Interval Probabilities Under Distributional Convergence

From CDF Convergence to Interval Probabilities

Distributional convergence controls the limiting probabilities of many events, but not every event automatically. For intervals, the key condition is that their endpoints are continuity points of the limiting CDF. If an endpoint carries probability mass in the limit, a small shift in the approximating distributions can move that mass across the interval boundary.

Write $F_n$ for the CDF of $X_n$, and $F$ for the CDF of $X$. The definition of $X_n\xrightarrow{d}X$, in its CDF form, is that $F_n(x)\to F(x)$ at every continuity point $x$ of $F$. The result below translates this pointwise convergence into convergence of interval probabilities.

Theorem: Suppose $X_n\xrightarrow{d}X$, and let $a<b$ be continuity points of the CDF $F$ of $X$. Then $$ P(a<X_n\le b)\longrightarrow P(a<X\le b)=F(b)-F(a). $$ The same limit holds if either or both endpoints are included instead: $P(a\le X_n\le b)$, $P(a<X_n<b)$, and $P(a\le X_n<b)$ all converge to $P(a<X\le b)$.

The half-open interval is the direct case: by the definition of a CDF, $P(a<X_n\le b)=F_n(b)-F_n(a)$. For other endpoint conventions, we also need to know that the approximating variables put vanishing probability on a continuity point. That fact is proved next.

Why Continuity-Point Endpoints Carry Vanishing Mass

A continuity point $a$ of $F$ satisfies $P(X=a)=0$. But this alone does not immediately say that $P(X_n=a)\to0$: the distributions of $X_n$ can have atoms even when the limit has none. CDF convergence supplies the needed control from nearby continuity points.

Lemma: If $X_n\xrightarrow{d}X$ and $a$ is a continuity point of $F$, then $P(X_n=a)\to0$.

Proof. Choose continuity points $c_k<a$ with $c_k\uparrow a$. Such points can be chosen because a CDF has at most countably many discontinuities, so its continuity points are dense. For each fixed $k$,

0\le P(X_n=a)\le P(c_k<X_n\le a)=F_n(a)-F_n(c_k).

Both $a$ and $c_k$ are continuity points of $F$, so taking the limit superior over $n$ gives

\limsup_{n\to\infty}P(X_n=a)\le F(a)-F(c_k).

As $k\to\infty$, $F(c_k)\to F(a-)$, the left limit of $F$ at $a$. Continuity at $a$ means $F(a-)=F(a)$, so the right-hand side tends to zero. Since the probabilities are nonnegative, $P(X_n=a)\to0$. $\square$

To finish the theorem, the half-open probability converges because $F_n(b)-F_n(a)\to F(b)-F(a)$. For the closed interval, add the lower-endpoint mass: $P(a\le X_n\le b)=P(a<X_n\le b)+P(X_n=a)$. The lemma shows that the added term tends to zero. For the open interval, subtract the upper-endpoint mass: $P(a<X_n<b)=P(a<X_n\le b)-P(X_n=b)$. The same lemma shows that this subtracted term tends to zero. Including the lower endpoint while excluding the upper one follows by both adjustments. Finally, $P(X=a)=P(X=b)=0$, so all these limiting interval probabilities equal $P(a<X\le b)$. $\square$

Worked Example: A Normal Sequence and a Fixed Interval

Suppose $X_n\sim N(0,1+1/n)$, where the second parameter is the variance, and consider the interval $(-1,2]$. Its endpoints are continuity points of the limiting CDF, which is that of $X\sim N(0,1)$.

For each $n$, standardizing by the standard deviation $\sqrt{1+1/n}$ gives

P(-1<X_n\le2) =\Phi\!\left(\frac{2}{\sqrt{1+1/n}}\right) -\Phi\!\left(\frac{-1}{\sqrt{1+1/n}}\right),

where $\Phi$ is the standard normal CDF. As $n\to\infty$, the two standardized endpoints tend to $2$ and $-1$. Continuity of $\Phi$ therefore gives

P(-1<X_n\le2)\longrightarrow \Phi(2)-\Phi(-1) \approx 0.97725-0.15866=0.81859.

The interval theorem gives the same result directly from distributional convergence. Because the limiting normal distribution has no endpoint atoms, changing either endpoint from open to closed does not change the limiting probability.

Worked Example: A Discrete Limit with Continuity-Point Endpoints

Let $X_n$ take the values $-2$ and $2$, with $P(X_n=-2)=1/3+1/(6n)$ and $P(X_n=2)=2/3-1/(6n)$. These are valid probabilities: both are positive for every $n\ge1$, and they sum to one. The limiting variable $X$ has probabilities $P(X=-2)=1/3$ and $P(X=2)=2/3$.

The CDFs converge at every continuity point of the limiting CDF: away from the two support points, the probabilities on the left of the point converge to the corresponding limiting probabilities. Thus $X_n\xrightarrow{d}X$. Now take the interval $(-3,0.5]$. Neither endpoint is a support point of $X$, so both are continuity points of its CDF. Directly,

P(-3<X_n\le0.5)=P(X_n=-2)=\frac13+\frac{1}{6n} \longrightarrow \frac13=P(-3<X\le0.5).

This example shows that a discrete limit causes no difficulty by itself. What matters for the stated result is whether the interval endpoints are atoms of the limit, not whether the distribution as a whole is discrete or continuous.

What Changes at an Endpoint Atom?

At a discontinuity of $F$, distributional convergence does not require $F_n$ to converge to $F$ there. More importantly for interval probabilities, mass approaching an endpoint can lie on one side of the boundary for every $n$, while the limiting atom lies exactly on the boundary. Whether that atom belongs to the interval then changes the limiting probability.

Worked Example: A Moving Point Mass at an Endpoint

Let $X_n=1/n$ with probability one. Then $X_n\xrightarrow{d}X$, where $X=0$ with probability one: at every $x<0$, the CDFs are zero, and at every $x>0$, they are eventually one. These are precisely the continuity points of the limiting CDF; $0$ is its jump point.

For the interval $(0,1]$, every $X_n$ lies inside, but the limit $X=0$ does not. Hence

P(0<X_n\le1)=1\quad\hbox{for every }n, \qquad P(0<X\le1)=0.

So the interval probabilities do not converge to the limiting interval probability. The endpoint $0$ is an atom, and the interval excludes it. This is not a contradiction to the theorem: its continuity-point condition fails.

Endpoint inclusion can matter in the other direction as well. If instead $X_n=-1/n$ and $X=0$, then for $[0,1)$, $P(0\le X_n<1)=0$ for every $n$, whereas $P(0\le X<1)=1$. By contrast, for $X_n=1/n$, the closed interval $[0,1]$ has probability one both before and after taking the limit. An endpoint atom does not force failure for every interval; it means the continuity-point theorem no longer guarantees the conclusion.

Using the Result Correctly

The interval theorem is useful whenever a distributional limit is used to approximate event probabilities, including tail or central-region probabilities. The practical check is simple: locate the interval endpoints on the limiting distribution. If the limiting CDF is continuous at both, the probability of the interval converges, regardless of whether the approximating distributions are continuous, discrete, or mixed.

For example, if $a$ and $b$ are continuity points, then the result applies to any of $(a,b]$, $[a,b]$, $(a,b)$, and $[a,b)$. At such endpoints, the choice of including or excluding the boundary contributes only a probability tending to zero. If an endpoint is an atom, however, do not substitute the interval endpoints into the CDF convergence criterion as though convergence there were assured.

Common pitfall: “$X_n\xrightarrow{d}X$, so the probability of every fixed interval converges to the probability of that interval under $X$.” This is false. The correction is to check both endpoints: interval-probability convergence is guaranteed when they are continuity points of the limiting CDF. An endpoint atom can make inclusion or exclusion of the boundary decisive.

Check Your Understanding

Use the continuity-point condition to decide which interval probabilities are guaranteed to converge.

If $X_n\xrightarrow{d}X$ and $a<b$ are continuity points of $F$, write $P(a<X_n\le b)$ in terms of $F_n$ and state its limit.
Why does the lemma about $P(X_n=a)$ require that $a$ be a continuity point of the limiting CDF?
For the two-point limit with support $\{-2,2\}$, are $-2$ and $0.5$ continuity points? Explain what this implies for using an interval with those endpoints.
For $X_n=1/n$ and $X=0$, compare the probabilities of $(0,1]$. Which endpoint condition fails?
Does an atom at an endpoint imply that interval probabilities must fail to converge? Use the closed interval in the moving-point-mass example to justify your answer.

Interval Probabilities Under Distributional Convergence

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