The Continuous Mapping Theorem For Probability Limits

Why Transformations Matter for Probability Limits

Asymptotic arguments often establish that a random quantity approaches a constant in probability, then use that quantity inside another expression. A sample estimate may be squared, exponentiated, or passed through a multivariate function. The continuous mapping theorem gives a precise condition under which the transformed quantity has the corresponding probability limit.

The condition concerns the transformation at the limiting value, not necessarily its behavior everywhere. There is also a practical requirement: the transformed random variables must actually be defined. This matters for functions such as the logarithm, which are not real-valued on all real inputs.

Definition: A scalar sequence $X_n$ converges in probability to a constant $c$, written $X_n\xrightarrow{p}c$, if for every $\varepsilon>0$, $$ P(|X_n-c|>\varepsilon)\longrightarrow 0. $$ For a random vector $X_n\in\mathbb{R}^k$, convergence in probability to $c\in\mathbb{R}^k$ means $P(\|X_n-c\|>\varepsilon)\to0$ for every $\varepsilon>0$, where $\|\cdot\|$ is the Euclidean norm.

The Continuous Mapping Theorem

Let $g$ transform scalar inputs to scalar outputs, or more generally map vectors in $\mathbb{R}^k$ to vectors in $\mathbb{R}^m$. The relevant continuity condition is local: for every output tolerance, inputs sufficiently close to the limiting value must produce outputs sufficiently close to the transformed limit.

Theorem (Continuous Mapping Theorem for convergence in probability): Suppose $X_n\xrightarrow{p}c$, and $g$ is continuous at $c$. If $g(X_n)$ is defined as a real-valued (or vector-valued) random variable for every $n$, then $$ g(X_n)\xrightarrow{p}g(c). $$ For a function defined only on a domain $D$, it is sufficient that $c$ is an interior point of $D$, $g$ is continuous at $c$, and $X_n\in D$ almost surely for each $n$. Alternatively, values on events where the original function is undefined must be assigned so that the transformed sequence is defined; if the assignment agrees with $g$ on a neighborhood of $c$, the same conclusion holds.

The domain condition is not merely technical. Convergence to a point inside a function's domain makes it increasingly likely that the input is near that point, but it does not guarantee that the input is in the domain with probability one for each $n$. For instance, a logarithm is not defined as a real number at a nonpositive input.

Why the Theorem Holds

The proof uses the definition of convergence in probability and continuity at the limit. It applies to scalar and vector inputs and outputs; absolute value is replaced by the appropriate norm when needed.

Choose an output tolerance.
Fix any $\varepsilon>0$. Continuity of $g$ at $c$ gives a $\delta>0$ such that, whenever $\|x-c\|<\delta$, $\|g(x)-g(c)\|<\varepsilon$.

Relate the two deviation events.
When $g(X_n)$ is defined, the event $\{\|g(X_n)-g(c)\|\geq\varepsilon\}$ is contained in $\{\|X_n-c\|\geq\delta\}$, apart from any exceptional events on which the transformation has been assigned values outside the neighborhood.

Take probabilities and limits.
Convergence in probability makes $P(\|X_n-c\|\geq\delta)\to0$. If an extension is used, $P(\|X_n-c\|\geq\delta)\to0$ also makes the probability of being outside the chosen neighborhood tend to zero. Thus the output deviation probability tends to zero for every $\varepsilon>0$.

More formally, if $g(X_n)$ is defined and $g$ is continuous on a neighborhood of $c$, choose a ball around $c$ contained in that neighborhood. Within the ball, continuity gives the event inclusion in Step 2; outside it, the probability tends to zero because $X_n\xrightarrow{p}c$. This proves the claim even when the function is only specified locally and values are assigned on exceptional events.

Worked Applications

Worked Example: Squaring a Scalar Sequence

Let $Z$ take the values $-1$ and $1$, each with probability $1/2$, and define $X_n=3+Z/n$. Find the probability limit of $X_n^2$.

Check the input limit. Since $|X_n-3|=1/n$ for either value of $Z$, for every $\varepsilon>0$, $$ P(|X_n-3|>\varepsilon)=0 $$ whenever $n\geq 1/\varepsilon$. Therefore $X_n\xrightarrow{p}3$.

Check the transformation. The function $g(x)=x^2$ is continuous at $3$, and it is defined for every real value of $X_n$. The theorem gives $X_n^2\xrightarrow{p}9$. Direct calculation confirms the result: $$ X_n^2-9=(3+Z/n)^2-9=\frac{6Z}{n}+\frac{1}{n^2}. $$ Because $|Z|=1$, the absolute deviation is at most $6/n+1/n^2$. For $n=100$, this bound is $0.0601$; as $n$ increases it tends to zero. Thus the squared quantity approaches $9$ in probability.

Worked Example: Taking a Logarithm With Its Domain Checked

Let $V$ have a uniform distribution on $(-1,1)$, and define $X_n=2+V/n$. Determine the probability limit of $\log(X_n)$, where $\log$ denotes the real natural logarithm.

Check convergence and definition. We have $|X_n-2|=|V|/n<1/n$, so $X_n\xrightarrow{p}2$. Also, for every $n\geq1$, $X_n>2-1/n\geq1$ almost surely. Consequently, $\log(X_n)$ is defined almost surely for every $n$; the limit $2$ lies in the interior of the logarithm's domain, $(0,\infty)$.

Apply the theorem and verify the bound. The logarithm is continuous at $2$, so $\log(X_n)\xrightarrow{p}\log(2)$. In fact, $$ |\log(X_n)-\log(2)|=\left|\log\left(1+\frac{V}{2n}\right)\right|. $$ For $|t|\leq1/2$, the mean value theorem gives $|\log(1+t)|\leq2|t|$, because the derivative $1/(1+t)$ has absolute value at most $2$ on that interval. Here $|V/(2n)|<1/2$, and hence $$ |\log(X_n)-\log(2)|\leq\frac{|V|}{n}<\frac{1}{n}. $$ For example, at $n=101$ the deviation is less than $1/101$, approximately $0.009901$, for every possible $V$. This confirms convergence to $\log(2)$, approximately $0.6931$.

Worked Example: A Continuous Function of a Random Vector

Let $R$ and $S$ each take values in $\{-1,1\}$; no independence assumption is needed. Define $$ X_n=\left(1+\frac{R}{n},-2+\frac{S}{n}\right),\qquad g(x,y)=x^2+3y. $$ Find the probability limit of $g(X_n)$.

Check vector convergence. The Euclidean distance from $X_n$ to $c=(1,-2)$ satisfies $$ \|X_n-c\|=\sqrt{(R/n)^2+(S/n)^2}=\frac{\sqrt{2}}{n}. $$ Thus for every $\varepsilon>0$, $P(\|X_n-c\|>\varepsilon)=0$ whenever $n\geq\sqrt{2}/\varepsilon$, so $X_n\xrightarrow{p}(1,-2)$.

Apply and check the transformation. The polynomial $g$ is continuous everywhere and is defined at every $X_n$. The theorem gives $$ g(X_n)\xrightarrow{p}g(1,-2)=1^2+3(-2)=-5. $$ Direct substitution yields $$ g(X_n)+5=\frac{2R+3S}{n}+\frac{1}{n^2}. $$ Since $|2R+3S|\leq5$, $|g(X_n)+5|\leq5/n+1/n^2$. At $n=100$, this is at most $0.0501$, and the bound tends to zero. The transformed scalar therefore converges in probability to $-5$.

Continuity Is Essential

A function need not preserve convergence in probability if it is discontinuous at the limiting value. For example, let $X_n$ equal $1/n$ or $-1/n$, each with probability $1/2$. Then $|X_n|=1/n$, so $X_n\xrightarrow{p}0$. Define $g(x)=1$ when $x>0$ and $g(x)=0$ when $x\leq0$. This function is discontinuous at $0$, and $g(0)=0$. But $g(X_n)$ equals $1$ with probability $1/2$ for every $n$. Therefore $$ P(|g(X_n)-g(0)|>1/2)=\frac{1}{2}, $$ which does not tend to zero. The transformed sequence does not converge in probability to $g(0)$.

Common Mistake: Ignoring the Function's Domain

A tempting but invalid argument is: “$X_n\xrightarrow{p}1$, and the logarithm is continuous at $1$, so $\log(X_n)\xrightarrow{p}0$.” Continuity at the limit is not enough if $\log(X_n)$ is not defined as a real random variable.

For a concrete counterexample, let $X_n=-1$ with probability $1/n$ and $X_n=1$ otherwise, for $n\geq2$. For every $\varepsilon>0$, $P(|X_n-1|>\varepsilon)\leq1/n$, so $X_n\xrightarrow{p}1$. Yet $\log(X_n)$ is undefined on the event $X_n=-1$, which has positive probability for every $n$. The claimed transformed sequence is therefore not a real-valued random variable as stated.

The correction is to check definition as well as continuity. One may require $X_n>0$ almost surely for every $n$, or explicitly define a real-valued extension on the events where $X_n\leq0$. If that extension agrees with the logarithm on a neighborhood of $1$, the probability of being outside that neighborhood tends to zero, and the continuous mapping argument applies to the extended sequence.

Key takeaway: To transfer convergence in probability through a transformation, verify that the input converges to a finite limit, the transformation is continuous at that limit, and the transformed random variables are defined. The theorem then gives $g(X_n)\xrightarrow{p}g(c)$.

Check Your Understanding

For each question, state any needed continuity or domain condition.

If $X_n\xrightarrow{p}4$, what probability limit does the continuous mapping theorem give for $X_n^3$?
Suppose $X_n\xrightarrow{p}c\in\mathbb{R}^2$ and $g:\mathbb{R}^2\to\mathbb{R}$ is continuous at $c$. State the probability limit of $g(X_n)$.
Why is it not sufficient, when applying the logarithm, to know only that $X_n\xrightarrow{p}2$? Give one condition that resolves the domain issue.
For $X_n\xrightarrow{p}0$, explain why applying $g(x)=1\{x>0\}$ is not automatically justified by the continuous mapping theorem.
In the vector example, what is the limiting value of $x^2+3y$ at $(1,-2)$, and what continuity property permits the theorem's use?