A Worked CDF Limit for Two-Point Distributions

Why a Two-Point Example Is Subtle

A random variable supported on two fixed values is simple to describe, but its CDF changes abruptly at those values. When the probabilities vary with $n$, the CDF limit is therefore not something to check indiscriminately at every real number. The key is to calculate the CDF exactly, identify the candidate limiting law, and then test convergence at the points where that law’s CDF is continuous.

This example makes the continuity-point criterion concrete. It also shows why the probabilities assigned to the two values determine the limit, and why a sequence of probabilities that keeps oscillating cannot settle on one fixed two-point distribution.

Definition and CDF Calculation

Definition: Let $a<b$ be fixed real numbers. For each positive integer $n$, suppose $X_n$ takes values only in $\{a,b\}$, with $P(X_n=a)=p_n$ and $P(X_n=b)=1-p_n$, where $0\leq p_n\leq 1$. Its CDF is $F_n(x)=P(X_n\leq x)$. For any real $x$, the CDF is determined by whether $x$ lies below $a$, between $a$ and $b$, or at or above $b$.

If $x<a$, neither possible value of $X_n$ is at most $x$, so $F_n(x)=0$. If $a\leq x<b$, the event $X_n\leq x$ occurs exactly when $X_n=a$, so its probability is $p_n$. If $x\geq b$, both possible values are at most $x$, and the probability is $1$. Thus

F_n(x)= \begin{cases} 0, & x<a,\\ p_n, & a\leq x<b,\\ 1, & x\geq b. \end{cases}

In particular, the interval in the middle includes $a$ but excludes $b$. This follows from the definition $F_n(x)=P(X_n\leq x)$: at $x=a$, the mass at $a$ is included; at $x=b$, the mass at $b$ is included as well.

Worked Example: Writing the CDF for Each $n$

Let $X_n$ take values $-2$ and $3$, with $P(X_n=-2)=p_n=1/4+1/(4n)$. Find $F_n(x)$ and calculate it at $x=-2$, $x=0$, and $x=3$.

The probabilities are valid: since $n\geq1$, $1/4<p_n\leq1/2$, and $P(X_n=3)=1-p_n=3/4-1/(4n)$. Applying the general formula with $a=-2$ and $b=3$,

F_n(x)= \begin{cases} 0, & x<-2,\\ \frac14+\frac{1}{4n}, & -2\leq x<3,\\ 1, & x\geq3. \end{cases}

At $x=-2$, $F_n(-2)=P(X_n=-2)=1/4+1/(4n)$. At $x=0$, the event $X_n\leq0$ again consists only of $X_n=-2$, so $F_n(0)=1/4+1/(4n)$. At $x=3$, both values are included, giving $F_n(3)=1$. The change from the middle value to $1$ happens at $3$, not just above $3$.

The Limiting Two-Point Law

Suppose the probabilities have a limit $p_n\to p$, where $p\in[0,1]$. The natural candidate is a random variable $X$ with probability $p$ at $a$ and probability $1-p$ at $b$. This definition includes the boundary cases: when $p=0$, $X=b$ with probability one; when $p=1$, $X=a$ with probability one.

Theorem: Let $a<b$ be fixed, and let $X_n$ have the two-point distribution specified above. If $p_n\to p\in[0,1]$, then $X_n\xrightarrow{d}X$, where $P(X=a)=p$ and $P(X=b)=1-p$. The limiting CDF is $$ F(x)= \begin{cases} 0, & x<a,\\ p, & a\leq x<b,\\ 1, & x\geq b. \end{cases} $$

Proof. The displayed expression for $F$ follows by the same event calculation used for $F_n$. We verify convergence at every continuity point of $F$. At any $x<a$, $F_n(x)=F(x)=0$ for every $n$. At any $a<x<b$, $F_n(x)=p_n\to p=F(x)$. At any $x>b$, $F_n(x)=F(x)=1$.

It remains to consider the endpoints, because either endpoint can be a continuity point in a boundary case. At $a$, the left-hand value of $F$ is $0$ and the value at and immediately to the right of $a$ is $p$. Thus $F$ is continuous at $a$ exactly when $p=0$. In that case, $F_n(a)=p_n\to0=F(a)$. At $b$, the value immediately to the left is $p$, while $F(b)=1$. Thus $F$ is continuous at $b$ exactly when $p=1$. In that case, $F_n(b)=1\to1=F(b)$. We have proved convergence at every continuity point, so the continuity-point criterion gives $X_n\xrightarrow{d}X$. $\square$

The endpoint check matters even though the formula for $F_n$ is short. If $0<p<1$, both $a$ and $b$ are jumps of $F$, so neither endpoint is required by the criterion. When $p=0$ or $p=1$, one endpoint becomes a continuity point, and convergence there must also be verified.

Worked Example: An Interior Limiting Probability

Let $X_n$ take values $-2$ and $3$, with $P(X_n=-2)=1/4+1/(4n)$. Find the limiting distribution and verify CDF convergence at continuity points.

Here $p_n=1/4+1/(4n)\to1/4$. The limiting random variable $X$ therefore satisfies $P(X=-2)=1/4$ and $P(X=3)=3/4$. Its CDF is

F(x)= \begin{cases} 0, & x<-2,\\ \frac14, & -2\leq x<3,\\ 1, & x\geq3. \end{cases}

For $x=-3$, which is below $-2$, both $F_n(-3)$ and $F(-3)$ equal $0$. For $x=0$, which lies strictly between the support points, $F_n(0)=1/4+1/(4n)\to1/4=F(0)$. For $x=4$, both CDFs equal $1$. These regions contain all continuity points: the limiting CDF jumps at $-2$ and $3$, so those two points are not required in the continuity-point check. The theorem gives $X_n\xrightarrow{d}X$.

Worked Example: A Degenerate Limiting Distribution

Let $X_n$ take values $1$ and $4$, with $P(X_n=1)=p_n=n/(n+1)$. Identify the limit and check the continuity-point criterion.

Since $p_n=1-1/(n+1)\to1$, the limiting variable equals $1$ with probability one. Its CDF is $F(x)=0$ for $x<1$ and $F(x)=1$ for $x\geq1$. For each $n$,

F_n(x)= \begin{cases} 0, & x<1,\\ \frac{n}{n+1}, & 1\leq x<4,\\ 1, & x\geq4. \end{cases}

For every $x<1$, $F_n(x)=F(x)=0$; for every $1<x<4$, $F_n(x)=n/(n+1)\to1=F(x)$; and for every $x>4$, both CDFs equal $1$. At $x=4$, $F$ is continuous, since it is already $1$ on both sides; $F_n(4)=1$ for every $n$. At $x=1$, $F$ jumps from $0$ to $1$, so it is not a continuity point. Thus convergence holds at every continuity point, and $X_n\xrightarrow{d}1$.

Why Convergence of the Probabilities Is the Decisive Check

For a proposed limiting two-point law on the same values $a<b$, the middle part of the CDF reads off the probability at $a$. In particular, at the midpoint $c=(a+b)/2$, the proposed limiting CDF is continuous and

F_n(c)=p_n,\qquad F(c)=p.

Consequently, convergence to that proposed law requires $p_n\to p$: the continuity-point criterion applied at $c$ requires the displayed CDF values to converge. Conversely, the theorem proves that $p_n\to p$ suffices. This gives an exact test for convergence to any specified two-point law on the same fixed support.

Worked Example: Oscillating Probabilities

Let $X_n$ take values $-1$ and $2$, with $P(X_n=-1)=p_n$, where $p_n=1/4$ for odd $n$ and $p_n=3/4$ for even $n$. Can this sequence converge in distribution to a two-point law on $\{-1,2\}$?

At the midpoint $c=1/2$, the CDF is $F_n(c)=p_n$. Its values alternate between $1/4$ and $3/4$, so they do not converge. For any proposed two-point limiting law on $\{-1,2\}$, the CDF is continuous at $1/2$; the continuity-point criterion would require $F_n(1/2)$ to converge to its value there. This is impossible. Therefore there is no distributional limit of the proposed two-point form on this support.

The calculation rules out a fixed two-point limit; the failure is not repaired by averaging the two alternating probabilities. A limit requires the CDF sequence itself to converge at the relevant continuity point.

Common Mistake: Testing the Wrong Points

A common mistake is to demand convergence of $F_n(x)$ at every $x$, then treat any failure at a jump of $F$ as evidence against convergence in distribution. The criterion requires convergence at the continuity points of the limiting CDF, not at its jumps. For example, if $0<p<1$, the proposed limit has jumps at both support points. The continuity-point check applies to points below, between, and above them, but not to the two jump locations.

The reverse mistake is to ignore an endpoint that is continuous in a degenerate limit. If $p=1$, the limit is concentrated at $a$, and $b$ is a continuity point of its CDF. A full check must include $b$, where $F_n(b)=1$. The theorem’s endpoint analysis handles these cases directly.

Key takeaway: For fixed support points $a<b$, the CDF of $X_n$ is $0$, $p_n$, or $1$ according to the location of $x$. If $p_n\to p$, the limiting masses are $p$ at $a$ and $1-p$ at $b$. Verify convergence at continuity points, paying particular attention to whether either endpoint becomes continuous when $p$ is $0$ or $1$.

Check Your Understanding

Use the CDF formula and continuity-point criterion to answer these questions.

If $X_n$ takes values $0$ and $5$, with $P(X_n=0)=p_n$, write $F_n(x)$ in its three regions.
If $p_n\to 2/5$, what are the two probabilities in the limiting distribution?
For a limiting two-point law with $0<p<1$, which two points are jumps of its CDF?
If $p_n\to0$, where is the limiting random variable concentrated, and is the CDF continuous at $a$ or $b$?
Why does an oscillating sequence $p_n$ fail the continuity-point test for convergence to a fixed two-point law on the same support?

A Worked CDF Limit for Two-Point Distributions

What You'll Learn

Why a Two-Point Example Is Subtle

Definition and CDF Calculation

Worked Example: Writing the CDF for Each \(n\)

The Limiting Two-Point Law

Worked Example: An Interior Limiting Probability

Worked Example: A Degenerate Limiting Distribution

Why Convergence of the Probabilities Is the Decisive Check

Worked Example: Oscillating Probabilities

Common Mistake: Testing the Wrong Points

Check Your Understanding

A Worked CDF Limit for Two-Point Distributions

What You'll Learn

Why a Two-Point Example Is Subtle

Definition and CDF Calculation

Worked Example: Writing the CDF for Each \(n\)

The Limiting Two-Point Law

Worked Example: An Interior Limiting Probability

Worked Example: A Degenerate Limiting Distribution

Why Convergence of the Probabilities Is the Decisive Check

Worked Example: Oscillating Probabilities

Common Mistake: Testing the Wrong Points

Check Your Understanding

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