Slutsky Problems with Dependent Perturbations

Why Dependence Is Not the Main Obstacle

A common Slutsky problem gives a sequence \(X_n\) with a weak limit and a second sequence \(R_n\) that becomes negligible in probability. The two variables may be strongly dependent: the error \(R_n\) might be constructed directly from \(X_n\). That dependence does not prevent the usual conclusions for sums and products.

The key distinction is not whether two variables are independent. It is whether one component is known to approach a constant in probability, or whether the pair has a specified joint limit. Marginal weak convergence of two nonconstant sequences, by itself, does not specify how they interact.

Convergence Notation

For random vectors in Euclidean space, \(X_n\Rightarrow X\) means that \(E[f(X_n)]\to E[f(X)]\) for every bounded continuous real-valued function \(f\). Convergence in probability \(R_n\to_p 0\) means that for every \(\varepsilon>0\), \(P(\lVert R_n\rVert>\varepsilon)\to0\). A sequence converging weakly to a finite-dimensional random vector is tight: for every \(\eta>0\), some bounded set contains \(X_n\) with probability at least \(1-\eta\), uniformly in \(n\).

The tightness fact is important in product arguments. Weak convergence prevents most of the probability mass of \(X_n\) from escaping to arbitrarily large values. A small error can therefore be controlled on the event that \(X_n\) is bounded, while the remaining event has small probability.

Joint Convergence with a Vanishing Component

Proof. Let \(f\) be any bounded continuous function on \(\mathbb{R}^{d+k}\), and write \(C=\sup |f|\). We show that

Fix \(\eta>0\). By tightness, choose \(M\geq1\) such that \(P(\lVert X_n\rVert>M)<\eta\) for every \(n\). On the compact set \(\{(x,r):\lVert x\rVert\leq M,\ \lVert r\rVert\leq1\}\), \(f\) is uniformly continuous. Thus, for any \(\delta>0\), there is a \(\rho\in(0,1]\) such that \(\lVert r\rVert\leq\rho\) and \(\lVert x\rVert\leq M\) imply \(|f(x,r)-f(x,0)|<\delta\). Splitting according to these two bounds gives

Taking the limit superior in \(n\) bounds this difference by \(\delta+2C\eta\), since \(R_n\to_p0\). As \(\delta\) and \(\eta\) can be made arbitrarily small, the difference tends to zero. Also, \(x\mapsto f(x,0)\) is bounded and continuous, so weak convergence gives \(E[f(X_n,0)]\to E[f(X,0)]\). Combining the two limits proves the asserted joint convergence. □

This theorem is the useful form of Slutsky’s result for dependent perturbations. Once the joint convergence is established, the continuous mapping theorem applies to continuous functions of the pair. For example, addition and multiplication are continuous maps on Euclidean spaces.

Adding a Vanishing Error

The additive conclusion follows immediately from the joint result and the continuous map \((x,r)\mapsto x+r\), when the dimensions agree.

The important point in an exercise is to identify the error itself. If the expression is \(X_n+Y_n\), and \(Y_n\to_p0\), then \(Y_n\) is the perturbation. There is no need to establish independence, nor to calculate a joint distribution.

Products: Two Useful Routes

For products, there are two closely related ways to apply the same principle. One can treat a factor converging in probability to a constant as the nearly constant component of the joint limit. Or one can show directly that a weakly convergent factor times a vanishing error converges in probability to zero.

Proof. Apply the joint-convergence theorem to \(X_n\) and \(Y_n-c\), which converges in probability to zero. It gives \((X_n,Y_n)\Rightarrow(X,c)\). The map \((x,y)\mapsto xy\) is continuous, so the continuous mapping theorem yields \(X_nY_n\Rightarrow Xc\). □

A related estimate makes explicit why multiplication by a vanishing error is safe.

Proof. Fix \(\varepsilon>0\) and \(\eta>0\). Tightness gives \(M\geq1\) such that \(P(|X_n|>M)<\eta\) for every \(n\). Then

The first term is less than \(\eta\), and the second tends to zero. Hence the limit superior of the left-hand side is at most \(\eta\). Since \(\eta\) is arbitrary, the probability tends to zero. □

When Marginal Convergence Is Not Enough

The preceding conclusions rely on a component converging in probability to a constant. It is a different situation if both \(X_n\) and \(Y_n\) merely converge weakly to nonconstant limits. Their marginal limits do not determine the joint limit, so they need not determine the limit of their product.

This comparison also clarifies the role of independence. Independence is not needed when the Slutsky conditions already give the joint limit. Conversely, marginal convergence alone does not become sufficient merely by naming the marginals; the dependence structure can still change the product.

A Tightness Pitfall

The product lemma uses tightness, which follows from \(X_n\Rightarrow X\). Without a suitable bound on \(X_n\), multiplying by an error that tends to zero need not produce a small product.

A Practical Problem-Solving Routine

For a sum or product involving dependent sequences, organize the argument around the available convergence information rather than attempting to factor probabilities or expectations.

Why This Distinction Matters

In applications, a perturbation often shares data, noise, or latent variables with the principal term. Treating dependence as an automatic obstruction can lead to unnecessary calculations or assumptions. Slutsky’s theorem says precisely when that concern can be set aside: weak convergence supplies tightness, and convergence in probability to a constant supplies the negligible component.

The converse caution is just as important. If neither variable approaches a constant, their marginal limits do not describe their joint behavior. A rigorous solution should state which joint-convergence result justifies the map being applied, rather than assuming that separately established marginal limits can simply be combined.