$. A continuous distribution has no jumps, so its CDF is continuous everywhere. A discrete distribution typically has jumps at its possible values.

The random variable $X$ in the definition identifies the limiting distribution. The sequence $X_n$ and $X$ do not need to be defined on the same probability space; the condition is about their CDFs. The definition concerns distributional behavior, not whether individual realizations of $X_n$ approach a particular realization of $X$.

How the Continuity-Point Condition Works

At any continuity point $x$ of $F$, the condition asks for the probability $P(X_n\leq x)$ to approach $P(X\leq x)$. It does not ask for equality at finite $n$, and it does not require convergence at every real number when $F$ has jumps. For a continuous limit, however, every real number is a continuity point, so the CDFs must converge everywhere.

CDF convergence also gives a useful consequence for intervals. If $a$ and $b$ are continuity points of $F$, with $a<b$, then

P(a<X_n\leq b)=F_n(b)-F_n(a) \longrightarrow F(b)-F(a)=P(a<X\leq b).

This follows by subtracting the two CDF limits. Thus, probabilities for intervals whose endpoints are continuity points converge to the corresponding limiting probabilities. The endpoints matter: if an endpoint is a jump point, this direct argument does not establish the same conclusion.

Worked Example: A Sequence Approaching a Continuous Normal Limit

Suppose $Z\sim N(0,1)$, and define $X_n=Z+1/n$. This is a hypothetical sequence with a small location shift that shrinks as $n$ increases. Determine whether $X_n$ converges in distribution to $Z$.

For any real $x$, $$ F_n(x)=P(X_n\leq x) =P(Z+1/n\leq x) =P(Z\leq x-1/n) =\Phi(x-1/n), $$ where $\Phi$ is the standard normal CDF. The proposed limiting CDF is $F(x)=\Phi(x)$. Since $\Phi$ is continuous at every real number and $x-1/n\to x$, $$ \lim_{n\to\infty}F_n(x) =\lim_{n\to\infty}\Phi(x-1/n) =\Phi(x)=F(x) $$ for every $x$. Therefore $X_n\xrightarrow{d}Z$.

For a numerical check, at $x=0.5$ and $n=10$, $F_{10}(0.5)=\Phi(0.4)\approx 0.6554$. The limiting CDF there is $F(0.5)=\Phi(0.5)\approx 0.6915$. The finite-$n$ probability is not yet equal to its limit, but the formula shows that it approaches the limit as $n$ grows.

Discrete Limiting Distributions

For a discrete limit, the CDF is flat between possible values and jumps at those values. The definition still applies, but only at continuity points. This is not a weaker version of the same pointwise requirement everywhere: it is the standard criterion for distributional convergence, including when the limit has atoms.

Worked Example: Convergence to a Bernoulli Distribution

Let $X_n$ be Bernoulli with success probability $p_n=0.4+0.1/n$. Thus $X_n=1$ with probability $p_n$ and $X_n=0$ with probability $1-p_n$. The probabilities are valid for every $n\geq1$, since $0.4<p_n\leq0.5$. Determine the limiting distribution.

The CDF is $$ F_n(x)= \begin{cases} 0, & x<0,\\ 1-p_n, & 0\leq x<1,\\ 1, & x\geq1. \end{cases} $$ As $n\to\infty$, $p_n\to0.4$. The CDF of a Bernoulli random variable $X$ with success probability $0.4$ is $$ F(x)= \begin{cases} 0, & x<0,\\ 0.6, & 0\leq x<1,\\ 1, & x\geq1. \end{cases} $$

The continuity points of $F$ are all real numbers except $0$ and $1$, where it jumps. If $x<0$, then $F_n(x)=0=F(x)$. If $0<x<1$, then $$ F_n(x)=1-p_n=1-(0.4+0.1/n)=0.6-0.1/n\longrightarrow0.6=F(x). $$ If $x>1$, then $F_n(x)=1=F(x)$. Therefore the CDFs converge at every continuity point, and $X_n\xrightarrow{d}X$, with $X$ Bernoulli with success probability $0.4$.

In this example, the CDFs happen to converge at the two jump points as well: $F_n(0)=1-p_n\to0.6=F(0)$ and $F_n(1)=1\to1=F(1)$. The definition does not require this extra convergence. A different sequence can converge in distribution while failing to converge at a jump.

Worked Example: A Jump Point Where CDF Convergence Fails

Let $X_n=1/n$ with probability one, and let $X=0$ with probability one. These are deterministic random variables. Check convergence in distribution, paying particular attention to $x=0$.

The limiting variable has CDF $F(x)=0$ for $x<0$ and $F(x)=1$ for $x\geq0$. It jumps at zero, so its continuity points are all $x\ne0$. For a fixed $x<0$, $F_n(x)=0=F(x)$, since $1/n>0$. For a fixed $x>0$, once $n$ is large enough that $1/n\leq x$, $F_n(x)=1=F(x)$. Thus $F_n(x)\to F(x)$ at every continuity point.

At $x=0$, however, $F_n(0)=P(1/n\leq0)=0$ for every $n$, while $F(0)=1$. This failure does not violate the definition because zero is a discontinuity point of the limiting CDF. Hence $X_n\xrightarrow{d}X$, even though the CDFs do not converge at zero.

This example also distinguishes convergence in distribution from exact agreement on probabilities at particular points. The mass of $X_n$ is at $1/n$, while the limiting mass is at zero; the locations approach each other, but the CDF values at the jump itself remain different.

A Short Justification: Why Interval Probabilities Converge

The interval consequence follows directly from the definition, without requiring a separate limit theorem. Take fixed $a<b$ that are both continuity points of $F$. The identity

P(a<X_n\leq b)=P(X_n\leq b)-P(X_n\leq a)=F_n(b)-F_n(a)

holds because the event $\{X_n\leq a\}$ is contained in $\{X_n\leq b\}$. By convergence in distribution, $F_n(a)\to F(a)$ and $F_n(b)\to F(b)$. Subtracting convergent real sequences gives

F_n(b)-F_n(a)\longrightarrow F(b)-F(a) =P(a<X\leq b).

The continuity-point restriction is used at both endpoints to invoke the definition. In particular, this argument should not be used to claim convergence of interval probabilities when an endpoint is a jump of the limiting CDF.

What the Limit Does—and Does Not—Say

Convergence in distribution describes how the entire CDF behaves at fixed thresholds. A statement such as $X_n\xrightarrow{d}X$ means that, for every continuity point $x$ of $F$, the chance that $X_n$ is at most $x$ approaches the corresponding chance for $X$. It does not say that $X_n=X$, that the two variables have the same distribution for finite $n$, or that their realized values must be close on a shared probability space.

The limit can be continuous or discrete. When it is continuous, checking the CDF limit at each fixed real $x$ verifies the definition. When it is discrete, first identify its jump points and check convergence at every other point. Convergence at jumps may occur, as in the Bernoulli example, but it is not part of the requirement.

The earlier material on convergence in probability introduced a different way to describe a sequence approaching a fixed value in probability. Keep the concepts distinct: this tutorial's definition is based on CDFs and permits a nondegenerate limiting distribution. The definition alone should not be used to assert convergence of means or other numerical summaries; those require additional conditions.

Common Mistakes

Checking only the jump points. A discrete CDF has jumps, but those points are precisely where the definition does not require convergence. Check every continuity point instead.
Demanding convergence at every real number. That is required when the limiting CDF is continuous everywhere, but not for a limit with jumps.
Confusing a limiting distribution with a limiting value. The limit may be a random variable with a nondegenerate distribution, not a single constant.
Concluding that finite-sample distributions equal the limit. Convergence means the CDF values approach their limits as $n$ grows; it does not imply equality at any particular finite $n$.
Inferring convergence of expectations from CDF convergence alone. Distributional convergence does not by itself guarantee that means converge. Additional assumptions are needed.

Key takeaway: To verify $X_n\xrightarrow{d}X$, write down the CDFs and show $F_n(x)\to F(x)$ at every continuity point of the limiting CDF. For a discrete limit, jumps are excluded from the required check; for a continuous limit, every real point must be checked.

Check Your Understanding

Use the continuity-point definition to answer the following questions.

What does $X_n\xrightarrow{d}X$ require of $F_n(x)$ and $F(x)$?
A limiting CDF jumps at $x=2$. Is convergence of $F_n(2)$ to $F(2)$ required? Explain.
If the limiting distribution has a continuous CDF, which points must be checked?
For continuity points $a<b$, express $P(a<X_n\leq b)$ using $F_n$, and state its limit.
In the deterministic example $X_n=1/n$, $X=0$, why does failure of CDF convergence at zero not contradict convergence in distribution?