Continuous Mapping and Slutsky’s Theorem

Y. Eddie Lu, Summer 2026

ECON 8002 course index · Lecture 6 of 9

Orientation

How to use this page

Answer every quiz aloud before opening it. For every limit argument, name the mode, the map, and the continuity set before manipulating symbols.

Primary source: ECON 8002 Lecture 6, pp. 38–40. The proofs and examples below expand and, where useful, clarify the source.

How should I think about Lecture 6?

A statistic has a known limit

A nuisance estimate becomes nearly constant

\(\rightarrow\)

Which operations preserve the limit, in which mode, and at which points?

\(\rightarrow\)

CMT: transform one convergent object

Slutsky: combine noise with a constant limit

Inference: transform and studentize estimators

Where this lecture sits

1. Measure spaces
2. Integration
3. Distribution properties
4. Conditioning
5. Convergence modes
6. Transform known limits (current lecture)
7. Laws of large numbers
8. Weak-convergence tools
9. Central limit theorems and the delta method

Prerequisite retrieval

Before proceeding, you should be able to state:

1. \(X_n\xrightarrow{p}X\) in an \(\varepsilon\)-probability sentence.
2. \(X_n\xrightarrow{d}X\) using c.d.f.s at continuity points.
3. \(X_n=O_p(1)\) and \(R_n=o_p(1)\).
4. The closed-set inequality in Portmanteau.

Example or application

Lecture 5 supplies the language. Lecture 6 supplies the legal operations.

The continuous mapping theorem

Quick quiz 1: preserve the mode

Suppose \(X_n\xrightarrow{d}X\) and \(g\) is continuous everywhere. Which conclusion is guaranteed?

Show answer

\[g(X_n)\xrightarrow{d}g(X).\]

CMT preserves distributional convergence. It does not strengthen it to probability convergence or imply convergence of expectations.

Why continuity appears

Definition and notation

A map \(g:\mathbb R^k\to\mathbb R^m\) is continuous at \(x\) if, for every sequence \(x_j\to x\), \[g(x_j)\to g(x).\] Equivalently, for every \(\varepsilon>0\) there is \(\delta>0\) such that \[\|y-x\|<\delta\implies \|g(y)-g(x)\|<\varepsilon.\]

Probability makes the input error small often. Continuity turns small input error into small output error.

A discontinuity can destroy a limit

Let \(X_n=1/n\) deterministically, so \(X_n\to0\) in every usual mode. Define

\[g(x)=\mathbf 1\{x=0\}.\]

Then \(g(X_n)=0\) for every \(n\), while \(g(0)=1\).

Counterexample or invalid inference

The problem is not randomness. The map fails to preserve even an ordinary deterministic limit at its discontinuity.

The limit decides where continuity matters

The reciprocal map \(g(x)=1/x\) is discontinuous only at \(0\).

• If \(X_n\xrightarrow{p}2\), then \(X_n^{-1}\xrightarrow{p}1/2\).
• If \(X_n\xrightarrow{d}X\) and \(P(X=0)=0\), then \(X_n^{-1}\xrightarrow{d}X^{-1}\), once the sample reciprocals are defined.
• If \(P(X=0)>0\), CMT does not license the conclusion.

The map may be bad elsewhere. Only values reached by the limiting law matter.

Quick quiz 2: continuity set

Let \(g(x)=\mathbf 1\{x\le t\}\) and \(X_n\xrightarrow{d}X\). What extra condition permits CMT?

Show answer

\(P(X=t)=0\). The indicator is discontinuous only at \(t\), so it is continuous at \(X\) almost surely exactly when the limiting law has no atom at \(t\).

Portmanteau supplies the weak-convergence bridge

For random vectors \(X_n,X\in\mathbb R^k\),

\[ X_n\xrightarrow{d}X \quad\Longleftrightarrow\quad \limsup_{n\to\infty}P(X_n\in F)\le P(X\in F) \]

for every closed \(F\subseteq\mathbb R^k\).

To study \(g(X_n)\), pull a target event back:

\[\{g(X_n)\in F\}=\{X_n\in g^{-1}(F)\}.\]

Continuous Mapping Theorem

Let \(X_n,X\) be random \(k\)-vectors, placed on a common probability space for the probability and almost-sure clauses. Let \(g:(\mathbb R^k,\mathcal B^k)\to(\mathbb R^m,\mathcal B^m)\) be measurable. Let \(D_g\) be its set of discontinuity points and suppose \(P(X\in D_g)=0\). Then

\[ \begin{aligned} X_n\xrightarrow{d}X&\implies g(X_n)\xrightarrow{d}g(X),\\ X_n\xrightarrow{p}X&\implies g(X_n)\xrightarrow{p}g(X),\\ X_n\xrightarrow{\mathrm{a.s.}}X&\implies g(X_n)\xrightarrow{\mathrm{a.s.}}g(X). \end{aligned} \]

Every assumption has a job

Assumption	Job
\(g:\mathbb R^k\to\mathbb R^m\)	Allows scalar, vector, and matrix-valued transformations.
\(g\) measurable	Makes \(g(X_n)\) and \(g(X)\) random vectors.
\(P(X\in D_g)=0\)	Ignores discontinuities the limit reaches with probability zero.
Continuity off \(D_g\)	Preserves the deterministic limit at relevant values.

Assumption check

CMT keeps the input convergence mode. It does not strengthen it.

Show proof

Almost-sure proof: work path by path

Let

\[A=\{\omega:X_n(\omega)\to X(\omega)\},\qquad B=\{\omega:X(\omega)\notin D_g\}.\]

Both events have probability one. Fix \(\omega\in A\cap B\).

1. The numerical sequence \(X_n(\omega)\) converges to \(X(\omega)\).
2. The point \(X(\omega)\) is a continuity point of \(g\).
3. Deterministic continuity gives \(g(X_n(\omega))\to g(X(\omega))\).

Thus \(g(X_n)\xrightarrow{\mathrm{a.s.}}g(X)\).

Probability proof: use the subsequence principle

A sequence \(Z_n\) converges in probability to \(Z\) if and only if every subsequence \(Z_{n_j}\) has a further subsequence \(Z_{n_{j_\ell}}\) satisfying

\[Z_{n_{j_\ell}}\xrightarrow{\mathrm{a.s.}}Z.\]

This turns a probability-convergence question into a pathwise question, where ordinary continuity can be applied directly.

Probability proof: apply pathwise CMT

Take any subsequence \(X_{n_j}\). Since \(X_n\xrightarrow{p}X\), there is a further subsequence with

\[X_{n_{j_\ell}}\xrightarrow{\mathrm{a.s.}}X.\]

The almost-sure part of CMT then gives

\[g(X_{n_{j_\ell}})\xrightarrow{\mathrm{a.s.}}g(X),\]

hence also convergence in probability. Every subsequence has such a further subsequence, so the converse half of the subsequence principle yields \(g(X_n)\xrightarrow{p}g(X)\).

Quick quiz 3: diagnose the proof

A proof says, “Since \(X_n\xrightarrow{p}X\), we have \(g(X_n)\xrightarrow{p}g(X)\).” What is missing?

Show answer

The proof must state that \(g\) is measurable and continuous at \(X\) almost surely. Convergence alone need not survive a discontinuous transformation.

Show proof

Weak proof: pull back and close

Let \(F\subseteq\mathbb R^m\) be closed and put \(C=D_g^c\). Since \(g\) is continuous on \(C\),

\[ \overline{g^{-1}(F)}\subseteq g^{-1}(F)\cup C^c. \]

Indeed, if \(x_j\in g^{-1}(F)\), \(x_j\to x\in C\), continuity gives \(g(x_j)\to g(x)\); closedness of \(F\) gives \(g(x)\in F\).

This is the deterministic core of the weak-convergence proof.

Weak proof: apply Portmanteau

\[ \begin{aligned} \limsup_n P(g(X_n)\in F) &=\limsup_nP(X_n\in g^{-1}(F))\\ &\le\limsup_nP(X_n\in\overline{g^{-1}(F)})\\ &\le P(X\in\overline{g^{-1}(F)})\\ &\le P(X\in g^{-1}(F))+P(X\in C^c)\\ &=P(g(X)\in F). \end{aligned} \]

The closed-set form of Portmanteau yields \(g(X_n)\xrightarrow{d}g(X)\).

Canonical example: normal to chi-square

Suppose \(X_n\xrightarrow{d}Z\) with \(Z\sim N(0,1)\). Since \(g(x)=x^2\) is continuous everywhere,

\[X_n^2\xrightarrow{d}Z^2\sim\chi_1^2.\]

Assumption check

CMT does not say the square is asymptotically normal. The limit is the law of the transformed variable \(Z^2\).

A reusable map library

Map	Continuity requirement
\(x\mapsto |x|\) or \(x^2\)	Everywhere
\(x\mapsto\log x\)	Limit is positive almost surely
\(x\mapsto1/x\)	Limit is nonzero almost surely
\((x,y)\mapsto x+y,xy,\max\{x,y\}\)	Everywhere, but needs joint convergence
\(x\mapsto a^\top x\)	Everywhere for fixed \(a\)
\(x\mapsto x^\top A x\)	Everywhere for fixed \(A\)
\(A\mapsto A^{-1}\)	On nonsingular matrices

Quick quiz 4: matrix inversion

Suppose \(A_n\xrightarrow{p}A\) for random \(k\times k\) matrices. What condition makes \(A_n^{-1}\xrightarrow{p}A^{-1}\) a valid CMT conclusion?

Show answer

\(P(\det A\ne0)=1\). Matrix inversion is continuous on the nonsingular matrices. The sample inverse must also be defined, perhaps arbitrarily on a vanishing exceptional event.

CMT does not imply expectation convergence

Let \(X_n=n\) with probability \(1/n\) and \(0\) otherwise. Then \(X_n\xrightarrow{p}0\), but

\[E[X_n]=1\quad\text{for every }n.\]

Counterexample or invalid inference

Convergence in probability controls typical errors. A rare large value can still carry a fixed expected contribution. Expectation convergence needs tail control, such as uniform integrability or domination.

Quick quiz 5: one map or two variables?

Suppose only \(X_n\xrightarrow{d}X\) and \(Y_n\xrightarrow{d}Y\). Can CMT be applied to \(g(x,y)=xy\)?

Show answer

Not from the marginal limits alone. CMT needs joint convergence \((X_n,Y_n)\xrightarrow{d}(X,Y)\). The product limit depends on the limiting dependence structure.

Slutsky’s theorem

Why Slutsky is needed

Econometric statistics often combine:

A CLT piece

\[X_n\xrightarrow{d}X.\]

Nondegenerate sampling noise remains.

An LLN piece

\[Y_n\xrightarrow{p}c.\]

The nuisance estimate becomes asymptotically constant.

Slutsky tells us how their sum, product, and ratio behave.

Slutsky’s theorem

Suppose \(X_n\xrightarrow{d}X\) and \(Y_n\xrightarrow{p}c\), where \(c\in\mathbb R\) is constant. Then

\[ \begin{aligned} X_n+Y_n&\xrightarrow{d}X+c,\\ Y_nX_n&\xrightarrow{d}cX,\\ Y_n^{-1}X_n&\xrightarrow{d}c^{-1}X,\qquad c\ne0. \end{aligned} \]

No finite-sample independence between \(X_n\) and \(Y_n\) is required.

Why a constant limit is special

The theorem hides a joint-convergence result:

\[ X_n\xrightarrow{d}X,\quad Y_n\xrightarrow{p}c \quad\Longrightarrow\quad (X_n,Y_n)\xrightarrow{d}(X,c). \]

The second coordinate carries no nontrivial limiting randomness. Therefore its limiting dependence with \(X\) does not need to be identified.

If \(Y_n\xrightarrow{d}Y\) with random \(Y\), marginal convergence is not enough.

Quick quiz 6: is independence required?

Suppose \(X_n\xrightarrow{d}X\) and \(Y_n\xrightarrow{p}3\), but \(X_n\) and \(Y_n\) are dependent for every \(n\). Find the product limit.

Show answer

\[X_nY_n\xrightarrow{d}3X.\]

Slutsky needs no independence because the second limiting coordinate is constant.

Show proof

Additive proof: bracket the c.d.f.

Fix \(t\in\mathbb R\) and \(\varepsilon>0\). On \(\{|Y_n-c|<\varepsilon\}\),

\[X_n\le t-c-\varepsilon\implies X_n+Y_n\le t \implies X_n\le t-c+\varepsilon.\]

Thus

\[ F_{X_n}(t-c-\varepsilon)-P(|Y_n-c|\ge\varepsilon) \le F_{X_n+Y_n}(t) \]

and the corresponding upper bound uses \(F_{X_n}(t-c+\varepsilon)\).

Additive proof: squeeze the limit

Fix \(t\) such that \(t-c\) is a continuity point of \(F_X\). Choose \(\varepsilon_m\downarrow0\) so that \(t-c\pm\varepsilon_m\) are also continuity points, which is possible because a c.d.f. has at most countably many discontinuities. For each \(m\),

\[ F_X(t-c-\varepsilon_m) \le \liminf_nF_{X_n+Y_n}(t) \le \limsup_nF_{X_n+Y_n}(t) \le F_X(t-c+\varepsilon_m). \]

Let \(m\to\infty\). Continuity at \(t-c\) gives

\[F_{X_n+Y_n}(t)\to F_X(t-c)=F_{X+c}(t).\]

Hence \(X_n+Y_n\xrightarrow{d}X+c\).

Show proof

The stochastic-order route

Two facts compress the product proof:

\[X_n\xrightarrow{d}X\implies X_n=O_p(1),\]

and

\[Y_n\xrightarrow{p}c\implies Y_n-c=o_p(1).\]

Then

\[O_p(1)o_p(1)=o_p(1).\]

This is the “bounded scale times diminishing error” rule.

Why the product rule is true

Let \(Z_n=O_p(1)\) and \(R_n=o_p(1)\). Fix \(\varepsilon,\eta>0\). Choose \(M\) so \(P(\|Z_n\|>M)<\eta/2\) eventually. Then

\[ P(\|Z_nR_n\|>\varepsilon) \le P(\|Z_n\|>M)+P(\|R_n\|>\varepsilon/M)<\eta \]

eventually. Thus \(Z_nR_n=o_p(1)\). For vectors and matrices, use compatible dimensions and a submultiplicative norm.

Why weak convergence gives \(O_p(1)\)

Fix \(\eta>0\). Choose a continuity point \(M\) of the law of \(|X|\) such that \(P(|X|>M)<\eta/2\). CMT gives

\[|X_n|\xrightarrow{d}|X|,\]

so \(P(|X_n|>M)\to P(|X|>M)<\eta/2\). Therefore \(P(|X_n|>M)<\eta\) eventually.

If the definition uses \(\sup_n\) rather than an eventual bound, enlarge \(M\) to absorb the finitely many earlier laws.

\(O_p(1)\) means that mass cannot drift arbitrarily far away. It does not impose an almost-sure finite cap.

Show proof

Product proof from the decomposition

Write

\[Y_nX_n=cX_n+(Y_n-c)X_n.\]

Now:

1. \(cX_n\xrightarrow{d}cX\) by CMT.
2. \((Y_n-c)X_n=o_p(1)\) because \(o_p(1)O_p(1)=o_p(1)\).
3. Additive Slutsky gives \(cX_n+(Y_n-c)X_n\xrightarrow{d}cX\).

This decomposition is a reusable proof pattern for feasible asymptotic statistics.

Quick quiz 7: repair the rate statement

Why is “\(X_n=O_p(1)\) means \(X_n\) is finitely capped” incorrect?

Show answer

\(O_p(1)\) is a high-probability bound: for every \(\eta>0\), there is a finite \(M\) such that \(P(|X_n|>M)<\eta\) for all sufficiently large \(n\). It is not an almost-sure deterministic bound.

Show proof

Ratio proof: reciprocal first

Assume \(c\ne0\). The reciprocal is continuous at \(c\), so CMT gives

\[Y_n^{-1}\xrightarrow{p}c^{-1}.\]

Then the product part of Slutsky gives

\[Y_n^{-1}X_n\xrightarrow{d}c^{-1}X.\]

The condition \(c\ne0\) supplies continuity at the limit. Defining \(Y_n^{-1}\) also requires \(Y_n\ne0\), at least outside an event whose probability vanishes.

Quick quiz 8: name both tools

Complete the proof and name each theorem:

\[Y_n\xrightarrow{p}c\ne0 \quad\Rightarrow\quad Y_n^{-1}\xrightarrow{p}c^{-1} \quad\Rightarrow\quad Y_n^{-1}X_n\xrightarrow{d}c^{-1}X.\]

Show answer

The first arrow is CMT with \(r(y)=1/y\). The second arrow is Slutsky’s product result together with \(X_n\xrightarrow{d}X\).

Applications and first-order refinements

Studentization

Suppose

\[\frac{\sqrt n(\bar X_n-\mu)}{\sigma}\xrightarrow{d}N(0,1), \qquad S_n\xrightarrow{p}\sigma>0.\]

CMT gives \(\sigma/S_n\xrightarrow{p}1\). Hence

\[ \frac{\sqrt n(\bar X_n-\mu)}{S_n} = \frac{\sqrt n(\bar X_n-\mu)}{\sigma}\frac{\sigma}{S_n} \xrightarrow{d}N(0,1). \]

This licenses replacing an unknown population scale by a consistent estimate.

Matrix Slutsky in regression

A common \(k\)-parameter expansion is

\[\sqrt n(\hat\beta_n-\beta_0)=Q_n^{-1}S_n,\]

where \(Q_n,Q\in\mathbb R^{k\times k}\), \(S_n,Z\in\mathbb R^k\), \(\beta_0\in\mathbb R^k\), \(Q_n\xrightarrow{p}Q\), \(Q\) is nonsingular, and \(S_n\xrightarrow{d}Z\sim N_k(0,\Omega)\) for \(\Omega\in\mathbb R^{k\times k}\).

CMT and multivariate Slutsky give

\[ \sqrt n(\hat\beta_n-\beta_0) \xrightarrow{d}Q^{-1}Z \sim N(0,Q^{-1}\Omega Q^{-\top}). \]

When Slutsky is not enough

Let \(Z\sim N(0,1)\).

• If \(X_n=Z\) and \(Y_n=Z\), then \(X_nY_n=Z^2\).
• If \(X_n=Z\) and \(Y_n=-Z\), both marginals are still standard normal, but \(X_nY_n=-Z^2\).

Counterexample or invalid inference

When both limits are random, the joint limiting law is the missing object. Marginal limits do not determine a product limit.

CMT versus the delta method

If \(\hat\theta_n\xrightarrow{p}\theta\) and \(g\) is continuous at \(\theta\), CMT gives

\[g(\hat\theta_n)\xrightarrow{p}g(\theta).\]

If also

\[\sqrt n(\hat\theta_n-\theta)\xrightarrow{d}Z\]

and \(g\) is differentiable at \(\theta\), the delta method gives

\[ \sqrt n\{g(\hat\theta_n)-g(\theta)\} \xrightarrow{d}g'(\theta)Z. \]

CMT gives consistency. The delta method gives a first-order distribution.

Worked chain: log transformation

Suppose \(\theta>0\), \(\hat\theta_n\xrightarrow{p}\theta\), and

\[\sqrt n(\hat\theta_n-\theta)\xrightarrow{d}N(0,V).\]

1. CMT: \(\log\hat\theta_n\xrightarrow{p}\log\theta\).
2. Derivative: \(g'(\theta)=1/\theta\).
3. Delta method:

\[ \sqrt n(\log\hat\theta_n-\log\theta) \xrightarrow{d}N\left(0,\frac{V}{\theta^2}\right). \]

Quick quiz 9: mixed theorem selection

Suppose \(Z_n\xrightarrow{d}N(0,1)\) and \(S_n\xrightarrow{p}2\). Find the limit of \(Z_n/S_n\).

Show answer

By CMT, \(S_n^{-1}\xrightarrow{p}1/2\). By Slutsky,

\[\frac{Z_n}{S_n}\xrightarrow{d}N\left(0,\frac14\right).\]

Synthesis and retrieval

Assumption audit

Before invoking either theorem, ask:

1. What is the exact convergence mode of every sequence?
2. Is the map fixed and measurable?
3. At which limiting values is it continuous?
4. For a multivariable map, do I have joint convergence?
5. If I divide or invert, is the limiting denominator nonsingular?
6. Am I claiming only convergence of random objects, or have I smuggled in an expectation or rate conclusion?

Show proof

Proof blueprints

Definition and notation

CMT in probability

Take an arbitrary subsequence, extract an almost-surely convergent further subsequence, apply pathwise continuity, then invoke the subsequence principle.

CMT in distribution

Take a closed target set. Pull it back, close the inverse image, apply Portmanteau, then remove null discontinuities.

Example or application

Slutsky

Separate the nondegenerate weak limit from the constant probability limit. Choose the sum, product, or reciprocal map and check continuity.

Common errors

• Concluding \(\xrightarrow{p}\) from a \(\xrightarrow{d}\) input.
• Applying a two-variable map from marginal convergence alone.
• Dividing when the limiting denominator can be zero.
• Reading \(O_p(1)\) as an almost-sure finite cap.
• Inferring expectation convergence from CMT.
• Treating a changing random map \(g_n\) as a fixed continuous map \(g\).
• Believing Slutsky needs independence.
• Believing separate random marginal limits determine a joint limit.

Quick quiz 10: theorem or no theorem?

For each claim, say CMT, Slutsky, or not enough information.

1. \(A_n\xrightarrow{p}A\), \(A\) nonsingular a.s.; find \(A_n^{-1}\).
2. \(X_n\xrightarrow{d}X\), \(Y_n\xrightarrow{d}Y\); find \(X_n+Y_n\).
3. \(X_n\xrightarrow{d}X\), \(Y_n\xrightarrow{p}0\); find \(X_n+Y_n\).

Show answer

1. CMT: \(A_n^{-1}\xrightarrow{p}A^{-1}\).
2. Not enough: joint convergence is missing.
3. Slutsky: \(X_n+Y_n\xrightarrow{d}X\).

Exit ticket 1

State CMT for distributional convergence, including dimensions, measurability, and the exact continuity requirement.

Show answer

If \(X_n\xrightarrow{d}X\) in \(\mathbb R^k\) and measurable \(g:\mathbb R^k\to\mathbb R^m\) is continuous at every \(x\in C\) as a map on \(\mathbb R^k\), with \(P(X\in C)=1\), then \(g(X_n)\xrightarrow{d}g(X)\).

Exit ticket 2

Rebuild the product proof from \(X_n\xrightarrow{d}X\) and \(Y_n\xrightarrow{p}c\).

Show proof

\(X_n=O_p(1)\) and \(Y_n-c=o_p(1)\), so \((Y_n-c)X_n=o_p(1)\). Also \(cX_n\xrightarrow{d}cX\). Since \(Y_nX_n=cX_n+(Y_n-c)X_n\), additive Slutsky yields the conclusion.

Exit ticket 3

Explain why Slutsky does not require independence.

Show answer

Because \(Y_n\xrightarrow{p}c\) makes the second limiting coordinate constant. The pair converges jointly to \((X,c)\), so no nontrivial limiting dependence remains to identify.

Mastery checklist

You should now be able to:

• state CMT in three modes and identify its continuity set;
• reconstruct its pathwise, probability, and Portmanteau proof strategies;
• distinguish a transformed limit from convergence of expectations;
• show \(X_n\xrightarrow{d}X\Rightarrow X_n=O_p(1)\);
• prove Slutsky’s product and ratio results using \(O_p/o_p\) and CMT;
• studentize an asymptotically normal statistic;
• recognize when joint convergence, rather than Slutsky, is required.

Close the page and retrieve

Name the mode. Name the map. Check continuity.

Then decide whether the problem needs CMT, Slutsky, or a first-order refinement such as the delta method.

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