Diagnosing Failed Limit Operations

Why Limit Calculations Need an Audit

An asymptotic calculation often combines several limits: a small error is multiplied by a random quantity, two convergent statistics are added, or a numerator is divided by a denominator. Each operation has conditions. A sequence that converges to zero does not remain negligible after multiplication by an arbitrary sequence, marginal limits do not determine the limit of every function of a pair, and a ratio is unstable near a zero denominator.

A useful diagnostic is to identify the exact operation being used and ask what controls it. For a product, check that the other factor is bounded in probability. For a function, check continuity at the limiting value. For a multivariate function, check joint convergence rather than only separate marginal convergence. For a ratio, check that the limiting denominator is nonzero, or at least nonzero with probability one in a distributional limit.

Definition: A sequence \(Y_n\) is tight if, for every \(\varepsilon>0\), there is a finite \(M\) such that \(\sup_n P(|Y_n|>M)<\varepsilon\). In asymptotic arguments, \(Y_n=O_p(1)\) means that \(Y_n\) is bounded in probability: for every \(\varepsilon>0\), some finite \(M\) satisfies \(\limsup_{n\to\infty}P(|Y_n|>M)<\varepsilon\). The notation \(X_n=o_p(1)\) means \(X_n\xrightarrow{p}0\).

For real-valued random variables, tightness and boundedness in probability express the same asymptotic control, up to finitely many initial terms. The key point is that a tight sequence puts arbitrarily high probability inside some fixed bounded interval, even though its values need not converge.

Products: A Vanishing Factor Needs a Tight Partner

The basic valid product rule is that a term converging to zero in probability can be multiplied by a tight term and still converge to zero. The tightness hypothesis prevents the second factor from growing fast enough, with substantial probability, to cancel the smallness of the first.

Theorem: If \(X_n\xrightarrow{p}0\) and \(Y_n=O_p(1)\), then \(X_nY_n\xrightarrow{p}0\).

Proof. Fix \(\varepsilon>0\) and \(\eta>0\). Since \(Y_n=O_p(1)\), choose \(M<\infty\) so that \(\limsup_n P(|Y_n|>M)<\eta\). For this fixed \(M\),

P(|X_nY_n|>\varepsilon) \leq P(|Y_n|>M)+P(|X_n|>\varepsilon/M).

Indeed, when \(|Y_n|\leq M\), the event \(|X_nY_n|>\varepsilon\) implies \(|X_n|>\varepsilon/M\). The second probability tends to zero because \(X_n\xrightarrow{p}0\). Thus \(\limsup_nP(|X_nY_n|>\varepsilon)\leq\eta\). Since \(\eta\) was arbitrary, the probability tends to zero. \(\square\)

Worked Example: When a Small Error Is Not Enough

Let \(X_n=1/n\) and \(Y_n=n\), both deterministic. Then \(X_n\to0\), but

X_nY_n=\frac{1}{n}\,n=1

for every \(n\), so the product does not tend to zero. The missing condition is tightness: for any fixed \(M\), \(Y_n=n>M\) for all sufficiently large \(n\), so \(Y_n\) is not bounded in probability.

For contrast, if \(Z\) is any finite real-valued random variable and \(Y_n=Z\), then \(Y_n\) is tight and \((1/n)Z\xrightarrow{p}0\). To see this directly, for each \(\varepsilon>0\), choose \(M\) with \(P(|Z|>M)<\varepsilon\). Once \(n>M/\varepsilon\), the event \(|Z|/n>\varepsilon\) is contained in \(|Z|>M\). Its probability is therefore less than \(\varepsilon\).

This distinction matters when simplifying remainders. A term \(o_p(1)\) can be discarded on its own, and its product with an \(O_p(1)\) term is still negligible. It cannot be multiplied by an unrestricted sequence and discarded without a further argument.

Functions: Continuity Is the Check

The continuous mapping theorem permits a function to be applied to a distributional limit, but continuity at the limit is essential. In its usual real-vector form, if \(X_n\xrightarrow{d}X\) in \(\mathbb{R}^k\) and \(g:\mathbb{R}^k\to\mathbb{R}^m\) is measurable and continuous at every point in a set having probability one under \(X\), then \(g(X_n)\xrightarrow{d}g(X)\). In particular, continuity everywhere is sufficient.

Worked Example: A Discontinuous Function Breaks the Conclusion

Take the deterministic sequence \(X_n=1/n\) and define \(g(x)=\mathbf{1}\{x>0\}\). Since \(X_n\to0\), it also converges in distribution to the constant zero. But \(g\) is discontinuous at zero, and direct calculation gives

g(X_n)=1\quad\text{for every }n, \qquad g(0)=0.

Thus \(g(X_n)\) does not converge in distribution to \(g(0)\): its distribution is always concentrated at 1, not at 0. The failed step is applying the continuous mapping theorem at a point where the function is discontinuous.

For a proposed transformation, identify the limiting random vector first, then locate the discontinuities of the transformation. It is not enough that a function is continuous at a typical finite-\(n\) value; the relevant condition concerns the limiting value. If the limit assigns positive probability to a discontinuity, the theorem does not provide the claimed conclusion. The conclusion may still hold for a particular model, but it needs a separate argument.

Joint Limits: Marginal Convergence Does Not Determine a Pair

A function of two statistics, such as their sum or ratio, depends on how the statistics vary together. Knowing \(X_n\xrightarrow{d}X\) and \(Y_n\xrightarrow{d}Y\) separately does not generally establish convergence of \((X_n,Y_n)\). The multivariate continuous mapping theorem requires joint convergence of that vector.

Worked Example: Identical Marginals, Different Sums

Let \(Z\) be Bernoulli with \(P(Z=0)=P(Z=1)=1/2\), and set \(X_n=Z\) for every \(n\). Define \(Y_n=Z\) when \(n\) is even and \(Y_n=1-Z\) when \(n\) is odd. Each of \(X_n\) and \(Y_n\) has the same Bernoulli distribution for every \(n\), so both marginal sequences converge in distribution to a Bernoulli random variable.

Their sums, however, alternate in distribution. If \(n\) is even, \(X_n+Y_n=2Z\), which equals 0 or 2 with probability \(1/2\) each. If \(n\) is odd, \(X_n+Y_n=Z+(1-Z)=1\) with probability one. These two distributions differ, so the sequence of sum distributions does not converge. In particular, the separate marginal limits do not justify a limit calculation for the sum.

The joint distributions alternate as well: for even \(n\), the pair is \((Z,Z)\); for odd \(n\), it is \((Z,1-Z)\). This example shows why the dependence structure, not just the two marginal laws, can be essential.

A valid route is to establish joint convergence and then apply a continuous function to the vector. When one component converges in probability to a constant, Slutsky’s theorem often supplies the joint convergence needed. But if both components have nonconstant distributional limits, marginal convergence alone is not a substitute for a joint-limit argument.

Ratios: Keep the Limiting Denominator Away From Zero

Division is a continuous operation where the denominator is nonzero and discontinuous where it is zero. For convergence in probability to constants, a nonzero limiting denominator is enough to justify the ratio limit. For distributional convergence, the corresponding condition is that the limiting denominator be nonzero with probability one.

Theorem: If \(A_n\xrightarrow{p}a\), \(B_n\xrightarrow{p}b\), and \(b\neq0\), then \(A_n/B_n\xrightarrow{p}a/b\), defining the ratio arbitrarily when \(B_n=0\). If \((A_n,B_n)\xrightarrow{d}(A,B)\) and \(P(B=0)=0\), then \(A_n/B_n\xrightarrow{d}A/B\), with the ratio again defined arbitrarily when \(B_n=0\).

Proof for convergence in probability. Because \(B_n\xrightarrow{p}b\neq0\), the event \(|B_n-b|<|b|/2\) has probability tending to one. On that event, \(|B_n|>|b|/2\), and

\left|\frac{A_n}{B_n}-\frac{a}{b}\right| \leq \frac{|A_n-a|}{|B_n|} +|a|\frac{|b-B_n|}{|bB_n|} \leq \frac{2|A_n-a|}{|b|} +\frac{2|a|\,|B_n-b|}{|b|^2}.

Both terms on the right converge to zero in probability. The complementary event has probability tending to zero, which proves the claim. For the distributional statement, define \(h(x,y)=x/y\) when \(y\neq0\) and \(h(x,0)=0\). This function is continuous at every point with \(y\neq0\). Since \(P(B=0)=0\), the continuous mapping theorem applied to \((A_n,B_n)\) gives \(h(A_n,B_n)\xrightarrow{d}h(A,B)\). The arbitrary definition on \(B_n=0\) does not affect the limit: joint convergence and \(P(B=0)=0\) imply \(P(B_n=0)\to0\). \(\square\)

Worked Example: A Joint Limit Supports a Ratio Approximation

Let \(Z\) and \(W\) be independent standard normal random variables, and define

A_n=1+\frac{Z}{\sqrt n}, \qquad B_n=2+\frac{W}{\sqrt n}.

The scaled deviations have the joint distribution of \((Z,W)\) for every \(n\), and \(B_n\xrightarrow{p}2\). Calculate the scaled ratio deviation:

\sqrt n\left(\frac{A_n}{B_n}-\frac12\right) = \frac{2Z-W}{2(2+W/\sqrt n)}.

Since \(W/\sqrt n\xrightarrow{p}0\), the denominator on the right converges in probability to 4. The numerator \(2Z-W\) is tight, so the product rule and the ratio result yield

\sqrt n\left(\frac{A_n}{B_n}-\frac12\right) \xrightarrow{d}\frac{2Z-W}{4}.

By independence, the limit is normal with mean zero and variance \((4\operatorname{Var}(Z)+\operatorname{Var}(W))/16=5/16\). The calculation uses joint information about \(Z\) and \(W\); its variance would not be determined by marginal variances alone without a covariance condition.

Worked Example: A Denominator Converging to Zero

Let \(A_n=1\) and \(B_n=1/n\). Then \(A_n\to1\) and \(B_n\to0\), but

\frac{A_n}{B_n}=n,

which diverges rather than converging to a finite ratio of limits. The denominator’s limit is zero, so the division step is not continuous at the limiting pair \((1,0)\). More generally, a denominator with a random limit can be used in a distributional ratio argument when that limit is nonzero almost surely; if it has positive probability of being zero, the standard continuous mapping justification is unavailable.

A Practical Diagnostic Sequence

When a proposed limit calculation is not immediate, check the operations in order rather than relying on a formal resemblance to an algebraic identity. The following questions isolate the common points of failure.

Identify the mode of convergence.
Record whether each component converges in probability or in distribution. Do not use a probability-convergence conclusion when only a distributional limit is known.

Check products and remainders.
If a term tends to zero in probability, check that the factor multiplying it is bounded in probability. Without this control, the product may not be negligible.

Check the map at the limit.
For a continuous mapping argument, locate the discontinuities and verify that the limiting random vector lands in them with probability zero.

Check joint convergence.
If the statistic is a function of several quantities, establish convergence of their vector. Separate marginal limits need not determine the function’s limit.

Check division separately.
Verify that the limiting denominator is a nonzero constant for a probability limit, or is nonzero almost surely for the usual distributional ratio argument.

Common Mistakes and Why These Checks Matter

Discarding a small term after multiplying it. The fact that \(X_n=o_p(1)\) does not imply \(X_nY_n=o_p(1)\) for arbitrary \(Y_n\); tightness is a sufficient control.
Applying a function without checking continuity. A function’s discontinuity at a possible limiting value can invalidate the continuous mapping step.
Combining marginal limits as if they were a joint limit. A sum, product, or ratio can depend on the relationship between its components.
Taking a ratio of limits when the denominator limit is zero. The ratio map is not continuous there, so the usual theorem does not apply.

These checks matter because an invalid intermediate limit can change the final asymptotic distribution, not just its presentation. A good repair is to state the missing condition explicitly, prove it from the model, or construct a counterexample to see why it cannot be omitted. The goal is not to reject every formal calculation, but to identify the exact theorem that licenses each step.

Check Your Understanding

For each question, identify the condition that controls the proposed limit operation.

If \(X_n\xrightarrow{p}0\) and \(Y_n=O_p(1)\), what can be concluded about \(X_nY_n\), and which condition makes the conclusion valid?
Give a deterministic sequence converging to zero and a second sequence for which their product does not converge to zero. Which hypothesis fails?
Why do separate distributional limits for \(X_n\) and \(Y_n\) not always determine the limit of \(X_n+Y_n\)?
State the condition on a function \(g\) that permits the usual continuous mapping theorem to be applied to \(X_n\xrightarrow{d}X\).
What denominator condition is needed for the standard distributional limit of \(A_n/B_n\) when \((A_n,B_n)\xrightarrow{d}(A,B)\)?