From Transform Limits to Probability Limits
Characteristic functions are useful partly because they often have simple formulas for sums and standard distributions. But a pointwise limit of characteristic functions does not automatically describe a probability distribution. The continuity theorem gives the exact additional condition that makes the limit meaningful: continuity at zero. It also explains why this condition lets us identify a limiting law, rather than merely a limiting formula.
For a real-valued random variable \(X\), its characteristic function is \(\phi_X(t)=\mathbb{E}[e^{itX}]\), for \(t\in\mathbb{R}\). Equivalently, if \(X\) has law \(\mu\), then \(\phi_X(t)=\int_{\mathbb{R}}e^{itx}\,\mu(dx)\). A sequence of probability laws \((\mu_n)\) converges weakly to \(\mu\), written \(\mu_n\Rightarrow\mu\), if \(\int f\,d\mu_n\to\int f\,d\mu\) for every bounded continuous real-valued function \(f\) on \(\mathbb{R}\).
The theorem combines two conclusions. Continuity at zero ensures that the limiting function is itself a characteristic function, so a probability law exists. The pointwise limit then identifies that law: every distributional limit must have characteristic function \(\phi\), and characteristic functions uniquely determine probability laws. The uniqueness fact is the standard uniqueness theorem for characteristic functions.
Why Continuity at Zero Gives a Limit Law
The key step in the proof is to show that the sequence of laws is tight. A family of laws on \(\mathbb{R}\) is tight if, for every \(\varepsilon>0\), there is a finite \(R\) such that every law in the family assigns probability at least \(1-\varepsilon\) to \([-R,R]\). Intuitively, tightness prevents probability mass from escaping arbitrarily far from the origin.
Proof. First suppose \(\mu_n\Rightarrow\mu\), where \(\mu\) is a probability law with characteristic function \(\phi\). For fixed \(t\), the functions \(x\mapsto\cos(tx)\) and \(x\mapsto\sin(tx)\) are bounded and continuous. The definition of weak convergence therefore gives convergence of their integrals, so \(\phi_n(t)\to\phi(t)\). Also, if \(X\) has law \(\mu\), then \(e^{itX}\to1\) as \(t\to0\), and \(|e^{itX}|=1\). Dominated convergence gives \(\phi(t)=\mathbb{E}[e^{itX}]\to1=\phi(0)\). This proves the reverse implication and the continuity of the limit at zero.
Now suppose \(\phi_n(t)\to\phi(t)\) for every \(t\), and \(\phi\) is continuous at zero. For any random variable \(X\) with law \(\mu_n\), define \(A_n(T)=\frac{1}{2T}\int_{-T}^{T}(1-\operatorname{Re}\phi_n(t))\,dt\), for \(T>0\). Since \(\operatorname{Re}\phi_n(t)=\mathbb{E}[\cos(tX)]\), integration and expectation can be interchanged (the integrand is bounded), giving \(A_n(T)=\mathbb{E}[1-\sin(TX)/(TX)]\), with the ratio interpreted as \(1\) at \(X=0\). If \(|X|\ge2/T\), then \(|\sin(TX)/(TX)|\le1/2\), so \(1-\sin(TX)/(TX)\ge1/2\). Consequently, \(\mu_n(\{|x|\ge2/T\})\le2A_n(T)\).
For fixed \(T>0\), pointwise convergence and the bound \(0\le1-\operatorname{Re}\phi_n(t)\le2\) allow dominated convergence in the integral defining \(A_n(T)\). Thus \(A_n(T)\to A(T)=\frac{1}{2T}\int_{-T}^{T}(1-\operatorname{Re}\phi(t))\,dt\). Because \(\phi(0)=1\) and \(\phi\) is continuous at zero, \(A(T)\to0\) as \(T\to0\). Given \(\varepsilon>0\), choose \(T\) small enough that \(A(T)<\varepsilon/8\). For all sufficiently large \(n\), \(A_n(T)<\varepsilon/4\), so \(\mu_n(\{|x|\ge2/T\})<\varepsilon/2\). The finitely many remaining laws each put probability tending to \(1\) on \([-R,R]\) as \(R\to\infty\); enlarging \(R\) handles all of them as well. This proves tightness.
We now use the standard Helly selection theorem: every tight sequence of probability laws on \(\mathbb{R}\) has a weakly convergent subsequence. Its proof is part of the later study of convergence via distribution functions. Tightness therefore gives a weakly convergent subsequence of any subsequence of \((\mu_n)\). If such a subsequence converges weakly to \(\nu\), the first part of the proof shows that its characteristic functions converge to the characteristic function of \(\nu\). Their pointwise limit is \(\phi\), so \(\nu\) has characteristic function \(\phi\). By uniqueness of characteristic functions, all these subsequential limits are the same law, say \(\mu\).
Finally, the whole sequence converges weakly to \(\mu\). If it did not, the definition of weak convergence would give a bounded continuous function \(f\), a positive \(\varepsilon\), and a subsequence along which \(\int f\,d\mu_n\) stays at least \(\varepsilon\) away from \(\int f\,d\mu\). Tightness gives a further weakly convergent subsequence. Its limit must be \(\mu\), contradicting that separation for \(f\). Hence \(\mu_n\Rightarrow\mu\). This proves the forward implication and the theorem. \(\square\)
Using the Theorem
Worked Example: A Small Amount of Mass Escapes
For each integer \(n\ge1\), let \(X_n=0\) with probability \(1-\frac1n\), and \(X_n=n\) with probability \(\frac1n\). Find the pointwise limit of the characteristic functions and use the continuity theorem to identify the limiting law.
From the two possible values of \(X_n\), its characteristic function is \(\phi_n(t)=(1-\frac1n)e^{it\cdot0}+\frac1n e^{itn}=1+\frac{e^{itn}-1}{n}\). For every fixed \(t\), \(|e^{itn}-1|\le2\), so \(|\phi_n(t)-1|\le2/n\to0\). Thus the pointwise limit is \(\phi(t)=1\) for every \(t\), which is continuous at zero.
The constant function \(1\) is the characteristic function of the point mass \(\delta_0\), the law assigning probability \(1\) to zero. The theorem gives \(\mu_n\Rightarrow\delta_0\). This example also shows why pointwise convergence is about distributions, not about every feature of the variables: the probability of the value \(n\) tends to zero even as that value moves farther away.
Check: The proposed limit law has characteristic function \(\int e^{itx}\,\delta_0(dx)=e^{it\cdot0}=1\), matching the pointwise limit.
Worked Example: Poisson Laws with a Convergent Parameter
Let \(X_n\) have a Poisson distribution with parameter \(\lambda_n=2+\frac1n\). Identify the weak limit.
The characteristic function of a Poisson variable with parameter \(\lambda_n\) is \(\phi_n(t)=\exp(\lambda_n(e^{it}-1))\). For each fixed \(t\), \(\lambda_n\to2\), so continuity of the exponential gives \(\phi_n(t)\to\phi(t)=\exp(2(e^{it}-1))\). The function \(\phi\) is continuous at zero because \(e^{i0}=1\) and \(\phi(0)=\exp(0)=1\).
The limiting function is the characteristic function of a Poisson variable with parameter \(2\), by the standard Poisson characteristic-function formula. The continuity theorem therefore gives \(X_n\Rightarrow X\), where \(X\) has that Poisson law.
Check: Substituting \(t=0\) into the limit gives \(\exp(2(1-1))=1\), as a characteristic function must at zero. The formula also agrees with the characteristic function of the stated limiting Poisson law.
Worked Example: A Pointwise Limit That Fails at Zero
Let \(X_n\) be uniform on the integers \(\{1,2,\ldots,n\}\), with each of the \(n\) values having probability \(1/n\). Determine the pointwise limit of its characteristic functions and explain why the continuity condition matters.
For \(t\notin2\pi\mathbb{Z}\), summing the finite geometric series gives \(\phi_n(t)=\frac1n\sum_{k=1}^{n}e^{ikt}=\frac{e^{it}(1-e^{int})}{n(1-e^{it})}\). For fixed such \(t\), the numerator has absolute value at most \(2\), while \(1-e^{it}\ne0\) is fixed. Therefore \(\phi_n(t)\to0\). If \(t\in2\pi\mathbb{Z}\), every summand equals \(1\), so \(\phi_n(t)=1\) for every \(n\). The pointwise limit is \(1\) on \(2\pi\mathbb{Z}\) and \(0\) elsewhere.
This limit is not continuous at zero: its value at zero is \(1\), but its value at \(t=1/m\) is \(0\) for every positive integer \(m\), since \(1/m\notin2\pi\mathbb{Z}\), and \(1/m\to0\). The continuity theorem does not apply. In fact, the laws have no weak probability limit: for any fixed \(R\), when \(n>R\), the number of values in \(\{1,\ldots,n\}\) exceeding \(R\) grows with \(n\), so the mass cannot remain concentrated in a fixed bounded interval.
Check: At \(t=0\), the defining sum has \(n\) terms equal to \(1\), and division by \(n\) gives \(\phi_n(0)=1\). Thus the discontinuity is in the pointwise limit near zero, not a failure of the characteristic functions to equal \(1\) at zero.
Common Mistake and Practical Significance
Common mistake: It is not enough to calculate a pointwise limit and call it a characteristic function. In the uniform-integers example, the pointwise limit exists at every \(t\), but it is discontinuous at zero and cannot be the characteristic function of a probability law. Every characteristic function is continuous at zero, as follows from dominated convergence. The correction is to check the continuity condition before identifying a proposed limiting distribution.
In applications, one often computes \(\phi_n(t)\), recognizes a candidate function \(\phi(t)\), and identifies a familiar law whose characteristic function is \(\phi\). The theorem then turns the calculation into a distributional convergence result, provided convergence holds for every real \(t\) and the limit is continuous at zero. Pointwise convergence alone is not enough, and continuity at zero is a condition on the limiting function—not a demand that the sequence converge uniformly in \(t\).
Check Your Understanding
For each question, distinguish the pointwise limit from the condition that allows it to identify a probability law.
- State the continuity theorem, including what pointwise convergence means and where continuity is required.
- Why does \(\mu_n\Rightarrow\mu\) imply convergence of the characteristic functions at every fixed \(t\)?
- In the escaping-mass example, what is the limiting characteristic function, and which probability law has it?
- For the uniform law on \(\{1,\ldots,n\}\), what is the pointwise limit at \(t=0\), and why does the limit fail the theorem's continuity condition?
- What role does tightness play in showing that a continuous-at-zero pointwise limit corresponds to a probability law?