Slutsky’s Theorem for Weak Limits and Vanishing Perturbations

Why a Constant Limit Changes the Picture

Weak convergence describes the limiting distribution of a random vector. Convergence in probability is stronger: it controls how close a random vector is to a specified value. Slutsky’s theorem combines these two kinds of convergence when one component approaches a constant in probability.

The constant is important. If two random vectors merely converge weakly on their own, their separate limits need not describe the limit of the pair. But if one component converges in probability to a fixed vector, it becomes a sufficiently small perturbation of that fixed value for continuous transformations of the pair.

The theorem requires no independence. The components may be dependent, and the examples below will make that dependence explicit.

Definition: Random vectors $X_n$ in $\mathbb{R}^d$ converge weakly to a random vector $X$, written $X_n \Rightarrow X$, if $\mathbb{E}[f(X_n)]\to\mathbb{E}[f(X)]$ for every bounded continuous function $f:\mathbb{R}^d\to\mathbb{R}$. Random vectors $Y_n$ converge in probability to a constant $c\in\mathbb{R}^k$, written $Y_n\to_p c$, if for every $\varepsilon>0$, $\mathbb{P}(\|Y_n-c\|>\varepsilon)\to0$.

All vectors in the theorem are understood to be defined on a common probability space, so convergence in probability has its usual meaning. The norm is the Euclidean norm in the relevant space.

A Useful Fact: Weak Convergence Gives Tightness

The proof of Slutsky’s theorem needs to control the weakly convergent component on a large bounded set. A sequence of random vectors $X_n$ in $\mathbb{R}^d$ is tight if, for every $\varepsilon>0$, there is a compact set $K\subseteq\mathbb{R}^d$ such that $\mathbb{P}(X_n\in K)>1-\varepsilon$ for every $n$.

Proposition: If $X_n\Rightarrow X$ in $\mathbb{R}^d$, then the sequence $(X_n)$ is tight.

Proof. Fix $\varepsilon>0$. Since $X$ is finite-valued almost surely, choose $R>0$ such that $\mathbb{P}(\|X\|<R)>1-\varepsilon/2$. The open-set part of the Portmanteau theorem gives $$ \liminf_{n\to\infty}\mathbb{P}(\|X_n\|<R) \geq \mathbb{P}(\|X\|<R)>1-\varepsilon/2. $$ Thus, for all sufficiently large $n$, $\mathbb{P}(\|X_n\|\leq R)>1-\varepsilon$. There are only finitely many remaining $n$. For each of them, $\|X_n\|$ is finite almost surely, so a sufficiently large radius makes the probability of lying in the corresponding closed ball greater than $1-\varepsilon$. Taking the largest of these finitely many radii and $R$, the resulting closed ball works for every $n$. It is compact in $\mathbb{R}^d$, proving tightness.

$\square$

The Perturbation Principle

Before stating the full theorem, consider a basic principle: if two sequences of random vectors are close in probability, then a weak limit for one transfers to the other. This is the step that turns a continuous transformation evaluated at $Y_n$ into one evaluated at its constant limit $c$.

Proposition: Suppose $U_n,V_n$ take values in $\mathbb{R}^m$, $U_n\Rightarrow U$, and $\|V_n-U_n\|\to_p0$. Then $V_n\Rightarrow U$.

Proof. Let $f:\mathbb{R}^m\to\mathbb{R}$ be bounded and Lipschitz, with $|f|\leq M$ and Lipschitz constant $L$. For any $\eta>0$, split according to whether $\|V_n-U_n\|\leq\eta$. On that event the difference in function values is at most $L\eta$; on its complement it is at most $2M$. Therefore $$ \left|\mathbb{E}[f(V_n)]-\mathbb{E}[f(U_n)]\right| \leq L\eta+2M\,\mathbb{P}(\|V_n-U_n\|>\eta). $$ For fixed $\eta$, the probability on the right tends to zero. Hence the limit superior of the left side is at most $L\eta$. Letting $\eta$ decrease to zero shows that the difference of expectations tends to zero. Since $U_n\Rightarrow U$, $\mathbb{E}[f(U_n)]\to\mathbb{E}[f(U)]$. Thus $\mathbb{E}[f(V_n)]\to\mathbb{E}[f(U)]$ for every bounded Lipschitz $f$. Bounded-Lipschitz tests determine weak convergence, so $V_n\Rightarrow U$.

$\square$

Slutsky’s Theorem

Theorem (Slutsky): Suppose $X_n\Rightarrow X$ in $\mathbb{R}^d$ and $Y_n\to_p c$ in $\mathbb{R}^k$, where $c$ is a constant. If $g:\mathbb{R}^d\times\mathbb{R}^k\to\mathbb{R}^m$ is continuous, then $$ g(X_n,Y_n)\Rightarrow g(X,c). $$ No independence assumption is required.

Proof. First, the continuous mapping theorem applied to $x\mapsto g(x,c)$ gives $g(X_n,c)\Rightarrow g(X,c)$. We show that $$ \|g(X_n,Y_n)-g(X_n,c)\|\to_p0. $$ Let $\varepsilon>0$ be a desired error in the output. By tightness, for any $\delta>0$ there is a compact $K\subseteq\mathbb{R}^d$ with $\mathbb{P}(X_n\notin K)<\delta$ for every $n$. Continuity of $g$ implies uniform continuity on the compact set $K\times\overline{B}(c,1)$, where $\overline{B}(c,1)$ is the closed unit ball about $c$. Consequently, there is $r\in(0,1]$ such that for every $x\in K$ and every $y$ with $\|y-c\|<r$, $$ \|g(x,y)-g(x,c)\|<\varepsilon. $$ It follows that $$ \mathbb{P}\big(\|g(X_n,Y_n)-g(X_n,c)\|\geq\varepsilon\big) \leq \mathbb{P}(X_n\notin K)+\mathbb{P}(\|Y_n-c\|\geq r). $$ The first term is less than $\delta$, while the second tends to zero. Thus the limit superior of the left side is at most $\delta$. Since $\delta>0$ was arbitrary, the displayed difference tends to zero in probability. Apply the perturbation proposition with $U_n=g(X_n,c)$ and $V_n=g(X_n,Y_n)$. It yields $g(X_n,Y_n)\Rightarrow g(X,c)$, as required.

$\square$

Familiar Forms of the Theorem

The general statement immediately gives several forms used frequently in applications. For real-valued $X_n$ and $Y_n$, addition and multiplication are continuous functions of the pair, so

X_n\Rightarrow X,\quad Y_n\to_p c \quad\Longrightarrow\quad X_n+Y_n\Rightarrow X+c,\qquad X_nY_n\Rightarrow cX.

If $X_n$ and $Y_n$ are vectors of the same dimension, the addition conclusion also holds with vector addition. If $Y_n$ is scalar and $X_n$ is a vector, the continuous map $(x,y)\mapsto yx$ gives $Y_nX_n\Rightarrow cX$.

A particularly useful form is the vanishing-perturbation principle. If $R_n\to_p0$ and $X_n\Rightarrow X$, then

X_n+R_n\Rightarrow X.

This follows by taking $Y_n=R_n$, $c=0$, and $g(x,y)=x+y$. More generally, if a sequence differs in probability by a quantity tending to zero from a sequence with a known weak limit, it has that same weak limit.

Worked Examples

Worked Example: Adding a Dependent Vanishing Perturbation

Let $Z$ have the standard normal distribution, and set $X_n=(1+1/n)Z$ and $Y_n=Z/n$. Both sequences are built from the same random variable, so they are dependent. Since $X_n\to Z$ almost surely, $X_n\Rightarrow Z$. Also $Y_n\to0$ almost surely, hence in probability. Slutsky’s theorem gives $X_n+Y_n\Rightarrow Z$. Directly, $$ X_n+Y_n=(1+2/n)Z\longrightarrow Z $$ almost surely. The example verifies that dependence does not prevent the theorem from applying.

Worked Example: Multiplying by a Component Approaching a Constant

With the same standard normal $Z$, let $X_n=Z$ and $Y_n=2+Z/n$. Then $X_n\Rightarrow Z$, while $Y_n\to2$ almost surely and therefore in probability. The product form of Slutsky’s theorem yields $X_nY_n\Rightarrow2Z$. Indeed, $$ X_nY_n=Z(2+Z/n)=2Z+Z^2/n\longrightarrow2Z $$ almost surely, because $Z$ is finite almost surely. Here too, $X_n$ and $Y_n$ are dependent.

Worked Example: A Nonlinear Continuous Transformation

Let $X_n=Z+1/n$, $Y_n=1+Z/n$, and define $g(x,y)=x^2+y$, a continuous function on $\mathbb{R}^2$. We have $X_n\Rightarrow Z$ and $Y_n\to_p1$. The theorem therefore gives $g(X_n,Y_n)\Rightarrow g(Z,1)=Z^2+1$. Checking the expression directly, $$ g(X_n,Y_n)=(Z+1/n)^2+(1+Z/n) =Z^2+1+\frac{3Z}{n}+\frac{1}{n^2} \longrightarrow Z^2+1 $$ almost surely. This illustrates why the theorem is stated for arbitrary continuous transformations, not only for addition and multiplication.

What the Theorem Does—and Does Not—Say

Slutsky’s theorem is useful because a component with a constant probability limit can be replaced, inside a continuous transformation, by that constant without changing the weak limit. It is not a claim that arbitrary weakly convergent components may be combined using only their marginal limits. The distinction is that convergence in probability to a constant controls the component’s deviation from that value, while weak convergence alone does not provide such control.

The theorem also does not require that the weak limit $X$ and the approximating variables have any particular dependence structure. Its proof controls the transformed difference $g(X_n,Y_n)-g(X_n,c)$ using tightness of $X_n$ and the fact that $Y_n$ is close to $c$ in probability. Independence is not part of that argument.

A common mistake is to treat $Y_n\to_p c$ as though $Y_n=c$ exactly for large $n$. That need not be true. The conclusion instead follows because the discrepancy becomes negligible in probability and continuity makes small changes in the input produce small changes in the output, on the bounded regions where $X_n$ lies with high probability.

Check Your Understanding

Use the theorem and its proof to answer the following questions.

State Slutsky’s theorem for a continuous function of a weakly convergent vector and a vector converging in probability to a constant.
Which part of the proof uses tightness of $X_n$, and why is continuity alone not enough to make that step uniform over all possible $X_n$?
If $X_n\Rightarrow X$ and $R_n\to_p0$, what is the weak limit of $X_n+R_n$?
Does Slutsky’s theorem require independence between $X_n$ and $Y_n$? Explain briefly.
Why does convergence in probability of $Y_n$ to a constant provide more useful information here than weak convergence of $Y_n$ alone?

Slutsky’s Theorem for Weak Limits and Vanishing Perturbations

What You'll Learn

Why a Constant Limit Changes the Picture

A Useful Fact: Weak Convergence Gives Tightness

The Perturbation Principle

Slutsky’s Theorem

Familiar Forms of the Theorem

Worked Examples

Worked Example: Adding a Dependent Vanishing Perturbation

Worked Example: Multiplying by a Component Approaching a Constant

Worked Example: A Nonlinear Continuous Transformation

What the Theorem Does—and Does Not—Say

Check Your Understanding

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