Sums and Products Under Slutsky Conditions

Why Constant Limits Simplify Weak-Limit Calculations

A weak limit describes the limiting distribution of a sequence, while convergence in probability to a constant describes a perturbation that becomes negligible in size. When these occur together, sums and products have predictable limits—even if the two sequences are dependent.

The assumptions matter. A sequence converging in probability to a constant does not, by itself, guarantee that its sum or product with another sequence has a weak limit. We need a weak limit for the other sequence as well. With that hypothesis in place, the calculation reduces to applying a continuous function to a jointly convergent pair.

Central principle: If $X_n\Rightarrow X$ and $Y_n\xrightarrow{p}c$, then a continuous operation applied to $(X_n,Y_n)$ has the corresponding weak limit obtained by applying that operation to $(X,c)$.

Set-Up and Notation

Throughout, $X_n$ and $Y_n$ are random vectors in $\mathbb{R}^d$, and $X$ is a random vector in $\mathbb{R}^d$. Weak convergence $X_n\Rightarrow X$ means convergence of expectations for every bounded continuous function $f:\mathbb{R}^d\to\mathbb{R}$. Convergence in probability $Y_n\xrightarrow{p}c$ means that for every $\varepsilon>0$,

\mathbb{P}\bigl(\|Y_n-c\|>\varepsilon\bigr)\longrightarrow 0.

Here $\|\cdot\|$ is the Euclidean norm. Addition of vectors is componentwise. We write $x\odot y$ for the coordinatewise product:

x\odot y=(x_1y_1,\ldots,x_dy_d), \qquad x,y\in\mathbb{R}^d.

This is distinct from a dot product, which is a scalar. For a scalar $a$ and vector $x$, $ax$ denotes ordinary scalar multiplication.

The Joint Limit Used in the Calculations

The relevant joint-convergence result is the following: if $X_n\Rightarrow X$ in $\mathbb{R}^d$ and $Y_n\xrightarrow{p}c$ in $\mathbb{R}^d$, then $(X_n,Y_n)\Rightarrow(X,c)$ in $\mathbb{R}^{2d}$. The two components at index $n$ need not be independent. The convergence-in-probability component has a constant limit, which is why this joint conclusion holds without an independence assumption.

We combine this result with the continuous mapping theorem: if $Z_n\Rightarrow Z$ and $g$ is continuous, then $g(Z_n)\Rightarrow g(Z)$. Addition and coordinatewise multiplication are continuous maps on Euclidean spaces, so they give the desired formulas.

Weak Limits for Sums

Theorem: Let $X_n\Rightarrow X$ in $\mathbb{R}^d$, and let $Y_n\xrightarrow{p}c$ in $\mathbb{R}^d$, where $c$ is a fixed vector. Then $$ X_n+Y_n\Rightarrow X+c. $$

Proof. By the joint-convergence result, $(X_n,Y_n)\Rightarrow(X,c)$. Define the continuous map $g:\mathbb{R}^d\times\mathbb{R}^d\to\mathbb{R}^d$ by $g(x,y)=x+y$. The continuous mapping theorem gives $$ g(X_n,Y_n)\Rightarrow g(X,c). $$ Since $g(X_n,Y_n)=X_n+Y_n$ and $g(X,c)=X+c$, the claimed convergence follows. $\square$

The result includes vanishing perturbations as the special case $c=0$. It also handles nonzero shifts: a term converging in probability to $c$ shifts the weak limit by $c$, rather than changing its shape.

Worked Example: Adding a Dependent Perturbation

Let $Z$ be a standard normal random variable, and set $X_n=Z$ and $Y_n=1+Z/n$. The two sequences are dependent because they are defined using the same $Z$. We have $X_n\Rightarrow Z$. Also, for every $\varepsilon>0$, $$ \mathbb{P}(|Y_n-1|>\varepsilon) =\mathbb{P}(|Z|>n\varepsilon)\longrightarrow 0, $$ since the tail probability of a fixed real-valued random variable tends to zero as its threshold tends to infinity. Thus $Y_n\xrightarrow{p}1$, and the sum theorem yields $X_n+Y_n\Rightarrow Z+1$, a normal distribution with mean $1$ and variance $1$. Directly, $X_n+Y_n=1+Z+Z/n$; the extra term $Z/n$ tends to zero in probability, in agreement with the theorem.

Weak Limits for Coordinatewise Products

Theorem: Let $X_n\Rightarrow X$ in $\mathbb{R}^d$, and let $Y_n\xrightarrow{p}c$ in $\mathbb{R}^d$, where $c$ is a fixed vector. Then $$ X_n\odot Y_n\Rightarrow X\odot c. $$

Proof. Again, $(X_n,Y_n)\Rightarrow(X,c)$. Define $h:\mathbb{R}^d\times\mathbb{R}^d\to\mathbb{R}^d$ by $h(x,y)=x\odot y$. Each coordinate of $h$ is the product of two real coordinates, so $h$ is continuous. The continuous mapping theorem gives $$ h(X_n,Y_n)\Rightarrow h(X,c). $$ The left side is $X_n\odot Y_n$, and the right side is $X\odot c$. This proves the result. $\square$

The constant $c$ can have zero coordinates. In those coordinates, the limiting product is zero; in the other coordinates, the limit is the corresponding coordinate of $X$ multiplied by the fixed coordinate of $c$. No division is involved in this conclusion.

Worked Example: A Vector Product with a Zero Coordinate

Let $U$ and $V$ be any real-valued random variables, and define $X_n=(U,V)$ and $Y_n=(2+U/n,V/n)$. Then $X_n\Rightarrow(U,V)$, since its distribution is the same for every $n$. For every $\varepsilon>0$, $$ \mathbb{P}\bigl(\|Y_n-(2,0)\|>\varepsilon\bigr) = \mathbb{P}\left(\sqrt{(U/n)^2+(V/n)^2}>\varepsilon\right) \longrightarrow 0. $$ Indeed, $\sqrt{U^2+V^2}$ is finite almost surely, so $\sqrt{U^2+V^2}/n\to0$ almost surely and hence in probability. Therefore $Y_n\xrightarrow{p}(2,0)$. The product is $$ X_n\odot Y_n=(2U+U^2/n,\;V^2/n). $$ The theorem gives $X_n\odot Y_n\Rightarrow(2U,0)$. The first coordinate is scaled by $2$, while the second tends to zero. The calculation requires neither independence of $U$ and $V$ nor any assumptions on their distributions beyond being real-valued random variables.

Scalar Multiplication of a Random Vector

The same reasoning covers a random scalar multiplying a random vector. This is a product operation with a scalar-valued component rather than a coordinatewise product of two vectors.

Corollary: Suppose $X_n\Rightarrow X$ in $\mathbb{R}^d$ and $A_n\xrightarrow{p}a$ for a fixed real number $a$. Then $A_nX_n\Rightarrow aX$.

To verify this, apply the joint-convergence result to $(X_n,A_n)$, whose limit is $(X,a)$. The map $(x,t)\mapsto tx$ from $\mathbb{R}^d\times\mathbb{R}$ to $\mathbb{R}^d$ is continuous. The continuous mapping theorem therefore gives $A_nX_n\Rightarrow aX$.

Worked Example: A Random Scalar Multiplier

Let $Z$ be standard normal, $X_n=Z$, and $A_n=3+|Z|/n$. Since $X_n\Rightarrow Z$ and $|Z|/n\to0$ almost surely, $A_n\xrightarrow{p}3$. The scalar-multiplication result gives $A_nX_n\Rightarrow3Z$, which is normal with mean $0$ and variance $9$. The exact product is $A_nX_n=3Z+Z|Z|/n$; its second term tends to zero almost surely, confirming the stated limit. Here the multiplier and vector are dependent, but the constant probability limit still gives the conclusion.

Assumptions, Dependence, and a Common Pitfall

The weak-limit assumption on $X_n$ is essential. Convergence in probability of $Y_n$ to a constant does not supply a weak limit for $X_n$. For example, in one dimension take $X_n=(-1)^n$ and $Y_n=1$. Then $Y_n\xrightarrow{p}1$, but $X_n+Y_n$ alternates between $0$ and $2$, and $X_nY_n$ alternates between $-1$ and $1$. Neither sequence has a weak limit: along the even and odd subsequences, the distributions are different point masses. Thus no limit can be concluded from the behavior of $Y_n$ alone.

A second common error is to treat dependence as an obstacle that must be removed by assuming independence. These results impose no independence condition. The joint-convergence result applies because one component converges in probability to a constant, and the continuous maps then give the sum and product limits. The dependence may affect the finite-$n$ joint distributions, as in the examples, without invalidating these conclusions.

Finally, be precise about which product is meant. For vectors, $X_n\odot Y_n$ denotes coordinatewise multiplication and has a vector limit. The dot product $X_n\cdot Y_n$ is scalar-valued and is also a continuous function of the pair, so the same joint-convergence argument can be used when that operation is the quantity of interest. Always specify the operation and its output space before stating a limit.

Check Your Understanding

For each question, identify the assumptions being used and the operation whose limit is being calculated.

If $X_n\Rightarrow X$ in $\mathbb{R}^d$ and $Y_n\xrightarrow{p}c$, what is the weak limit of $X_n+Y_n$?
Under the same assumptions, write the $j$-th coordinate of the limit of $X_n\odot Y_n$.
Does either conclusion require $X_n$ and $Y_n$ to be independent? Explain why or why not.
What hypothesis is missing if one knows only that $Y_n\xrightarrow{p}c$ and wants to conclude a weak limit for $X_n+Y_n$?
If $A_n\xrightarrow{p}0$ and $X_n\Rightarrow X$, what is the weak limit of $A_nX_n$?