Weak Convergence of Distributions
Y. Eddie Lu, Summer 2026
ECON 8002 course index · Lecture 8 of 9
Orientation
How to use this page
Read every theorem in three passes: identify the objects, separate assumptions from conclusions, then rebuild the proof map without looking.
Answer each quiz aloud before opening it. Exposure is not retrieval.
How should I think about Lecture 8?
Mass must not drift to infinity.
Sets and test functions detect laws.
\(\longrightarrow\)
Weak convergence asks whether probability laws approach a genuine limiting law.
\(\longrightarrow\)
Tightness produces subsequences.
Characteristic functions identify limits.
The delta method transforms fluctuations.
Course map
- Measure spaces
- Integration and Radon–Nikodym derivatives
- Distribution properties
- Conditioning
- Convergence modes
- Continuous mapping and Slutsky
- Laws of large numbers
- Weak limits and the delta method (current lecture)
- Central limit theorems and inference
Lecture 8 supplies the compactness and identification tools used to prove the central limit theorems in Lecture 9.
Prerequisite retrieval
Before proceeding, retrieve these facts:
- \(X_n\xrightarrow{d}X\) means c.d.f. convergence at continuity points.
- \(X_n\xrightarrow{d}c\) iff \(X_n\xrightarrow{p}c\) for a constant \(c\).
- CMT preserves the convergence mode under continuity at the limiting value.
- \(R_n=o_p(1)\) and \(Z_n=O_p(1)\) imply \(R_nZ_n=o_p(1)\).
- A characteristic function is \(\varphi_X(t)=E[e^{it^\top X}]\).
Tightness and subsequential limits
Weak convergence compares laws
For random vectors \(X_n,X\in\mathbb R^k\),
\[ X_n\xrightarrow{d}X \iff F_{X_n}(x)\to F_X(x) \]
at every continuity point \(x\) of \(F_X\).
The variables need not share a probability space. Weak convergence compares the probabilities they assign, not outcome-by-outcome numerical errors.
Retrieval check.
Must \(X_n\) and \(X\) be defined on one probability space for \(X_n\xrightarrow{d}X\)?
No. Their laws are being compared. A common space is required for \(X_n\xrightarrow{p}X\) or \(X_n\xrightarrow{\mathrm{a.s.}}X\) as ordinarily defined.
Motivation: a valid c.d.f. sequence can lose all mass
Let \(X_n=n\) surely. Its c.d.f. is
\[F_n(x)=\mathbf 1\{x\ge n\}.\]
For every fixed \(x\), \(F_n(x)\to0\). But the zero function is not a c.d.f. on \(\mathbb R\): its limit at \(+\infty\) is not one.
Pointwise limits of c.d.f.s need not be probability distributions. The mass can move beyond every fixed bounded region.
Definition 31: tightness
A sequence of probability measures \(\{P_n\}\) on \((\mathbb R^k,\mathcal B^k)\) is tight if
\[ \forall\varepsilon>0\ \exists K_\varepsilon\subset\mathbb R^k\text{ compact}: \inf_n P_n(K_\varepsilon)\ge1-\varepsilon. \]
One compact set must contain almost all mass for every member of the sequence.
Tightness is stochastic boundedness
For random vectors \(X_n\in\mathbb R^k\),
\[ \{P_{X_n}\}\text{ tight} \iff \forall\varepsilon>0\ \exists M_\varepsilon<\infty: \sup_nP(\|X_n\|>M_\varepsilon)<\varepsilon. \]
This is the uniform form of \(\|X_n\|=O_p(1)\). Closed Euclidean balls are compact.
Tightness plays the role that boundedness plays for deterministic sequences. It prevents escape, but it does not choose a unique limit.
Examples, nonexamples, and a boundary case
| Sequence | Tight? | Reason |
|---|---|---|
| \(X_n\sim N(0,1+n^{-1})\) | yes | one large interval controls all normal tails |
| \(X_n=n\) surely | no | every compact set eventually misses all mass |
| \(X_n\sim N((-1)^n,1)\) | yes | means and variances remain bounded |
The last sequence is tight but does not converge. Its even and odd subsequences have different limits.
Retrieval check.
Let \(X_n=0\) surely for even \(n\) and \(X_n=1\) surely for odd \(n\). Is the sequence tight? Does it converge in distribution?
It is tight because all mass lies in \([0,1]\). It does not converge: the even and odd subsequences converge to different point masses.
Proposition 8.1: tightness produces subsequences
For probability measures \(\{P_n\}\) on \(\mathbb R^k\):
- Every subsequence has a further weakly convergent subsequence iff \(\{P_n\}\) is tight.
- If \(\{P_n\}\) is tight and every weakly convergent subsequence has the same limit \(P\), then \(P_n\Rightarrow P\).
Part 1 is the finite-dimensional form of Prokhorov’s theorem.
Why one subsequential limit is enough
Suppose \(P_n\) did not converge weakly to \(P\). Then some weak neighborhood \(U\) of \(P\) would contain infinitely few \(P_n\).
- Extract a subsequence entirely outside \(U\).
- Tightness gives a weakly convergent further subsequence.
- Its limit must be \(P\) by hypothesis.
- A sequence converging to \(P\) must eventually lie in \(U\), a contradiction.
Lecture 9 will use this pattern: prove tightness, identify every possible subsequential limit, then conclude convergence of the full sequence.
Theorem 8.2: Helly selection
For every sequence of one-dimensional c.d.f.s \(\{F_n\}\), there is a subsequence \(\{F_{n_j}\}\) and a nondecreasing, right-continuous function \(F:\mathbb R\to[0,1]\) such that
\[F_{n_j}(x)\to F(x)\]
at every continuity point of \(F\).
The multivariate compactness result requires the corresponding multivariate extension, not scalar Helly alone.
A Helly limit may be defective
For \(F_n(x)=\mathbf1\{x\ge n\}\), every subsequence has the same pointwise limit \(F(x)=0\).
The limit is nondecreasing and right-continuous, but
\[\lim_{x\to\infty}F(x)=0\ne1.\]
Tightness supplies the missing tail condition and forces the selected limit to be a genuine c.d.f.
Retrieval check.
Which c.d.f. property can fail in Helly’s theorem when the sequence is not tight?
Total mass can be lost: the selected limit need not approach one as \(x\to+\infty\) or zero as \(x\to-\infty\).
The Portmanteau equivalences
Portmanteau: seven views of one limit
For random vectors \(X_n,X\in\mathbb R^k\), the following are equivalent:
- \(X_n\xrightarrow{d}X\).
- \(E[f(X_n)]\to E[f(X)]\) for every bounded continuous \(f\).
- The same for every bounded Lipschitz \(f\).
- \(\liminf_nE[f(X_n)]\ge E[f(X)]\) for every nonnegative continuous \(f\).
The remaining set formulations come next.
Portmanteau: the set formulations
The preceding statements are also equivalent to:
- For every open \(G\), \[\liminf_nP(X_n\in G)\ge P(X\in G).\]
- For every closed \(C\), \[\limsup_nP(X_n\in C)\le P(X\in C).\]
- For every Borel \(B\) with \(P(X\in\partial B)=0\), \[P(X_n\in B)\to P(X\in B).\]
Why inequalities point in opposite directions
Open set. Points on the boundary are excluded. Approximating laws may place less mass inside than the limit, hence a lower bound through \(\liminf\).
Closed set. Boundary points are included. Approximating laws may carry extra mass there, hence an upper bound through \(\limsup\).
Equality is recovered when the limiting law assigns zero probability to the boundary.
Boundary mass is the obstruction
Let \(X_n=1/n\) and \(X=0\) surely.
- For \(B=(-1,1)\), \(P(X\in\partial B)=0\) and both probabilities equal one eventually.
- For \(B=\{0\}\), \(P(X\in\partial B)=1\) but \[P(X_n\in B)=0\not\to1=P(X\in B).\]
The geometry of \(B\) alone does not decide convergence. The limiting probability of its boundary does.
Retrieval check.
If \(X\sim N(0,1)\), is \(B=(-\infty,2]\) a continuity set of \(P_X\)?
Yes. Its boundary is \(\{2\}\), and a continuous distribution assigns zero probability to every singleton.
Complete Portmanteau proof map
One complete implication architecture is
\[ (1)\Rightarrow(2)\Rightarrow(3)\Rightarrow(5) \Longleftrightarrow(6)\Rightarrow(7)\Rightarrow(1), \]
with
\[ (2)\Rightarrow(4)\Rightarrow(5). \]
The source proves only selected arrows. The next sections supply the missing links.
From weak convergence to test functions
Because \(\mathbb R^k\) is Polish, Skorokhod’s representation theorem gives random vectors \(Y_n,Y\) on one probability space such that
\[P_{Y_n}=P_{X_n},\qquad P_Y=P_X,qquad Y_n\xrightarrow{\mathrm{a.s.}}Y.\]
For bounded continuous \(f\), CMT and bounded convergence yield
\[E[f(X_n)]=E[f(Y_n)]\to E[f(Y)]=E[f(X)].\]
Bounded Lipschitz functions are a subclass, so \((2)\Rightarrow(3)\) is immediate.
Bounded Lipschitz functions detect open sets
For open \(G\), define
\[f_m(x)=\min\{1,m\,d(x,G^c)\}.\]
Then \(f_m\) is bounded and Lipschitz, \(0\le f_m\le\mathbf1_G\), and \(f_m(x)\uparrow\mathbf1_G(x)\).
Thus, for fixed \(m\),
\[E[f_m(X)]=\lim_nE[f_m(X_n)]\le\liminf_nP(X_n\in G).\]
Let \(m\to\infty\) and use monotone convergence to obtain the open-set inequality.
Nonnegative functions and truncation
If \(f\ge0\) is continuous, then \(f_M=f\wedge M\) is bounded and continuous. Hence
\[E[f_M(X)]=\lim_nE[f_M(X_n)]\le\liminf_nE[f(X_n)].\]
Let \(M\to\infty\). Monotone convergence gives
\[E[f(X)]\le\liminf_nE[f(X_n)].\]
Conversely, applying this result to continuous approximations of \(\mathbf1_G\) recovers the open-set inequality.
Open and closed sets are complements
For closed \(C\), its complement \(C^c\) is open. Therefore
\[ \begin{aligned} \limsup_nP(X_n\in C) &=1-\liminf_nP(X_n\in C^c)\\ &\le1-P(X\in C^c)=P(X\in C). \end{aligned} \]
The same calculation in reverse shows \((5)\Longleftrightarrow(6)\).
Closed and open bounds give continuity sets
For a Borel set \(B\),
\[B^\circ\subseteq B\subseteq\overline B.\]
Portmanteau gives
\[ P(X\in B^\circ)\le\liminf_nP(X_n\in B) \le\limsup_nP(X_n\in B)\le P(X\in\overline B). \]
If \(P(X\in\partial B)=0\), the two outside probabilities equal \(P(X\in B)\), so the middle sequence converges.
Continuity rectangles recover c.d.f.s
Let \(B_x=(-\infty,x_1]\times\cdots\times(-\infty,x_k]\). Then \(F_X(x)=P(X\in B_x)\).
If \(x\) is a continuity point of the multivariate c.d.f., then \(P_X(\partial B_x)=0\). Thus \(B_x\) is a \(P_X\)-continuity set. Starting from condition (7), convergence on continuity sets gives
\[F_{X_n}(x)\to F_X(x).\]
This is exactly c.d.f. convergence at every continuity point, condition (1), so it proves the promised arrow \((7)\Rightarrow(1)\) and closes the cycle.
Retrieval check.
If \(C\) is closed, complete the statement: \(\limsup_nP(X_n\in C)\ \square\ P(X\in C)\).
The symbol is \(\le\). Closed sets may retain boundary mass that disappears in the limit, so Portmanteau gives an upper bound.
Why boundedness cannot be dropped
Let \(X_n=n\) with probability \(1/n\) and \(X_n=0\) otherwise. Then \(X_n\xrightarrow{p}0\), hence \(X_n\xrightarrow{d}0\), but
\[E[X_n]=1\not\to0.\]
Weak convergence controls bounded continuous test functions. Unbounded payoffs need additional tail control, such as uniform integrability.
Characteristic functions and vector limits
Characteristic functions encode laws
For \(X\in\mathbb R^k\),
\[\varphi_X(t)=E[e^{it^\top X}],\qquad t\in\mathbb R^k.\]
They satisfy \(|\varphi_X(t)|\le1\), \(\varphi_X(0)=1\), and continuity at zero. The law of \(X\) uniquely determines \(\varphi_X\), and the characteristic function uniquely determines the law.
For \(Y=A^\top X+v\),
\[\varphi_Y(u)=e^{iu^\top v}\varphi_X(Au).\]
Theorem 8.4: Lévy-Cramér continuity theorem
For random vectors \(X_n,X\in\mathbb R^k\),
\[ X_n\xrightarrow{d}X \iff \varphi_{X_n}(t)\to\varphi_X(t) \quad\forall t\in\mathbb R^k. \]
Necessity is an application of Portmanteau. Sufficiency requires showing that pointwise Fourier convergence cannot hide mass escaping to infinity.
Lévy necessity: use real test functions
For fixed \(t\),
\[e^{it^\top x}=\cos(t^\top x)+i\sin(t^\top x).\]
Both real functions are bounded and continuous. Portmanteau therefore gives
\[E[\cos(t^\top X_n)]\to E[\cos(t^\top X)],\] \[E[\sin(t^\top X_n)]\to E[\sin(t^\top X)].\]
Combining real and imaginary parts yields \(\varphi_{X_n}(t)\to\varphi_X(t)\).
Lévy sufficiency: proof map
- Use a symmetric Fourier average near zero to prove tightness.
- Extract a weakly convergent further subsequence.
- Apply the necessity direction to its characteristic functions.
- Use uniqueness to identify the subsequential limit as \(P_X\).
- Use Proposition 8.1(ii) to recover convergence of the full sequence.
The delicate step is uniform tail control.
The Fourier average controls tails
For the scalar proof and \(u>0\), Fubini gives
\[ \frac1u\int_{-u}^{u}[1-\varphi_{X_n}(t)]\,dt =2\int\left(1-\frac{\sin(ux)}{ux}\right)dP_{X_n}(x). \]
If \(|x|>2/u\), then the integrand on the right is at least \(1/2\). Hence
\[ P(|X_n|>2/u) \le \frac1u\int_{-u}^{u}[1-\varphi_{X_n}(t)]\,dt. \]
The symmetric integral is real even though each characteristic function is complex.
For \(\mathbb R^k\), apply the scalar bound to each coordinate projection \(e_\ell^\top X_n\), then use a union bound to control the complement of one compact box.
Repairing the uniform-tightness step
Continuity of \(\varphi_X\) at zero lets us choose \(u\) so the corresponding integral for \(X\) is below \(\varepsilon/2\).
Pointwise convergence and boundedness permit DCT on \([-u,u]\), so the same bound holds for all \(n\ge N\). It does not directly hold uniformly over every \(n\).
Enlarge the compact interval to control the finitely many laws \(P_{X_1},\ldots,P_{X_{N-1}}\). This produces one compact set controlling all \(n\) and proves tightness.
Identify every subsequential limit
Tightness gives \(P_{X_{n_j}}\Rightarrow P\) along a further subsequence. Let \(\varphi_P\) be the characteristic function of \(P\).
By Lévy necessity,
\[\varphi_{X_{n_j}}(t)\to\varphi_P(t).\]
The original hypothesis gives the same limit \(\varphi_X(t)\). Thus \(\varphi_P=\varphi_X\), and uniqueness implies \(P=P_X\).
Every convergent subsequence has limit \(P_X\), so \(X_n\xrightarrow{d}X\).
Quick quiz 6: locate tightness
Retrieval check.
Why does pointwise convergence of characteristic functions not finish the sufficiency proof immediately?
One must first ensure that the sequence has a weakly convergent subsequence. The Fourier bound supplies tightness, which supplies that subsequence.
Theorem 8.5: Cramér-Wold device
For random vectors \(X_n,X\in\mathbb R^k\),
\[ X_n\xrightarrow{d}X \iff c^\top X_n\xrightarrow{d}c^\top X \quad\forall c\in\mathbb R^k. \]
Every one-dimensional direction must converge. Checking only the coordinates does not identify the dependence structure.
Cramér-Wold proof
For \(u\in\mathbb R\) and \(c\in\mathbb R^k\),
\[\varphi_{c^\top X_n}(u)=\varphi_{X_n}(uc).\]
- Vector convergence implies projection convergence by CMT.
- Conversely, projection convergence and Lévy give convergence of \(\varphi_{X_n}(uc)\) for every \(u,c\).
- Set \(u=1\) and \(c=t\) to obtain convergence at every \(t\in\mathbb R^k\).
- Lévy-Cramér then gives \(X_n\xrightarrow{d}X\).
Coordinates alone are not enough
Let \(Z\sim N(0,1)\). Compare
\[X=(Z,Z),\qquad Y=(Z,-Z).\]
Both vectors have standard-normal first and second coordinates. But for \(c=(1,1)^\top\),
\[c^\top X=2Z,\qquad c^\top Y=0.\]
Marginal convergence of each coordinate is weaker than joint convergence.
Retrieval check.
To apply Cramér-Wold in \(\mathbb R^3\), is it enough to check \(c=e_1,e_2,e_3\)?
No. Those directions identify only the marginals. The theorem requires every fixed \(c\in\mathbb R^3\) to recover the joint law.
Density convergence and the delta method
Theorem 8.6: Scheffé’s theorem
Let \(f_n\) and \(f\) be densities with respect to the same measure \(\lambda\). If
\[f_n(x)\to f(x)\quad\lambda\text{-a.e.}\]
and \(f\) is itself a density, then
\[\int|f_n-f|\,d\lambda\to0.\]
The condition that \(f\) integrate to one prevents mass from disappearing.
Scheffé: the corrected proof
Since \((f-f_n)^+\le f\) and \((f-f_n)^+\to0\) almost everywhere, DCT gives
\[\int(f-f_n)^+\,d\lambda\to0.\]
Both densities integrate to one, so
\[\int(f_n-f)^+\,d\lambda=\int(f-f_n)^+\,d\lambda.\]
Therefore
\[ \int|f_n-f|\,d\lambda =2\int(f-f_n)^+\,d\lambda\to0. \]
Scheffé is stronger than weak convergence
For any Borel set \(B\),
\[ |P_{X_n}(B)-P_X(B)| \le\int_B|f_n-f|\,d\lambda \le\int|f_n-f|\,d\lambda\to0. \]
Thus probabilities converge uniformly over all Borel sets, not only continuity sets. In particular, \(X_n\xrightarrow{d}X\).
If \(f_n(x)\to0\) pointwise while each \(f_n\) integrates to one, Scheffé does not apply because zero is not a density.
Quick quiz 8: conserved mass
Retrieval check.
Why is the assumption “\(f\) is a density” essential in Scheffé’s theorem?
Pointwise convergence can lose mass. Equality of the total integrals forces the positive and negative errors to balance and makes \(L^1\) convergence possible.
CMT versus the delta method
If \(X_n\xrightarrow{p}c\) and \(g\) is continuous, CMT gives only
\[g(X_n)\xrightarrow{p}g(c).\]
If a scaled error has a nondegenerate limit,
\[a_n(X_n-c)\xrightarrow{d}Y,\]
the delta method finds the first-order limit of \(a_n[g(X_n)-g(c)]\).
CMT transforms a level. The delta method transforms a local fluctuation.
Theorem 8.7(i): first-order delta method
Let \(X_n,Y\in\mathbb R^k\), \(c\in\mathbb R^k\), and \(a_n>0\) with \(a_n\to\infty\). If
\[a_n(X_n-c)\xrightarrow{d}Y\]
and \(g:\mathbb R^k\to\mathbb R\) is differentiable at \(c\), then
\[ a_n[g(X_n)-g(c)] \xrightarrow{d}\nabla g(c)^\top Y. \]
Differentiability supplies the exact remainder
Differentiability at \(c\) means
\[g(c+h)-g(c)=\nabla g(c)^\top h+r(h),\]
where
\[\frac{r(h)}{\|h\|}\to0\qquad\text{as }h\to0.\]
Substitute \(h=X_n-c\) and multiply by \(a_n\):
\[ a_n[g(X_n)-g(c)] =\nabla g(c)^\top a_n(X_n-c)+a_nr(X_n-c). \]
Delta proof: the remainder vanishes
From \(a_n(X_n-c)\xrightarrow{d}Y\),
\[a_n\|X_n-c\|=O_p(1).\]
Since \(a_n^{-1}\to0\), Slutsky gives \(X_n\xrightarrow{p}c\). Therefore
\[ \frac{r(X_n-c)}{\|X_n-c\|}=o_p(1). \]
Define the quotient as zero when \(X_n=c\).
Hence
\[a_nr(X_n-c)=O_p(1)o_p(1)=o_p(1).\]
Cramér-Wold handles the linear term; Slutsky adds the negligible remainder.
Quick quiz 9: first-order transformation
Retrieval check.
If \(\sqrt n(\hat\theta_n-\theta)\xrightarrow{d}N(0,V)\) and \(\theta>0\), find the limit of \(\sqrt n(\log\hat\theta_n-\log\theta)\).
Since \(g'(\theta)=1/\theta\), \(\sqrt n(\log\hat\theta_n-\log\theta)\xrightarrow{d}N(0,V/\theta^2)\).
Theorem 8.7(ii): higher-order delta method
Suppose \(g\) admits an order-\(m\) Taylor expansion near \(c\) with
\[D^jg(c)=0\quad(j=1,\ldots,m-1),\]
and remainder \(o(\|h\|^m)\). Then
\[ a_n^m[g(X_n)-g(c)] \xrightarrow{d} \frac1{m!}D^mg(c)[Y,\ldots,Y]. \]
Continuous \(m\)th partial derivatives on a neighborhood are a standard sufficient condition for the required remainder. A nonzero \(D^mg(c)\) makes the leading term nontrivial.
Higher-order remainder proof
Put \(H_n=a_n(X_n-c)\). Weak convergence gives \(H_n=O_p(1)\), and \(X_n\xrightarrow{p}c\). With the quotient defined as zero at \(h=0\),
\[ a_n^m r(X_n-c) =\|H_n\|^m \frac{r(X_n-c)}{\|X_n-c\|^m} =O_p(1)o_p(1)=o_p(1). \]
CMT sends the continuous \(m\)-linear leading term to \(m!^{-1}D^mg(c)[Y,\ldots,Y]\). Slutsky adds the negligible remainder.
Zero gradient: first order can say nothing
Let \(g(x)=x^2\), \(c=0\), and
\[\sqrt nX_n\xrightarrow{d}Z.\]
The first-order derivative is \(g'(0)=0\), so the ordinary delta method gives the degenerate limit zero. The second-order result gives
\[nX_n^2=(\sqrt nX_n)^2\xrightarrow{d}Z^2.\]
If \(Z\sim N(0,1)\), the limit is \(\chi_1^2\).
Quick quiz 10: choose the order
Retrieval check.
If \(g'(c)=0\) but \(g''(c)\ne0\), what scaling is suggested when \(\sqrt n(X_n-c)=O_p(1)\)?
Use \(n\), because \((X_n-c)^2=O_p(n^{-1})\). The second-order delta method produces \(\tfrac12g''(c)Y^2\).
Corollary 8.8: normal delta method
If
\[a_n(X_n-c)\xrightarrow{d}Y,\qquad Y\sim N_k(\mu,\Sigma),\]
then
\[ a_n[g(X_n)-g(c)] \xrightarrow{d} N\!\left(\nabla g(c)^\top\mu, \nabla g(c)^\top\Sigma\nabla g(c)\right). \]
The covariance matrix may be singular. If the asymptotic variance is zero, the first-order limit is degenerate.
Econometric use: a ratio estimator
Let \(g(\mu_1,\mu_2)=\mu_1/\mu_2\) with \(\mu_2\ne0\). Then
\[ \nabla g(\mu_1,\mu_2) =\begin{pmatrix}\mu_2^{-1}\\-\mu_1\mu_2^{-2}\end{pmatrix}. \]
If
\[\sqrt n(\hat\mu-\mu)\xrightarrow{d}N_2(0,\Sigma),\]
then the asymptotic variance of \(\hat\mu_1/\hat\mu_2\) is \(\nabla g(\mu)^\top\Sigma\nabla g(\mu)\).
This covers ratios, elasticities, and scaled performance measures.
Denominator audit. The condition \(\mu_2\ne0\) is structural: the ratio is undefined at zero, and the derivative needed by the delta method is not finite there.
Synthesis and retrieval
Assumption audit
| Result | Assumption doing the work | Failure without it |
|---|---|---|
| Tightness | one compact set controls all laws | mass can escape |
| Portmanteau equality | \(P(X\in\partial B)=0\) | boundary probabilities may jump |
| Lévy sufficiency | continuity at zero and uniqueness | candidate need not be a law |
| Scheffé | \(f\) is a density | total mass may disappear |
| Delta method | differentiability plus a valid remainder | linear approximation may fail |
| Ratio delta method | denominator limit nonzero | statistic may explode |
Common errors and false converses
- Tightness gives subsequential limits, not a unique limit.
- Weak convergence does not imply convergence of unbounded expectations.
- Coordinatewise convergence does not imply joint convergence.
- Pointwise density convergence does not imply \(L^1\) convergence unless the limiting function is a density.
- CMT does not supply a rate or asymptotic variance.
- A zero gradient does not prove the transformed statistic has no uncertainty.
- Nonsingularity of \(\Sigma\) is not required for a linear normal image.
Proof blueprints
Portmanteau
test functions \(\to\) open/closed bounds \(\to\) continuity sets \(\to\) c.d.f.s
Lévy
Fourier tail bound \(\to\) tightness \(\to\) subsequence \(\to\) uniqueness
Delta
Taylor expansion \(\to\) \(O_p(1)o_p(1)\) remainder \(\to\) Slutsky
Quick quiz 12: theorem selection
Retrieval check.
Match each task to a theorem: identify a vector limit from every projection; upgrade density convergence to uniform probability approximation; transform a \(\sqrt n\)-scaled estimator through a smooth nonlinear map.
Cramér-Wold; Scheffé; the delta method, respectively.
Exit ticket 1: state Portmanteau
State c.d.f. convergence, bounded-continuous expectation convergence, the open-set \(\liminf\) inequality, and the closed-set \(\limsup\) inequality. For probability convergence on a Borel set, add \(P(X\in\partial B)=0\).
Exit ticket 2: rebuild Lévy sufficiency
Fourier averaging proves eventual tail control; finitely many early laws are absorbed into a larger compact set; tightness gives a convergent subsequence; characteristic-function uniqueness identifies its limit; the subsequence criterion gives convergence of the full sequence.
Exit ticket 3: prove the delta remainder is negligible
Write
\[a_nr(X_n-c)=a_n\|X_n-c\|\frac{r(X_n-c)}{\|X_n-c\|}.\]
The first factor is \(O_p(1)\), the second is \(o_p(1)\) by differentiability and \(X_n\xrightarrow{p}c\), so their product is \(o_p(1)\).
Mastery checklist
You should now be able to:
- define tightness and distinguish it from convergence;
- state all seven Portmanteau equivalents and prove the main implication cycle;
- explain the tightness and uniqueness steps in Lévy-Cramér;
- use Cramér-Wold without confusing marginals with joint convergence;
- prove Scheffé with positive and negative parts;
- derive first- and second-order delta limits;
- identify the assumption that fails in each counterexample.