A Worked CDF Limit for Expanding Uniform Supports

Why Expanding Supports Need an Endpoint Check

A sequence of distributions can change its support with $n$, so its CDF must be read from the support before taking a limit. For this example the support expands by a small amount at the right endpoint: the $n$th law is uniform on $[0,1+1/n]$. The natural candidate limit is uniform on $[0,1]$, but the comparison needs care at $x=1$, where the candidate support ends, and at the moving endpoint $1+1/n$.

Distributional convergence is assessed by comparing CDF values at fixed real numbers. The continuity-point criterion says that a sequence with CDFs $F_n$ converges in distribution to a law with CDF $F$ if and only if $F_n(x)\to F(x)$ at every continuity point $x$ of $F$. We will calculate these CDFs directly, check all fixed $x$, and then quantify their maximum discrepancy.

The CDFs and Their Endpoints

Definition: Let $X_n$ have the uniform distribution on $[0,1+1/n]$, and let $X$ have the uniform distribution on $[0,1]$. Write $F_n(x)=P(X_n\leq x)$ and $F(x)=P(X\leq x)$. A uniform distribution on an interval of length $a>0$ assigns probability to a subinterval in proportion to that subinterval's length; its density is $1/a$ on the interval and zero outside it.

Set $a_n=1+1/n=(n+1)/n$. The event $X_n\leq x$ has probability zero if $x<0$. If $0\leq x\leq a_n$, the portion of the support that lies at or to the left of $x$ has length $x$, out of total length $a_n$. If $x>a_n$, the event has probability one. Thus

F_n(x)= \begin{cases} 0, & x<0,\\[4pt] \dfrac{x}{a_n}=\dfrac{n}{n+1}x, & 0\leq x\leq 1+\dfrac1n,\\[8pt] 1, & x>1+\dfrac1n. \end{cases}

The candidate limit is uniform on an interval of length one, so its CDF is

F(x)= \begin{cases} 0, & x<0,\\ x, & 0\leq x\leq 1,\\ 1, & x>1. \end{cases}

These formulas include their support endpoints. In particular, $F_n(0)=0$, $F_n(a_n)=1$, $F(0)=0$, and $F(1)=1$. Each CDF is continuous at both endpoints of its support: at zero, its linear segment tends to zero from the right; at the upper endpoint, that segment tends to one, matching the value to the right. Consequently, $F$ is continuous at every real number, and the continuity-point criterion requires convergence at every fixed $x$.

Worked Example: Checking the Formula for a Particular Support

Take $n=4$. The support is $[0,1+1/4]=[0,5/4]$, and the density on that interval is $4/5$. Find $F_4(0)$, $F_4(1)$, and $F_4(5/4)$.

At $x=0$, the event $X_4\leq 0$ has probability zero, since a single endpoint has zero probability under a continuous uniform law. Thus $F_4(0)=0$.

Since $1$ is inside $[0,5/4]$, the part of the support up to $1$ has length $1$. Therefore $F_4(1)=(4/5)(1)=4/5$.

At the upper endpoint, the entire support is included, so $F_4(5/4)=(4/5)(5/4)=1$. The endpoint value agrees with the value on the right of the support; there is no jump there.

Convergence at Every Fixed Real Number

The limit is determined by fixing $x$ first and then letting $n$ increase. The cases below cover the real line, including the two endpoints of the candidate support.

For $x<0$.
Both distributions assign probability zero to $(-\infty,x]$, so $F_n(x)=F(x)=0$ for every $n$.

At $x=0$.
The endpoint has probability zero under either continuous uniform law, and $F_n(0)=F(0)=0$ for every $n$.

For $0<x<1$.
Since $x\leq 1+1/n$ for every $n$, $F_n(x)=nx/(n+1)$. Hence $F_n(x)\to x=F(x)$.

At $x=1$.
The point lies in every $[0,1+1/n]$, so $F_n(1)=n/(n+1)\to 1=F(1)$.

For $x>1$.
Choose an integer $N$ large enough that $1/N<x-1$. For every $n\geq N$, $1+1/n\leq 1+1/N<x$. Therefore $F_n(x)=1=F(x)$ for all $n\geq N$.

It follows that $F_n(x)\to F(x)$ for every real $x$. Since $F$ is continuous everywhere, the continuity-point criterion gives $X_n\xrightarrow{d}X$, with $X$ uniform on $[0,1]$. The argument at $x=1$ is worth noticing: the $n$th CDF does not equal one there, but its value $n/(n+1)$ approaches one. The limiting CDF at $1$ is indeed one.

Worked Example: A Fixed Point Above the Limiting Support

Evaluate the CDFs at the fixed point $x=6/5$. The candidate CDF is $F(6/5)=1$, because $6/5>1$. The $n$th support ends at $1+1/n$, which is at or below $6/5$ exactly when $1/n\leq 1/5$, or $n\geq 5$. Thus $F_n(6/5)=1$ for every $n\geq 5$.

For comparison, when $n=4$, the support ends at $5/4$, which exceeds $6/5$. The point is then inside the support, and $F_4(6/5)=(4/5)(6/5)=24/25$. For every $n\geq5$, the CDF value is exactly $1$, so convergence at this fixed point is not merely asymptotic: it eventually becomes exact.

The Exact Maximum CDF Difference

The pointwise calculation can be strengthened by finding the largest difference between the two CDFs over the whole real line. This shows how close the distributions are in CDF terms for each $n$, while making the special role of $x=1$ explicit.

Theorem: For every positive integer $n$, $$ \sup_{x\in\mathbb{R}}|F_n(x)-F(x)|=\frac{1}{n+1}. $$

Proof. If $x<0$, the difference is zero. If $0\leq x\leq1$, both CDFs are on their linear portions, including their endpoint values, and

|F_n(x)-F(x)| =\left|\frac{n}{n+1}x-x\right| =\frac{x}{n+1} \leq\frac{1}{n+1}.

Equality holds at $x=1$, because $F_n(1)=n/(n+1)$ while $F(1)=1$. Next, if $1<x<1+1/n$, the candidate CDF is one and the $n$th CDF is still on its linear portion. Therefore

|F_n(x)-F(x)| =1-\frac{n}{n+1}x <1-\frac{n}{n+1} =\frac{1}{n+1}.

At $x=1+1/n$, both CDFs equal one, and for $x>1+1/n$ both remain one. These cases cover every real $x$. The difference is nowhere greater than $1/(n+1)$, and it equals that bound at $x=1$, proving the stated supremum. $\square$

Because $1/(n+1)\to0$, this also proves uniform convergence of these CDFs: the largest absolute difference over all $x$ tends to zero. In this particular example, the discrepancy decreases at the explicit rate $1/(n+1)$. Pointwise convergence alone would not generally supply such a uniform bound; here it is available because the CDFs have simple linear pieces and matching constant tails.

Worked Example: Locating the Largest Difference

Let $n=9$, so the expanding support ends at $10/9$. At $x=1$, the CDF values are $F_9(1)=9/10$ and $F(1)=1$, giving an absolute difference of $1/10$.

For $0\leq x\leq1$, the difference is $x/10$, which increases from zero to $1/10$. For $1<x<10/9$, the difference is $1-(9/10)x$, which decreases from values below $1/10$ toward zero. At and beyond $10/9$, it is zero. Therefore the largest difference is $1/10$, attained at $x=1$, exactly as the general formula predicts.

Moving Endpoints and a Common Mistake

The upper support endpoint $a_n=1+1/n$ moves with $n$. At that moving point, $F_n(a_n)=1$ for every $n$. This is a correct endpoint calculation, but it is not itself a pointwise convergence test: the argument of $F_n$ changes with $n$. The continuity-point criterion concerns $F_n(x)$ at one fixed $x$ as $n$ varies.

For example, the fixed point $x=1$ is not the moving endpoint for any finite $n$. Its CDF value is $n/(n+1)$, not one, and convergence follows by taking the limit of those values. Conversely, for a fixed $x>1$, the moving endpoint eventually falls below $x$, after which $F_n(x)=1$. Keeping fixed and moving arguments distinct prevents a common error: inferring convergence from a calculation made only at the changing support boundary.

Key takeaway: Derive the CDF with its endpoint conventions, then compare $F_n(x)$ and $F(x)$ at fixed $x$. Here the candidate limit is uniform on $[0,1]$, convergence holds at every real number, and the exact maximum CDF difference is $1/(n+1)$.

Check Your Understanding

Use the displayed CDF formulas to answer the following.

For $n=5$, calculate the upper support endpoint and the values $F_5(1)$ and $F_5(6/5)$.
Why does checking $F_n(1+1/n)=1$ for every $n$ not, by itself, establish pointwise CDF convergence?
At the fixed point $x=1$, what is $F_n(1)$, and to what value does it converge?
For fixed $x>1$, explain why $F_n(x)$ eventually equals one.
Where is the maximum of $|F_n(x)-F(x)|$ attained, and what is its value?