Continuous Mapping and Slutsky Theorems

Transforming a Limit

Limits of random variables are often easier to identify before a calculation than after it. If \(X_n\) converges in distribution to \(X\), a continuous function \(g\) should carry that limit to \(g(X)\). If a second sequence \(Y_n\) converges in probability to a constant, it can often be combined with \(X_n\) even when the two sequences are dependent. These principles give a reliable route to limits of sums, products, and ratios, while also indicating exactly when a calculation is not justified.

Throughout, the random variables are defined on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Convergence in probability, written \(Y_n\xrightarrow{\mathbb{P}}c\), means that for every \(\varepsilon>0\), \(\mathbb{P}(|Y_n-c|>\varepsilon)\to0\). Convergence in distribution, written \(X_n\xrightarrow{d}X\), means convergence of the distribution functions at every continuity point of the distribution function of \(X\). We use the standard equivalent characterization: \(X_n\xrightarrow{d}X\) if and only if \(\mathbb{E}[f(X_n)]\to\mathbb{E}[f(X)]\) for every bounded continuous real function \(f\). The same characterization applies to random vectors in finite-dimensional Euclidean space.

Proof. First consider convergence in distribution. Let \(h:\mathbb{R}^m\to\mathbb{R}\) be any bounded continuous function. Then \(h\circ g\) is bounded and continuous on \(\mathbb{R}^k\). By the bounded-continuous-function characterization, \(\mathbb{E}[h(g(X_n))]\to\mathbb{E}[h(g(X))]\). Applying that characterization again gives \(g(X_n)\xrightarrow{d}g(X)\).

Now suppose \(X_n\xrightarrow{\mathbb{P}}X\), where both take values in \(\mathbb{R}^k\). Fix \(\varepsilon>0\). For each \(x\in\mathbb{R}^k\), continuity of \(g\) at \(x\) gives a radius \(\delta_x>0\) such that \(\|y-x\|<\delta_x\) implies \(\|g(y)-g(x)\|<\varepsilon\). To handle the random value \(X\), first choose a compact set \(K\) with \(\mathbb{P}(X\notin K)\) as small as desired. Cover \(K\) with finitely many balls centered at points \(x_1,\ldots,x_r\), each small enough that whenever \(X\) and \(X_n\) are both sufficiently close to the same center \(x_j\), their images under \(g\) differ by less than \(\varepsilon\). Such a finite cover exists by compactness and continuity. The event \(\{\|g(X_n)-g(X)\|>\varepsilon\}\) is therefore contained in \(\{X\notin K\}\) together with events where \(X_n\) fails to be sufficiently close to \(X\). The latter probabilities tend to zero by convergence in probability. Letting \(\mathbb{P}(X\notin K)\) be arbitrarily small proves the result. \(\square\)

For the second part, the compact-set argument is useful when \(g\) is not uniformly continuous on all of its domain. In common scalar applications, continuity at the possible limit values is the key condition. For example, the reciprocal function is continuous at every nonzero number, but not at zero.

Slutsky’s Theorem and Its Proof

The continuous mapping theorem handles a transformation of one convergent sequence. Slutsky’s theorem handles several sequences when at least one has a constant probability limit. It is especially useful because the sequence converging in probability need not be independent of the sequence converging in distribution.

Proof. We verify convergence using bounded continuous test functions. First, \(X_n\xrightarrow{d}X\) implies that the sequence \((X_n)\) is tight: for every \(\eta>0\), there is a sufficiently large compact set \(K\) such that \(\mathbb{P}(X_n\notin K)<\eta\) for all sufficiently large \(n\). In one dimension this follows by choosing continuity points \(-M,M\) of the distribution of \(X\) with small tail probabilities and using convergence of the distribution functions there. The finite-dimensional version follows by applying the same tail control to the coordinates.

Let \(f\) be bounded and continuous on \(\mathbb{R}^{k+\ell}\), and write \(|f|\le C\). Fix a small error tolerance. Tightness lets us restrict \(X_n\) to a compact set \(K\), except on an event of probability at most \(\eta\). Convergence in probability lets us restrict \(Y_n\) to a fixed compact neighborhood of \(c\), with \(\mathbb{P}(\|Y_n-c\|>\delta)\) tending to zero for each \(\delta>0\). On the product of these compact sets, \(f\) is uniformly continuous. Hence, for any \(\varepsilon>0\), \(\delta\) can be chosen so that \(x\in K\) and \(\|y-c\|\le\delta\) imply \(|f(x,y)-f(x,c)|<\varepsilon\). Splitting the expectation according to these events gives

For fixed \(\eta\) and \(\delta\), the last probability tends to zero. Since \(\eta\) and \(\varepsilon\) can be made arbitrarily small, the difference of expectations tends to zero. Finally, \(x\mapsto f(x,c)\) is bounded and continuous, so \(\mathbb{E}f(X_n,c)\to\mathbb{E}f(X,c)\). Thus \(\mathbb{E}f(X_n,Y_n)\to\mathbb{E}f(X,c)\) for every such \(f\), proving the asserted joint convergence. \(\square\)

Taking continuous functions of the joint pair gives the familiar computational rules. If \(X_n\xrightarrow{d}X\) and \(Y_n\xrightarrow{\mathbb{P}}c\), then

For ratios, the additional condition \(c\ne0\) is essential. The map \((x,y)\mapsto x/y\) is continuous where \(y\ne0\), so Slutsky’s theorem and continuous mapping give

The ratio is defined whenever \(Y_n\ne0\). Under \(Y_n\xrightarrow{\mathbb{P}}c\ne0\), the probability that \(Y_n=0\) tends to zero: for example, \(\{Y_n=0\}\subseteq\{|Y_n-c|\ge |c|\}\). If a ratio is undefined on a positive-probability event for a particular \(n\), its definition must be specified there; the limit conclusion is unchanged by redefining it on events whose probabilities tend to zero.

Convergence in Probability to a Constant

A constant limit has a useful special property: convergence in distribution to that constant already implies convergence in probability. This is why Slutsky’s hypothesis can equivalently be phrased as \(Y_n\xrightarrow{d}c\) when \(c\) is constant.

Proof. Suppose first that \(Y_n\xrightarrow{\mathbb{P}}c\). For any bounded continuous \(f\), continuity at \(c\) and the bound on \(f\) imply \(\mathbb{E}[f(Y_n)]\to f(c)\): for a chosen \(\varepsilon>0\), choose \(\delta>0\) such that \(|y-c|<\delta\) gives \(|f(y)-f(c)|<\varepsilon\), then split the expectation over \(\{|Y_n-c|<\delta\}\) and its complement. The complement probability tends to zero. Thus \(Y_n\xrightarrow{d}c\).

Conversely, suppose \(Y_n\xrightarrow{d}c\). Fix \(\varepsilon>0\). Choose continuity points \(a\) and \(b\) of the distribution function of the constant \(c\) such that \(c-\varepsilon<a<c<b<c+\varepsilon\). Such points exist because the constant distribution has no discontinuities except at \(c\). Distributional convergence gives \(\mathbb{P}(Y_n\le a)\to0\) and \(\mathbb{P}(Y_n\le b)\to1\). Therefore \(\mathbb{P}(Y_n<a)\to0\) and \(\mathbb{P}(Y_n>b)\to0\), and \(\{|Y_n-c|>\varepsilon\}\subseteq\{Y_n<a\}\cup\{Y_n>b\}\). Its probability tends to zero. Hence \(Y_n\xrightarrow{\mathbb{P}}c\). \(\square\)

When a Limit Rule Does Not Apply

The conditions in these theorems are substantive. A common error is to use the ratio rule when the denominator tends to zero. For instance, take deterministic random variables \(X_n=1/n\) and \(Y_n=1/n\). Both converge in probability to zero, but \(X_n/Y_n=1\) for every \(n\). The expression \(X/Y\) for the proposed limits would be \(0/0\), which is undefined. It is therefore incorrect to conclude that the ratio tends to a ratio of the limits. A zero denominator limit does not provide continuity of division, and the ratio can have other behaviors.

Another error is to infer a joint limit from marginal distributional limits alone. Slutsky’s theorem avoids that problem only because the second limit is a constant in probability. If both coordinates have nonconstant distributional limits, their dependence can affect the limit of their sum or product. Additional information, such as joint convergence, is then needed before applying a continuous transformation to the pair.

Common mistake. A wrong calculation is: “\(X_n\xrightarrow{\mathbb{P}}0\) and \(Y_n\xrightarrow{\mathbb{P}}0\), so \(X_n/Y_n\xrightarrow{\mathbb{P}}0/0=0\).” The correction is that \(0/0\) is not defined and the ratio map is discontinuous there. In the concrete example \(X_n=Y_n=1/n\), direct calculation gives \(X_n/Y_n=(1/n)/(1/n)=1\) for every \(n\). Always check that a ratio’s denominator has a nonzero limit before applying Slutsky’s theorem.