Choosing Convergence Tests for Sequence Problems

A Question Before a Test

A sequence can be close to its proposed limit in one sense and fail to converge in another. A tail probability asks whether a deviation is unlikely at one index; an almost-sure statement asks what happens eventually along a whole sample path. A moment bound controls the size of deviations on average, while a CDF comparison concerns distributions rather than individual outcomes. The first step in a sequence problem is therefore to identify the requested mode of convergence. The next is to choose a test whose assumptions match the sequence.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \(X_n\) and \(X\) be real-valued random variables on it. We use convergence in probability, almost-sure convergence, convergence in \(L^r\), and convergence in distribution as distinct claims. In particular, convergence in distribution does not require the variables to be defined on the same probability space, whereas almost-sure convergence does.

Definition: The sequence \(X_n\) converges in probability to \(X\) if, for every \(\varepsilon>0\), \(\mathbb{P}(|X_n-X|>\varepsilon)\to0\). It converges almost surely to \(X\) if \(\mathbb{P}(\{\omega:X_n(\omega)\to X(\omega)\})=1\). For \(r>0\), it converges in \(L^r\) if \(\mathbb{E}|X_n-X|^r\to0\). It converges in distribution to \(X\) if its CDFs \(F_n(x)=\mathbb{P}(X_n\le x)\) converge to \(F(x)=\mathbb{P}(X\le x)\) at every continuity point \(x\) of \(F\).

These definitions suggest different checks. For convergence in probability, calculate or bound the probability of a deviation larger than a fixed \(\varepsilon\). For almost-sure convergence, control the events of such deviations across all indices, often by a summability argument. For \(L^r\) convergence, calculate the moment error. For convergence in distribution, compare CDFs only at continuity points of the proposed limit. These are tests, not interchangeable labels.

Tail Tests and Sample-Path Tests

A tail calculation can settle convergence in probability directly. For almost-sure convergence, one useful sufficient test is the first Borel–Cantelli lemma: if events \(A_n\) satisfy \(\sum_{n=1}^{\infty}\mathbb{P}(A_n)<\infty\), then \(\mathbb{P}(A_n\text{ infinitely often})=0\). This result does not require independence. Applied to deviation events, it gives the following criterion.

Criterion (summable deviations): If, for every \(\varepsilon>0\), \(\sum_{n=1}^{\infty}\mathbb{P}(|X_n-X|>\varepsilon)<\infty\), then \(X_n\to X\) almost surely.

Proof. For each positive integer \(m\), let \(A_{n,m}=\{|X_n-X|>1/m\}\). The assumption makes \(\sum_n\mathbb{P}(A_{n,m})\) finite, so the first Borel–Cantelli lemma implies that, with probability one, only finitely many \(A_{n,m}\) occur. Intersect these probability-one events over the countable set \(m=1,2,\ldots\); the intersection still has probability one. On any sample point in that intersection, for each \(m\), there is an index after which \(|X_n-X|\le1/m\). Given any \(\varepsilon>0\), choose \(m\) with \(1/m<\varepsilon\). From that index onward, \(|X_n-X|<\varepsilon\). This is exactly convergence to \(X\) on the sample point, proving the criterion. \(\square\)

Summability is sufficient, not necessary. Also, convergence in probability alone does not establish almost-sure convergence: it tests one index at a time, whereas a sample-path claim concerns infinitely many indices together.

Worked Example: Probability Convergence Without Almost-Sure Convergence

Let \(A_n\) be independent events on a probability space with \(\mathbb{P}(A_n)=1/n\), and set \(X_n=\mathbf{1}_{A_n}\) and \(X=0\). The probabilities are valid, including \(\mathbb{P}(A_1)=1\). For \(0<\varepsilon<1\), the strict tail event \(\{|X_n|>\varepsilon\}\) is exactly \(A_n\), so

\mathbb{P}(|X_n|>\varepsilon)=\frac1n\longrightarrow0.

For \(\varepsilon\ge1\), the strict event \(\{|X_n|>\varepsilon\}\) is empty because \(X_n\) takes only the values 0 and 1. Thus \(X_n\to0\) in probability for every \(\varepsilon>0\). But \(\sum_n\mathbb{P}(A_n)=\sum_n1/n=\infty\), and the events are independent. The second Borel–Cantelli lemma therefore gives \(\mathbb{P}(A_n\text{ infinitely often})=1\). On that probability-one event, \(X_n=1\) infinitely often, so \(X_n\) does not converge to 0. The independence hypothesis is essential for this use of the second lemma.

Check: the same sequence has tail probability \(1/n\) for each fixed \(0<\varepsilon<1\), which vanishes, but its tail probabilities are not summable. The two conclusions are consistent: convergence in probability holds, while almost-sure convergence fails.

Moment Bounds: Useful Sufficient Conditions

Moment calculations can turn into tail bounds. For a nonnegative random variable \(Y\) and \(a>0\), Markov’s inequality gives \(\mathbb{P}(Y>a)\le\mathbb{E}[Y]/a\). Taking \(Y=|X_n-X|^r\) and \(a=\varepsilon^r\) shows that a vanishing \(r\)-th moment error is enough for convergence in probability.

Proposition: If \(r>0\) and \(\mathbb{E}|X_n-X|^r\to0\), then \(X_n\to X\) in probability.

Proof. Fix any \(\varepsilon>0\). Since \(|X_n-X|^r\) is nonnegative, Markov’s inequality applies, and because \(r>0\), the event \(\{|X_n-X|>\varepsilon\}\) equals \(\{|X_n-X|^r>\varepsilon^r\}\). Therefore

0\le \mathbb{P}(|X_n-X|>\varepsilon) \le \frac{\mathbb{E}|X_n-X|^r}{\varepsilon^r} \longrightarrow 0.

This is the definition of convergence in probability. \(\square\)

The proposition is one-way: small moments guarantee small tails, but convergence in probability does not by itself imply convergence of moments. The earlier material on moment error and on convergence that does not preserve moments develops these distinctions. When a problem supplies a variance, Chebyshev’s inequality is a convenient special case: for a random variable \(Y\) with finite mean and variance, \(\mathbb{P}(|Y-\mathbb{E}Y|>\varepsilon)\le\operatorname{Var}(Y)/\varepsilon^2\).

Worked Example: A Summable Tail Bound Gives Almost-Sure Convergence

Suppose \(A_n\) are events with \(\mathbb{P}(A_n)=1/n^2\), and let \(X_n=n\mathbf{1}_{A_n}\), with proposed limit \(X=0\). No independence assumption is needed. Fix \(\varepsilon>0\). The strict event \(\{|X_n|>\varepsilon\}\) equals \(A_n\) when \(n>\varepsilon\), and is empty when \(n\le\varepsilon\), since then \(n\le\varepsilon\). Thus its probabilities have a finite initial segment of zeros followed by \(1/n^2\). For every integer \(N\ge1\), the tail is bounded by

\sum_{n=N}^{\infty}\frac1{n^2} \le \frac1{N^2}+\int_N^\infty\frac{dx}{x^2} =\frac1{N^2}+\frac1N<\infty.

Here the inequality follows because \(x^{-2}\) decreases, so \(1/n^2\le\int_{n-1}^{n}x^{-2}\,dx\) for \(n\ge N+1\); the first term is kept separately. The summable-deviations criterion now gives \(X_n\to0\) almost surely. As a moment check, \(\mathbb{E}|X_n|=n\mathbb{P}(A_n)=1/n\to0\), which also implies convergence in probability by the proposition with \(r=1\).

Check: the tail-event sum is finite for each fixed positive threshold, and the first moment is exactly \(n(1/n^2)=1/n\). Both computations support convergence, but the almost-sure conclusion comes from summability and Borel–Cantelli, not from the moment calculation alone.

CDF Tests and Distributional Limits

When the question asks for a distributional limit, compute \(F_n(x)\) and check convergence at every continuity point of the proposed limiting CDF \(F\). It is not necessary to have \(F_n(x)\to F(x)\) at a discontinuity of \(F\). This continuity-point characterization is the standard CDF criterion for convergence in distribution, developed in the earlier material on distributional convergence through CDFs.

Worked Example: Checking a CDF Limit at the Right Points

For \(n\ge1\), let \(X_n\) have probability \(1-1/n\) at 0 and probability \(1/n\) at 1. The proposed limit \(X=0\) has CDF \(F(x)=0\) for \(x<0\), and \(F(x)=1\) for \(x\ge0\). Directly from the two possible values of \(X_n\),

F_n(x)= \begin{cases} 0,&x<0,\\ 1-\frac1n,&0\le x<1,\\ 1,&x\ge1. \end{cases}

For \(x<0\), \(F_n(x)=0=F(x)\). For \(0<x<1\), \(F_n(x)=1-1/n\to1=F(x)\); for \(x>1\), \(F_n(x)=1=F(x)\). These are all continuity points of \(F\). At \(x=0\), which is a discontinuity of \(F\), \(F_n(0)=1-1/n\to0\), while \(F(0)=1\). This discrepancy does not invalidate convergence in distribution, because the criterion requires convergence only at continuity points. At \(x=1\), both CDFs equal 1, although this additional check is not needed.

Check: each \(F_n\) is nondecreasing, has values between 0 and 1, and tends to 1 as \(x\) increases beyond 1. The limiting CDF is the point mass at 0, as proposed, and the sole failure of pointwise convergence noted above is at its jump.

Averages: Choosing Between an LLN and a CLT

For averages of independent, identically distributed observations, the law of large numbers and the central limit theorem answer different questions. The weak law gives convergence in probability of the sample mean to its population mean when the observations have finite variance. The strong law gives almost-sure convergence under the same finite-variance assumption. The central limit theorem instead describes the distribution of centered, scaled fluctuations around that mean; it does not claim that the standardized quantity converges to a constant. These are standard results established in the earlier material on laws of large numbers and the central limit theorem.

Worked Example: Bernoulli Averages and Their Fluctuations

Let \(Y_1,Y_2,\ldots\) be independent Bernoulli random variables with \(\mathbb{P}(Y_k=1)=\mathbb{P}(Y_k=0)=1/2\), and let \(\overline{Y}_n=(Y_1+\cdots+Y_n)/n\). Each variable has mean \(1/2\) and variance \(1/4\). Independence gives

\mathbb{E}\overline{Y}_n=\frac12,\qquad \operatorname{Var}(\overline{Y}_n) =\frac1{n^2}\sum_{k=1}^n\operatorname{Var}(Y_k) =\frac{n(1/4)}{n^2} =\frac1{4n}.

Chebyshev’s inequality therefore yields, for every \(\varepsilon>0\), \(\mathbb{P}(|\overline{Y}_n-1/2|>\varepsilon)\le1/(4n\varepsilon^2)\to0\), proving convergence in probability. For example, at \(n=100\) and \(\varepsilon=1/10\), the bound is \(1/(4\cdot100\cdot(1/10)^2)=1/4\). The strong law also applies: the variables are iid with finite variance, so \(\overline{Y}_n\to1/2\) almost surely.

For fluctuations, the central limit theorem applies because the variables are iid with mean \(1/2\) and finite, nonzero variance \(1/4\). It gives

\frac{\sum_{k=1}^nY_k-n/2}{\sqrt{n/4}} \ \xrightarrow{d}\ N(0,1).

Check: the numerator centers the sum at \(n(1/2)=n/2\), and its variance is \(n(1/4)=n/4\), so the denominator is its standard deviation. The LLN concerns \(\overline{Y}_n\to1/2\); the CLT concerns the centered, standardized sum and its limiting distribution. The two conclusions are not competing answers to the same question.

A Practical Selection Rule

Before calculating, write down the exact mode of convergence the problem asks for and the candidate limit. Then choose the matching route:

Probability: fix \(\varepsilon>0\) and evaluate or bound \(\mathbb{P}(|X_n-X|>\varepsilon)\).
Almost surely: examine deviations along sample paths. If the deviation probabilities are summable at every positive threshold, apply the first Borel–Cantelli lemma. If using the second lemma to show infinitely many deviations, verify independence as well as divergence of the probability sum.
Moment convergence: calculate \(\mathbb{E}|X_n-X|^r\). A moment tending to zero implies convergence in probability, but the reverse implication requires additional conditions.
Distribution: calculate CDFs and compare them at continuity points of the proposed limiting CDF.
Sample averages: use an LLN for convergence of averages to a mean; use the CLT for the limiting distribution of centered, scaled sums or averages. Check the hypotheses before applying either result.

A common mistake is to reason that \(\mathbb{P}(A_n)\to0\) means \(A_n\) occurs only finitely often almost surely. In the independent-event example above, \(\mathbb{P}(A_n)=1/n\to0\), but \(\sum_n1/n=\infty\), and the second Borel–Cantelli lemma says that \(A_n\) occurs infinitely often almost surely. The correction is to distinguish a vanishing single-index probability from a statement about infinitely many events. Conversely, summability is only a sufficient test for almost-sure convergence: for nested events \(A_n=\{U\le1/n\}\) with \(U\) uniform on \((0,1)\), \(\mathbb{P}(A_n)=1/n\) is not summable, but every outcome with \(U>0\) belongs to only finitely many \(A_n\). Thus \(\mathbf{1}_{A_n}\to0\) almost surely.

Check Your Understanding

For each question, identify the mode of convergence being tested before choosing a calculation.

If \(\mathbb{P}(|X_n-X|>\varepsilon)\to0\) for every \(\varepsilon>0\), which mode of convergence has been established?
Suppose \(\sum_n\mathbb{P}(|X_n-X|>1/m)<\infty\) for every positive integer \(m\). Which lemma gives almost-sure convergence, and why is a countable set of thresholds sufficient?
For independent events with probabilities \(1/n\), do the probabilities tend to zero, have finite sum, or both? What does the second Borel–Cantelli lemma conclude?
When checking convergence in distribution to a CDF with a jump at \(x_0\), must \(F_n(x_0)\to F(x_0)\)?
For an iid sample with finite, nonzero variance, which result concerns the sample mean’s limit and which concerns the centered, standardized sum’s limiting distribution?