Common Errors In Reasoning About Convergence

Why Convergence Arguments Go Wrong

Asymptotic arguments often compress several steps into a line such as “the estimator converges, so transform it” or “this term tends to a constant, so replace it.” Those steps are safe only when the mode of convergence and the conditions of the relevant result match the claim. Confusing convergence in probability with convergence in distribution, overlooking a discontinuity, or treating a probability limit as an exact equality can invalidate an otherwise plausible argument.

The aim here is not to repeat the laws of large numbers or central limit theorems. It is to make their conclusions harder to misuse: state exactly what converges, check the condition for transforming it, and distinguish an asymptotic approximation from a finite-sample identity.

First Check: Which Mode of Convergence?

Definition: A sequence $X_n$ converges in probability to $X$, written $X_n \xrightarrow{p} X$, if for every $\varepsilon>0$, $$ P(|X_n-X|>\varepsilon)\longrightarrow 0. $$ A sequence $X_n$ converges in distribution to $X$, written $X_n \xrightarrow{d} X$, if its cumulative distribution functions converge to the cumulative distribution function of $X$ at every continuity point of that limiting distribution function.

Convergence in probability describes how often $X_n$ lies a non-negligible distance from $X$ on the probability space where both are defined. Convergence in distribution describes the limiting distributions; it does not, by itself, describe whether $X_n$ and $X$ are close on any particular shared probability space. Convergence in probability implies convergence in distribution, but the reverse implication generally fails.

There is an important exception to remember: if $X_n \xrightarrow{d} c$, where $c$ is a constant, then $X_n \xrightarrow{p} c$. A constant has a degenerate distribution, and convergence to that distribution forces the probability of being outside any fixed neighborhood of $c$ to vanish.

Worked Example: Distributional Convergence Is Not Always Probability Convergence

For every $n$, let $X_n$ equal $1$ with probability $1/2$ and $-1$ with probability $1/2$. Its distribution does not change with $n$, so $X_n$ converges in distribution to a random variable with that same two-point distribution.

Can this sequence converge in probability to a constant $c$? For any real $c$, at least one of $-1$ and $1$ is at distance at least $1$ from $c$. That value occurs with probability $1/2$. Since a distance of at least $1$ is strictly greater than $1/2$, $$ P(|X_n-c|>1/2)\geq 1/2 $$ for every $c$ and every $n$. This probability cannot tend to zero, so $X_n$ does not converge in probability to any constant.

The conclusion is not that convergence in distribution can never imply convergence in probability. In particular, convergence in distribution to a constant does imply convergence in probability to that constant. The example instead shows why a nondegenerate distributional limit cannot be treated as a probability limit to a constant.

Second Check: Is the Transformation Continuous Where It Matters?

A function applied to a convergent sequence need not preserve its limit when the function is discontinuous at the limiting value. The relevant conditions differ slightly by mode of convergence.

Theorem: If $X_n \xrightarrow{p} X$ and $g$ is continuous at $X$ with probability one, then $g(X_n) \xrightarrow{p} g(X)$. If $X_n \xrightarrow{d} X$ and the set of discontinuities of $g$ has probability zero under the distribution of $X$, then $g(X_n) \xrightarrow{d} g(X)$.

For a probability limit $X=c$ that is constant, continuity at $c$ is the usual straightforward condition. For a random limit $X$, continuity is required with probability one at the values $X$ can take. For a distributional limit, a function may be discontinuous somewhere and still be usable: what matters is whether the limit distribution assigns positive probability to its discontinuity set.

Worked Example: A Discontinuous Indicator Can Break a Probability Limit

Let $X_n=1/n$, a deterministic random variable, and define $g(x)=1$ if $x>0$, and $g(x)=0$ if $x\leq 0$. First, $X_n \xrightarrow{p}0$: for any $\varepsilon>0$, $P(|X_n|>\varepsilon)$ is zero once $1/n\leq\varepsilon$.

But $g$ is discontinuous at $0$, and the transformed sequence is $$ g(X_n)=1 \quad\text{for every }n,\qquad g(0)=0. $$ Thus $g(X_n)$ does not converge in probability to $g(0)$; for example, $P(|g(X_n)-g(0)|>1/2)=1$ for every $n$. The original convergence is correct, but applying the probability version of the continuous mapping theorem is not justified because continuity fails at the limit.

Worked Example: A Discontinuity That Does Not Block a Distributional Limit

Let $X_n=X$ for every $n$, where $X$ takes values $-1$ and $1$, each with probability $1/2$. Let $g(x)=1$ if $x>0$, and $g(x)=0$ if $x\leq0$. The function is discontinuous at $0$, but $P(X=0)=0$. Since $X_n$ has the same distribution as $X$ for every $n$, $X_n \xrightarrow{d} X$, and $g(X_n)$ has the same distribution as $g(X)$ for every $n$. Therefore $g(X_n)\xrightarrow{d}g(X)$.

This example illustrates the precise condition: for the distributional mapping result, discontinuity at points that have zero probability under the limiting distribution is not itself an obstacle. It would be wrong either to ignore discontinuity altogether or to reject every function that has any discontinuity.

Third Check: A Limit Is Not an Exact Replacement

A probability limit says that deviations become small in probability as $n$ grows. It does not say the deviations are zero at any finite sample size. Slutsky’s theorem provides a valid way to simplify limiting distributions while retaining this distinction.

Theorem: If $X_n \xrightarrow{p} c$ for a constant $c$ and $Y_n \xrightarrow{d} Y$, then $X_n+Y_n \xrightarrow{d} c+Y$ and $X_nY_n \xrightarrow{d} cY$. These results do not require independence between $X_n$ and $Y_n$.

The theorem permits replacement of $X_n$ by $c$ in the limiting distribution of the sum or product. It does not assert that $X_n=c$ in the sample, nor does it permit replacing a random variable with a nondegenerate distributional limit by a constant.

Worked Example: Using a Probability Limit Without Erasing Randomness

Let $R_n$ take values $-1$ and $1$, and let $Z$ have a standard normal distribution. Set $X_n=2+R_n/\sqrt{n}$ and $Y_n=Z$. The construction may place $R_n$ and $Z$ on the same probability space; no independence assumption is needed.

For any fixed $\varepsilon>0$, $|X_n-2|=1/\sqrt{n}$. Consequently, $P(|X_n-2|>\varepsilon)=0$ for all sufficiently large $n$, so $X_n\xrightarrow{p}2$. Also $Y_n\xrightarrow{d}Z$, since its distribution is standard normal for every $n$.

Slutsky’s theorem therefore gives $X_nY_n\xrightarrow{d}2Z$, a normal limit with mean $0$ and variance $4$. To see why the random part vanishes on the relevant scale, subtract $2Y_n$: $$ X_nY_n-2Y_n=\frac{R_nZ}{\sqrt{n}}, \qquad P\left(\left|\frac{R_nZ}{\sqrt{n}}\right|>\varepsilon\right) =P(|Z|>\varepsilon\sqrt{n})\longrightarrow 0. $$ Here $|R_n|=1$, and the normal tail probability tends to zero as its threshold grows. The factor $X_n$ is not exactly $2$; its deviation becomes negligible, while the random $Z$ remains in the limiting distribution.

A Short Justification: Why a Constant Distributional Limit Is Special

Suppose $X_n\xrightarrow{d}c$, with $c$ constant, and fix $\varepsilon>0$. The points $c-\varepsilon$ and $c+\varepsilon$ are continuity points of the degenerate distribution function at $c$. Convergence of distribution functions therefore gives

P(X_n\leq c-\varepsilon)\longrightarrow 0, \qquad P(X_n\leq c+\varepsilon)\longrightarrow 1.

The event $\{|X_n-c|>\varepsilon\}$ is contained in the union of $\{X_n\leq c-\varepsilon\}$ and $\{X_n>c+\varepsilon\}$. The probabilities of both events tend to zero by the displayed limits. Hence $P(|X_n-c|>\varepsilon)\to0$ for every $\varepsilon>0$, which is exactly convergence in probability to $c$.

Common Mistakes and a Reliable Diagnostic

Calling every distributional limit a probability limit. A nondegenerate limiting distribution does not make the sequence close in probability to a constant. State the mode actually established.
Applying a mapping theorem without checking continuity. Identify the limit and check continuity at its value, or check whether the distributional limit assigns probability zero to the discontinuity set.
Assuming any discontinuity invalidates a distributional transformation. The distributional mapping condition permits discontinuities that occur with probability zero under the limiting distribution.
Replacing a random quantity by its limit at finite $n$. $X_n\xrightarrow{p}c$ does not mean $X_n=c$. Use a result such as Slutsky’s theorem to justify the replacement in an asymptotic distribution.
Discarding a random limiting term. If $Y_n\xrightarrow{d}Y$ for a nondegenerate $Y$, the randomness generally remains; Slutsky’s theorem replaces only the factor or summand with a constant probability limit.

A practical check is to write four items before simplifying an asymptotic expression: the sequence, its mode of convergence, the proposed transformation, and the condition that makes the transformation valid. If a constant probability limit and a distributional limit appear together, state explicitly which term Slutsky’s theorem allows you to replace in the limiting distribution.

Check Your Understanding

For each question, identify the convergence claim and the condition that makes it valid or invalid.

If $X_n\xrightarrow{d}c$ for a constant $c$, what stronger mode of convergence also follows?
Let $X_n=1/n$ and $g(x)=1$ for $x>0$, $g(x)=0$ otherwise. Why does the probability version of the continuous mapping theorem not apply at the limit?
For a distributional transformation, must $g$ be continuous everywhere? State the relevant condition.
If $A_n\xrightarrow{p}3$ and $B_n\xrightarrow{d}B$, what is the distributional limit of $A_nB_n$, and does independence need to be assumed?
Explain why $X_n\xrightarrow{p}c$ does not justify asserting $X_n=c$ for a finite sample size.