Why Transform a Distributional Limit?
A limit theorem often gives a useful approximation for a random variable or vector, but the statistic of interest may be a transformation of it. If a centered and scaled statistic converges in distribution, for example, we may want the limit of its square, norm, or several linear combinations at once. The continuous mapping principle gives a general rule for carrying a distributional limit through a fixed continuous function.
The rule is simple to state, but its scope matters: the function must be continuous on its domain, and the transformed limit is the function applied to the limiting random vector. The resulting distribution need not have the same form as the original one.
Definition and Theorem
A random vector is measurable as a map into its Euclidean state space. If \(g:\mathbb{R}^k\to\mathbb{R}^m\) is continuous, then \(g(X_n)\) and \(g(X)\) are also random vectors: continuous functions are Borel measurable, so composing them with measurable random vectors preserves measurability.
The map may change dimension: \(g\) can turn a vector into a scalar, a scalar into a vector, or one vector into another of a different dimension. It does not need to be one-to-one or have an inverse. It must, however, be the same fixed function for every \(n\).
Proof of the Continuous Mapping Theorem
The definition of convergence in distribution makes the proof direct. It is important that the test function in the definition can be composed with \(g\) without losing boundedness or continuity.
Let \(h:\mathbb{R}^m\to\mathbb{R}\) be any bounded continuous function. To prove \(g(X_n)\xrightarrow{d}g(X)\), we must show convergence of expectations for every such \(h\).
Because \(g\) and \(h\) are continuous, \(h\circ g:\mathbb{R}^k\to\mathbb{R}\) is continuous. Because \(h\) is bounded, \(h\circ g\) is bounded as well.
The definition of \(X_n\xrightarrow{d}X\), applied to the bounded continuous function \(h\circ g\), gives the desired expectation convergence.
In detail, $$ E[h(g(X_n))]=E[(h\circ g)(X_n)] \longrightarrow E[(h\circ g)(X)] =E[h(g(X))]. $$ This holds for every bounded continuous \(h:\mathbb{R}^m\to\mathbb{R}\). By the definition of convergence in distribution, \(g(X_n)\xrightarrow{d}g(X)\). \(\square\)
Worked Applications
Worked Example: Squaring a Standard Normal Limit
Suppose \(X_n\xrightarrow{d}Z\), where \(Z\sim N(0,1)\). Find the limiting distribution of \(X_n^2\).
Take \(g(x)=x^2\), which is continuous on \(\mathbb{R}\). The continuous mapping theorem gives \(X_n^2\xrightarrow{d}Z^2\). To identify this limit, for \(t<0\) we have \(P(Z^2\leq t)=0\). For \(t\geq0\), $$ P(Z^2\leq t) =P(-\sqrt{t}\leq Z\leq \sqrt{t}) =\Phi(\sqrt{t})-\Phi(-\sqrt{t}) =2\Phi(\sqrt{t})-1, $$ where \(\Phi\) is the standard normal distribution function and the last equality uses the symmetry of the standard normal distribution. Thus \(Z^2\) has a chi-square distribution with one degree of freedom.
For \(t>0\), differentiating its distribution function gives the density $$ \frac{d}{dt}\{2\Phi(\sqrt{t})-1\} =2\phi(\sqrt{t})\frac{1}{2\sqrt{t}} =\frac{e^{-t/2}}{\sqrt{2\pi t}}, $$ where \(\phi(z)=e^{-z^2/2}/\sqrt{2\pi}\). The density integrates to one because it was obtained from the distribution function, which increases from zero to one. The limit is not normal: squaring has changed the support to \([0,\infty)\) and produced a right-skewed distribution.
Worked Example: A Vector-Valued Linear Transformation
Suppose \((X_{n1},X_{n2})\xrightarrow{d}(Z_1,Z_2)\), where \(Z_1,Z_2\) are independent \(N(0,1)\) random variables. Find the joint limit of $$ \bigl(X_{n1}+X_{n2},\;X_{n1}-X_{n2}\bigr). $$
Define \(g(x_1,x_2)=(x_1+x_2,x_1-x_2)\), a continuous map from \(\mathbb{R}^2\) to \(\mathbb{R}^2\). The theorem gives $$ \bigl(X_{n1}+X_{n2},\;X_{n1}-X_{n2}\bigr) \xrightarrow{d} \bigl(Z_1+Z_2,\;Z_1-Z_2\bigr). $$ To identify the joint law, set \(U=Z_1+Z_2\) and \(V=Z_1-Z_2\). Since \(Z_1,Z_2\) are independent standard normals, their joint characteristic function at \((s,t)\) after transformation is $$ E[e^{i(sU+tV)}] =E[e^{i((s+t)Z_1+(s-t)Z_2)}] =\exp\left(-\frac{(s+t)^2+(s-t)^2}{2}\right) =\exp(-s^2-t^2). $$ This factors as \(\exp(-s^2)\exp(-t^2)\), the product of the characteristic functions of independent \(N(0,2)\) variables. Therefore \(U\) and \(V\) are independent, each with distribution \(N(0,2)\), and the displayed vector converges to that joint distribution.
Worked Example: The Norm of a Bivariate Normal Limit
Suppose \((X_{n1},X_{n2})\xrightarrow{d}(Z_1,Z_2)\), where \(Z_1,Z_2\) are independent standard normal variables. Find the limit distribution of \(\sqrt{X_{n1}^2+X_{n2}^2}\).
The Euclidean norm \(g(x_1,x_2)=\sqrt{x_1^2+x_2^2}\) is continuous everywhere, including at \((0,0)\). Hence $$ \sqrt{X_{n1}^2+X_{n2}^2} \xrightarrow{d} R=\sqrt{Z_1^2+Z_2^2}. $$ For \(r<0\), \(P(R\leq r)=0\). For \(r\geq0\), the joint density of \((Z_1,Z_2)\) is \((2\pi)^{-1}e^{-(z_1^2+z_2^2)/2}\). Integrating over the disk of radius \(r\), and using polar coordinates with radial coordinate \(\rho\), gives $$ P(R\leq r) =\int_0^{2\pi}\int_0^r \frac{1}{2\pi}e^{-\rho^2/2}\rho\,d\rho\,d\theta =\int_0^r e^{-\rho^2/2}\rho\,d\rho =1-e^{-r^2/2}. $$ The derivative for \(r>0\) is \(r e^{-r^2/2}\), so \(R\) has the Rayleigh distribution with this density. The norm has converted a two-dimensional limiting vector into a nonnegative scalar limit.
What the Principle Does—and Does Not—Say
The continuous mapping theorem is useful because it converts a known joint or scalar limit into a limit for a statistic built from that vector. Common continuous maps include polynomials, norms, and fixed linear or affine transformations. In particular, if \(A\) is an \(m\times k\) matrix and \(b\in\mathbb{R}^m\), then \(g(x)=Ax+b\) is continuous, so \(X_n\xrightarrow{d}X\) implies \(AX_n+b\xrightarrow{d}AX+b\).
This principle gives convergence to the distribution of the transformed limiting random variable, not necessarily to a distribution in the same family as the original limit. In the first example, a normal limit became chi-square; in the third, a bivariate normal limit became Rayleigh. The map can also discard information or combine coordinates, so the transformed limit may have fewer dimensions than the original one.
The theorem as stated requires continuity of \(g\) throughout its Euclidean domain. It is a sufficient condition, not a claim that every useful transformation must be continuous everywhere. A function with a discontinuity requires additional care; the theorem above alone does not settle such a case.
Common Mistake
- Wrong reasoning: “Since \(X_n\) converges to a normal distribution, any function of \(X_n\) must also have a normal limit.”
- Correction: The theorem identifies the limit as \(g(X)\). Its distribution depends on the function. For \(g(x)=x^2\) and \(X\sim N(0,1)\), the limit is chi-square with one degree of freedom, not normal.
- Another error: applying the result without checking continuity. The theorem applies directly when the chosen fixed map is continuous on the whole state space under consideration.
Check Your Understanding
Answer each question by stating the relevant map and, where possible, identifying the transformed limit.
- If \(X_n\xrightarrow{d}Z\sim N(0,1)\), what is the limit distribution of \(e^{X_n}\)?
- Why does the proof require \(h\circ g\) to be bounded and continuous?
- If \(X_n\) is a random vector in \(\mathbb{R}^k\) and \(A\) is a fixed matrix, what does the theorem imply about \(AX_n\)?
- For independent standard normal \(Z_1,Z_2\), what is the distribution function of \(\max(Z_1,Z_2)\)? Identify the continuous mapping argument that gives convergence to this limit.
- Does the continuous mapping theorem say that \(g(X)\) has the same distributional family as \(X\)? Give a counterexample.