Slutsky's Theorem and Its Proof

Why Slutsky’s Theorem Is Useful

Asymptotic calculations often produce one quantity whose distribution approaches a nondegenerate random variable and another quantity that settles near a constant. For example, a normalized statistic may have a limiting distribution while an estimated scale converges in probability to its target. Slutsky’s theorem makes precise how these pieces combine. Its key feature is that the pieces do not need to be independent.

The theorem’s central step is joint convergence: if $X_n$ converges in distribution to $X$ and $Y_n$ converges in probability to a constant $c$, then the pair $(X_n,Y_n)$ converges in distribution to $(X,c)$. Once this is established, continuous transformations give the usual addition and multiplication rules.

Definitions: For random vectors $X_n$ and $X$ in $\mathbb{R}^k$, $X_n\xrightarrow{d}X$ means that $E[h(X_n)]\to E[h(X)]$ for every bounded continuous function $h:\mathbb{R}^k\to\mathbb{R}$. For random vectors $Y_n$ and a constant $c\in\mathbb{R}^m$, $Y_n\xrightarrow{p}c$ means that $P(\|Y_n-c\|>\varepsilon)\to0$ for every $\varepsilon>0$. Any fixed norm may be used in these finite-dimensional spaces.

The Joint-Convergence Step

Convergence of the two marginal distributions alone would not generally determine the limiting joint distribution. Here the second coordinate has the special feature that its probability limit is a constant. That concentration is enough to establish the joint limit, regardless of any dependence between the coordinates.

Theorem (joint-convergence form of Slutsky’s theorem): Suppose $X_n\xrightarrow{d}X$ in $\mathbb{R}^k$ and $Y_n\xrightarrow{p}c$ in $\mathbb{R}^m$, where $c$ is nonrandom. Then $$ (X_n,Y_n)\xrightarrow{d}(X,c). $$ No independence assumption is required.

Proof. We use the bounded-continuous-function definition of convergence in distribution. Let $h:\mathbb{R}^{k+m}\to\mathbb{R}$ be bounded and continuous, with $|h|\le M$. We will show that $E[h(X_n,Y_n)]\to E[h(X,c)]$.

First, the sequence $X_n$ is tight: for every $\eta>0$, there is a compact set $K\subset\mathbb{R}^k$ such that $\sup_n P(X_n\notin K)<\eta$. In this setting, tightness follows from $X_n\xrightarrow{d}X$. Indeed, choose an open ball $B_R$ with $P(X\in B_R)>1-\eta/2$. The Portmanteau theorem gives $\liminf_n P(X_n\in B_R)\ge P(X\in B_R)$, so all sufficiently large $n$ have probability at least $1-\eta$ in the closed ball of radius $R$. Enlarging the radius to handle the finitely many earlier $n$ gives the claimed compact set.

On the compact set $K\times\{y:\|y-c\|\le1\}$, the function $h$ is uniformly continuous. Given $\varepsilon>0$, choose $0<\delta\le1$ so that $|h(x,y)-h(x,c)|<\varepsilon$ whenever $x\in K$ and $\|y-c\|\le\delta$. Splitting the expectation according to whether $X_n\in K$ and $\|Y_n-c\|\le\delta$ gives

\left|E[h(X_n,Y_n)]-E[h(X_n,c)]\right| \le \varepsilon+2M\left\{P(X_n\notin K)+P(\|Y_n-c\|>\delta)\right\}.

The first probability in braces is less than $\eta$, uniformly in $n$, and the second tends to zero because $Y_n\xrightarrow{p}c$. Since $\varepsilon$ and $\eta$ can be made arbitrarily small, the difference of expectations tends to zero. Also, $x\mapsto h(x,c)$ is bounded and continuous, so $X_n\xrightarrow{d}X$ implies $E[h(X_n,c)]\to E[h(X,c)]$. Combining these limits proves $E[h(X_n,Y_n)]\to E[h(X,c)]$, as required. $\square$

Continuous Transformations

The joint result supplies the input for a continuous transformation. The continuous mapping theorem says that if $Z_n\xrightarrow{d}Z$ and $g$ is measurable and continuous at $Z$ with probability one, then $g(Z_n)\xrightarrow{d}g(Z)$. In particular, it applies when $g$ is continuous everywhere.

Theorem (Slutsky’s theorem): If $X_n\xrightarrow{d}X$ in $\mathbb{R}^k$ and $Y_n\xrightarrow{p}c$ in $\mathbb{R}^m$, then for any measurable $g:\mathbb{R}^{k+m}\to\mathbb{R}^r$ that is continuous at $(X,c)$ with probability one, $$ g(X_n,Y_n)\xrightarrow{d}g(X,c). $$

Derivation. The joint-convergence result gives $(X_n,Y_n)\xrightarrow{d}(X,c)$. Apply the continuous mapping theorem to the function $g$ to obtain the stated limit. $\square$

Taking $g(x,y)=x+y$ gives the addition rule; taking $g(x,y)=xy$ for real-valued variables gives the product rule. These functions are continuous everywhere, so

X_n+Y_n\xrightarrow{d}X+c, \qquad X_nY_n\xrightarrow{d}cX.

For real-valued $X_n,Y_n$, if $c\ne0$, the function $g(x,y)=x/y$ is continuous wherever $y\ne0$, including at every $(x,c)$. Thus

\frac{X_n}{Y_n}\xrightarrow{d}\frac{X}{c}.

If a ratio is undefined on an event where $Y_n=0$, it may be assigned any fixed value there, provided that the probability of this event tends to zero. Since $Y_n\xrightarrow{p}c\ne0$, $P(Y_n=0)\le P(|Y_n-c|\ge |c|)\to0$, so such a convention cannot change the limiting distribution.

Worked Applications

Worked Example: Adding a Dependent Perturbation

Let $Z\sim N(0,1)$, and for each $n\ge1$ define $X_n=Z$ and $Y_n=2+\mathbf{1}_{\{Z>0\}}/n$. Find the limiting distribution of $X_n+Y_n$.

The distribution of $X_n$ is $N(0,1)$ for every $n$, so $X_n\xrightarrow{d}Z$. Also, $|Y_n-2|\le1/n$ for every outcome, hence $Y_n\xrightarrow{p}2$. Slutsky’s addition rule gives

X_n+Y_n\xrightarrow{d}Z+2\sim N(2,1).

The variables are dependent: $Y_n$ records whether $Z$ is positive. That dependence does not affect the result because the second term approaches the constant 2.

Worked Example: Multiplying by a Convergent Scale

Let $Z\sim N(0,1)$, set $X_n=2Z$, and let $Y_n=3+\mathbf{1}_{\{Z\ge0\}}/n$. Determine the limit of $X_nY_n$.

For every $n$, $X_n\sim N(0,4)$, so $X_n\xrightarrow{d}X$, where $X\sim N(0,4)$. Since $|Y_n-3|\le1/n$, $Y_n\xrightarrow{p}3$. The product rule therefore yields $X_nY_n\xrightarrow{d}3X$. Multiplying a normal variable by 3 multiplies its variance by $3^2$, so

3X\sim N(0,9\cdot4)=N(0,36).

The finite-$n$ product also makes the vanishing perturbation visible: $X_nY_n=6Z+2Z\mathbf{1}_{\{Z\ge0\}}/n$. The second term tends to zero almost surely because $Z$ is finite almost surely. The limit is therefore consistent with the Slutsky calculation.

Worked Example: A Nonlinear Transformation

Let $Z\sim N(0,1)$, $X_n=Z$, $Y_n=1/n$, and $g(x,y)=x^2+y$. Find the limiting distribution of $g(X_n,Y_n)$.

We have $X_n\xrightarrow{d}Z$ and $Y_n\xrightarrow{p}0$. The function $g$ is continuous everywhere, so Slutsky’s theorem gives $g(X_n,Y_n)\xrightarrow{d}g(Z,0)=Z^2$. By definition, the square of a standard normal variable has a chi-square distribution with one degree of freedom. In particular, its density is

f(t)=\frac{1}{\sqrt{2\pi t}}e^{-t/2},\qquad t>0.

Here the transformed statistic is exactly $Z^2+1/n$. Its added term vanishes, while its random part retains the nondegenerate chi-square limit.

Common Pitfall and Why the Constant Matters

A common mistake is to assume that separate convergence statements automatically imply joint convergence, or to multiply marginal limiting distributions as though the variables were independent. Marginal limits alone do not generally specify dependence. Slutsky’s theorem avoids this problem because one coordinate converges in probability to a constant; the proof shows that this coordinate can be replaced by its constant inside any bounded continuous test function, with an error that vanishes.

It would be a different claim to replace $Y_n\xrightarrow{p}c$ by $Y_n\xrightarrow{d}Y$ for a nonconstant $Y$. Separate distributional limits do not, in general, establish a joint limit or determine the dependence between $X$ and $Y$. The constant probability limit is the ingredient that makes Slutsky’s joint-convergence step work without an independence condition.

Key takeaway: Combine a distributional limit with a probability limit to a constant by first establishing joint convergence. Then apply the continuous mapping theorem; do not impose independence, and do not confuse two marginal limits with a joint limit.

Check Your Understanding

Answer each question using the joint-convergence and transformation results above.

If $X_n\xrightarrow{d}X$ and $Y_n\xrightarrow{p}4$, what is the limit in distribution of $X_n+Y_n$?
Does Slutsky’s theorem require $X_n$ and $Y_n$ to be independent? Identify the condition that makes independence unnecessary.
If $X_n\xrightarrow{d}X$ and $Y_n\xrightarrow{p}c\ne0$, what is the limit of $X_n/Y_n$?
Why does convergence of both marginal distributions, by itself, not generally determine convergence of a pair?
Suppose $X_n\xrightarrow{d}X$ and $Y_n\xrightarrow{p}0$. What limit follows for $X_n^2+Y_n$, and which theorem justifies it?