The Continuous Mapping Theorem For Distributional Limits

Why Transforming a Distributional Limit Requires Care

A distributional limit describes how the probabilities associated with a sequence of random variables settle down. Econometric arguments often need limits for a transformed quantity instead: a squared statistic, an absolute deviation, or an indicator that a statistic crosses a threshold. The continuous mapping theorem gives a precise condition under which the transformation preserves convergence in distribution.

The key check is not simply whether a transformation is continuous everywhere. It may have discontinuities and still be usable, provided the limiting random variable has probability zero of landing at any of them. Conversely, a single discontinuity can matter when the limit places positive probability there.

Definition: Write \(X_n \xrightarrow{d} X\) when the random variables \(X_n\) converge in distribution to \(X\). For a measurable function \(g:\mathbb{R}\to\mathbb{R}\), its discontinuity set \(D_g\) is the set of points at which \(g\) is not continuous.

The Continuous Mapping Theorem

For real-valued random variables, the distributional version of the theorem is:

Theorem: If \(X_n \xrightarrow{d} X\), \(g:\mathbb{R}\to\mathbb{R}\) is measurable, and \(P(X\in D_g)=0\), then \(g(X_n)\xrightarrow{d}g(X)\). In particular, if \(g\) is continuous everywhere, the condition on its discontinuity set holds automatically.

The condition concerns the distribution of the limit \(X\), not whether each \(X_n\) happens to avoid the discontinuities. A discontinuity is harmless when it occurs only at values assigned zero probability by \(X\). For an everywhere-continuous transformation such as \(g(x)=x^2\), no additional check is needed. For a threshold indicator, the cutoff is a discontinuity, so one must check whether the limit has an atom at that cutoff.

The theorem gives a limiting distribution, not an exact finite-sample distribution. In general, the distribution of \(g(X_n)\) need not equal the distribution of \(g(X)\); the claim is that the former converges to the latter.

Why the Theorem Holds

A short justification uses the closed-set characterization of convergence in distribution: if \(X_n\xrightarrow{d}X\), then for every closed set \(C\), \(\limsup_{n\to\infty}P(X_n\in C)\leq P(X\in C)\). Let \(F\) be any closed set in the range of \(g\), and write \(A=g^{-1}(F)\). The set \(A\) need not be closed, so the characterization cannot be applied to it directly. Instead, consider its closure \(\overline{A}\).

Relate the closure to continuity.
If \(x\in\overline{A}\), \(x\notin D_g\), and \(g(x)\notin F\), continuity at \(x\) and the fact that \(F\) is closed give a neighborhood of \(x\) whose values under \(g\) all lie outside \(F\). That neighborhood cannot intersect \(A\), contradicting \(x\in\overline{A}\). Hence \(\overline{A}\subseteq A\cup D_g\).

Apply the closed-set bound.
Since \(A\subseteq\overline{A}\) and \(\overline{A}\) is closed, \(\limsup_{n\to\infty}P(g(X_n)\in F)=\limsup_{n\to\infty}P(X_n\in A)\leq P(X\in\overline{A})\).

Use the zero-probability condition.
The inclusion from Step 1 gives \(P(X\in\overline{A})\leq P(X\in A)+P(X\in D_g)=P(g(X)\in F)\). Thus the closed-set bound holds for \(g(X_n)\) and \(g(X)\), which establishes \(g(X_n)\xrightarrow{d}g(X)\).

This argument also shows why the condition is stated in terms of probability at discontinuities. The closed-set bound can absorb points in \(D_g\) only when their total probability under \(X\) is zero.

Worked Applications

Worked Example: Squaring a Normal Limit

Suppose \(X_n\sim N(1.5+1/n,4)\). The means \(1.5+1/n\) tend to \(1.5\), and the variance is \(4\) for every \(n\), so \(X_n\xrightarrow{d}X\), where \(X\sim N(1.5,4)\). Find the distributional limit of \(X_n^2\), and calculate the limiting probability that the square is at most \(4\).

Take \(g(x)=x^2\). It is continuous at every real number, so \(P(X\in D_g)=0\). The theorem therefore gives

X_n^2 \xrightarrow{d} X^2,\qquad X\sim N(1.5,4).

The limit is the square of a normal random variable with mean \(1.5\) and standard deviation \(2\). To calculate the requested probability, \(X^2\leq4\) is equivalent to \(-2\leq X\leq2\). Standardizing both endpoints gives

P(X^2\leq4) =P(-2\leq X\leq2) =\Phi\left(\frac{2-1.5}{2}\right)-\Phi\left(\frac{-2-1.5}{2}\right) =\Phi(0.25)-\Phi(-1.75) \approx 0.5987-0.0401 =0.5586.

Thus the limiting probability that the squared variable is at most \(4\) is approximately \(0.5586\). The units of \(X^2\), if \(X\) has units, are the square of the units of \(X\).

Worked Example: An Indicator at a Continuity Point in Probability

Let \(X_n\sim N(1/n,1)\), so \(X_n\xrightarrow{d}X\) for \(X\sim N(0,1)\). Define \(g(x)=\mathbf{1}\{x\leq0\}\), which equals \(1\) when \(x\leq0\) and \(0\) otherwise. Does the theorem apply, and what is the limiting distribution of \(g(X_n)\)?

The function is discontinuous only at \(0\). A normal random variable has zero probability at any individual point, so \(P(X=0)=0\) and \(P(X\in D_g)=0\). The theorem applies. The transformed variable takes values \(0\) and \(1\), and

P(g(X_n)=1)=P(X_n\leq0) =\Phi\left(\frac{0-1/n}{1}\right) =\Phi(-1/n) \longrightarrow \Phi(0)=0.5.

For example, at \(n=20\), this probability is \(\Phi(-0.05)\approx0.4801\), and it tends to \(0.5\). Therefore \(g(X_n)\) converges in distribution to a Bernoulli random variable that equals \(1\) with probability \(0.5\) and \(0\) with probability \(0.5\). The jump in \(g\) causes no problem because the limiting normal distribution puts no probability mass at the jump.

Worked Example: A Discontinuity at an Atom Can Break the Conclusion

Let \(X_n=1/n\) with probability \(1\). Then \(X_n\xrightarrow{d}0\), since the sequence is deterministic and \(1/n\to0\). Use \(g(x)=\mathbf{1}\{x>0\}\). What happens to the transformed sequence?

The only discontinuity of \(g\) is at \(0\), and the limiting variable \(X=0\) has \(P(X=0)=1\). Thus the theorem’s condition fails. Direct calculation confirms that the proposed conclusion would be false:

g(X_n)=g(1/n)=1\ \text{with probability }1,\qquad g(X)=g(0)=0\ \text{with probability }1.

The transformed sequence is constantly \(1\), so it converges in distribution to the constant \(1\), not to \(g(X)=0\). The issue is not that indicator functions are unusable; it is that this cutoff coincides with a point carrying all the limiting probability.

Practical Checks and a Common Pitfall

When applying the theorem, write down both the transformation and its discontinuity set. Then assess the probability that the limiting variable lies in that set. For example, if \(g(x)=\mathbf{1}\{x\leq c\}\), its only discontinuity is \(c\), and the required condition is \(P(X=c)=0\). A continuous limiting distribution satisfies this condition for every fixed \(c\), while a discrete limit may not.

A common mistake is to conclude that \(g(X_n)\xrightarrow{d}g(X)\) merely because \(X_n\xrightarrow{d}X\) and \(g\) is a function. That reasoning omits the continuity check. In the third example, the input converges in distribution, but the indicator changes abruptly exactly where the limiting variable lies. The correction is to check \(P(X\in D_g)=0\) before applying the theorem, and to avoid asserting the result when this condition fails.

Failure of the condition means the theorem does not guarantee the conclusion; it does not mean the conclusion must fail in every such case. When the limit has positive probability at a discontinuity, examine the transformed sequence directly or use another argument. The continuous mapping theorem is a sufficient condition, not a claim that every transformation with a discontinuity must disrupt convergence.

Key takeaway: From \(X_n\xrightarrow{d}X\), infer \(g(X_n)\xrightarrow{d}g(X)\) when \(g\) is measurable and \(P(X\in D_g)=0\). Always check continuity under the limiting distribution, especially for threshold indicators and other jump functions.

Check Your Understanding

For each question, identify the transformation’s discontinuities and decide whether the stated condition for the theorem holds.

If \(X_n\xrightarrow{d}X\) and \(g(x)=|x|\), what is the distributional limit of \(g(X_n)\), and why does the theorem apply?
Suppose \(X\sim N(2,1)\) and \(g(x)=\mathbf{1}\{x\leq2\}\). What is \(P(X\in D_g)\), and what is the limiting probability that \(g(X_n)=1\) when \(X_n\xrightarrow{d}X\)?
Let \(X_n=2+1/n\) with probability \(1\), and let \(g(x)=\mathbf{1}\{x>2\}\). State the limits of \(X_n\) and \(g(X_n)\), and explain why the theorem’s condition does not hold.
Does an everywhere-continuous function require the limiting variable to have a continuous distribution? Explain.
Does \(P(X\in D_g)>0\) prove that \(g(X_n)\) cannot converge in distribution to \(g(X)\)? Explain the distinction between a failed sufficient condition and a counterexample.