From a Pointwise Limit to a Distributional Limit
Characteristic functions turn questions about a sequence of distributions into questions about functions of a real argument. The continuity theorem makes this useful: under a specific condition at the origin, pointwise convergence of characteristic functions guarantees convergence in distribution. This tutorial focuses on two especially recognizable limits: the characteristic function of a point mass and that of a normal distribution. The condition at zero matters; pointwise convergence alone is not enough.
For each \(n\), let \(X_n\) be a real-valued random variable on a probability space \((\Omega_n,\mathcal{F}_n,\mathbb{P}_n)\). Its characteristic function is \(\phi_n(t)=\mathbb{E}_n[e^{itX_n}]\), for \(t\in\mathbb{R}\). We write \(X_n\Rightarrow X\) for convergence in distribution: the distribution functions of \(X_n\) converge to that of \(X\) at every continuity point of the latter.
To use the theorem, check both its inputs. First, the pointwise limit must exist for every real \(t\), not merely for a selected collection of arguments. Second, the resulting function must be continuous at zero. When those conditions hold, identify the limiting characteristic function using a known distribution. The characteristic function of the point mass at \(a\) is \(e^{iat}\). For a normal distribution with mean \(m\) and variance \(\sigma^2>0\), it is \(e^{imt-\sigma^2t^2/2}\). Both functions equal \(1\) at zero and are continuous there.
Proof. In the first case, \(e^{iat}\) is the characteristic function of the constant random variable equal to \(a\), and it is continuous at zero because \(e^{ia\cdot 0}=1\) and the exponential function is continuous. In the second case, \(e^{imt-\sigma^2t^2/2}\) is the characteristic function of \(N(m,\sigma^2)\), and it is continuous at zero by continuity of the exponential function. In each case the assumed pointwise convergence and continuity at zero are exactly the hypotheses of the continuity theorem. Its conclusion gives the stated convergence in distribution. \(\square\)
The corollary is a recognition rule, not a license to match only a few values. A limiting function that resembles a normal characteristic function near zero is not enough: the pointwise limit must equal that function at every real argument.
Worked Applications
Worked Example: Rare Events Converge to a Point Mass
For each integer \(n\ge1\), let \(X_n\) equal \(1\) with probability \(1/n\) and \(0\) with probability \(1-1/n\). Find the pointwise limit of its characteristic functions and identify the distributional limit.
By evaluating \(e^{itX_n}\) at the two possible values of \(X_n\),
Fix any real \(t\). The number \(e^{it}-1\) is then fixed, so \((e^{it}-1)/n\to0\) as \(n\to\infty\). Thus \(\phi_n(t)\to1\) for every real \(t\). The constant function \(1\) is \(e^{i0t}\), the characteristic function of the point mass at zero, and is continuous at zero. The corollary gives \(X_n\Rightarrow0\).
Check: At \(t=0\), the original formula gives \(\phi_n(0)=1+(1-1)/n=1\) for every \(n\), agreeing with the limit. The probabilities are valid even at \(n=1\): they are \(1\) for \(X_1=1\) and \(0\) for \(X_1=0\).
Worked Example: Normal Variables with Shrinking Variance
Let \(X_n\) have distribution \(N(m,\sigma^2/n)\), where \(m\in\mathbb{R}\) and \(\sigma>0\) are fixed. Find its limiting distribution.
The characteristic function of \(N(m,v)\), for \(v>0\), is \(e^{imt-vt^2/2}\). Substituting \(v=\sigma^2/n\) gives
For each fixed real \(t\), the term \(\sigma^2t^2/(2n)\) tends to zero. Therefore \(\phi_n(t)\to e^{imt}\). This is the characteristic function of the point mass at \(m\), and it is continuous at zero. Hence \(X_n\Rightarrow m\).
Check: At \(t=0\), \(\phi_n(0)=\exp(0)=1\), and the limiting function also equals \(e^{im\cdot0}=1\). The limiting characteristic function is that of a constant, not of a normal distribution with positive variance: its exponent has no quadratic term.
Worked Example: Standardized Binomial Sums Have a Normal Limit
Let \(B_1,\ldots,B_n\) be independent Bernoulli random variables with \(\mathbb{P}(B_k=1)=\mathbb{P}(B_k=0)=1/2\), and put \(Y_n=\sum_{k=1}^n B_k\). Define \(Z_n=(2Y_n-n)/\sqrt n\). Show by characteristic functions that \(Z_n\Rightarrow N(0,1)\).
For a single \(B_k\), the contribution \((2B_k-1)/\sqrt n\) equals \(1/\sqrt n\) or \(-1/\sqrt n\), each with probability \(1/2\). Its characteristic function at \(t\) is therefore
Independence makes the characteristic function of the sum the product of the individual characteristic functions. Since \(Z_n=\sum_{k=1}^n(2B_k-1)/\sqrt n\), this gives \(\phi_{Z_n}(t)=\cos(t/\sqrt n)^n\).
Fix \(t\). For all sufficiently large \(n\), \(t/\sqrt n\) is near zero, where the cosine is positive, so the logarithm below is defined. Taylor's formula at zero gives \(\cos x=1-x^2/2+O(x^4)\). Also, Taylor's formula for \(\log(1+u)\) gives \(\log(1+u)=u+O(u^2)\) for \(u\) near zero. Taking \(u=\cos x-1=-x^2/2+O(x^4)\) yields \(\log(\cos x)=-x^2/2+O(x^4)\). Substituting \(x=t/\sqrt n\), with \(t\) fixed, gives
Exponentiating shows \(\phi_{Z_n}(t)\to e^{-t^2/2}\) for each real \(t\). This limit is the characteristic function of \(N(0,1)\) and is continuous at zero. The continuity theorem gives \(Z_n\Rightarrow N(0,1)\).
Check: At \(t=0\), the exact characteristic function is \(\cos(0)^n=1\), and the limiting function is \(e^0=1\). The standardization also centers and scales correctly: \(\mathbb{E}Y_n=n/2\) and \(\operatorname{Var}(Y_n)=n/4\), so \(\mathbb{E}Z_n=0\) and \(\operatorname{Var}(Z_n)=4(n/4)/n=1\).
When Pointwise Convergence Is Not Enough
Continuity at zero is not an optional detail. A characteristic function of a real-valued random variable is continuous at zero: if \(X\) is such a variable, then \(e^{itX}\to1\) as \(t\to0\), and \(|e^{itX}|=1\), so bounded convergence gives \(\mathbb{E}[e^{itX}]\to1\). Consequently, a pointwise limit that is discontinuous at zero cannot itself be a characteristic function.
Worked Example: Uniform Distributions Give a Discontinuous Limit
Let \(U_n\) be uniform on \([-n,n]\). Compute the pointwise limit of its characteristic functions and check whether the continuity theorem applies.
For \(t\ne0\), integrating the density \(1/(2n)\) gives
At \(t=0\), the defining expectation gives \(\phi_n(0)=1\). For fixed \(t\ne0\), \(|\sin(nt)/(nt)|\le1/(n|t|)\), which tends to zero. Thus the pointwise limit is \(1\) at zero and \(0\) at every nonzero \(t\). This function is discontinuous at zero, so the continuity theorem's condition fails. It is not the characteristic function of any real-valued random variable, since every such characteristic function is continuous at zero.
Check: The limit at zero is obtained by evaluating \(\phi_n(0)=1\) for every \(n\); the limit away from zero follows from the displayed bound at each fixed nonzero \(t\). These two values do not approach one another as \(t\to0\), confirming the discontinuity.
Common mistake. It is wrong to argue, “Each \(\phi_n\) is a characteristic function, so its pointwise limit must be one too.” In the uniform example, every \(\phi_n\) is a characteristic function, but the pointwise limit is \(1\) at zero and \(0\) elsewhere. That limit is discontinuous at zero and cannot be a characteristic function. The correction is to check the theorem's continuity condition before assigning a limiting distribution.
In applications, keep the reasoning in order: calculate the characteristic function, take the pointwise limit for an arbitrary fixed real \(t\), check continuity at zero, and only then identify the limiting law. For the two common targets here, \(e^{iat}\) signals a point mass at \(a\), while \(e^{imt-\sigma^2t^2/2}\) with \(\sigma^2>0\) signals a normal distribution.
Check Your Understanding
For each question, identify the pointwise limit and state whether the continuity theorem applies.
- If \(\phi_n(t)=e^{it(3+1/n)}\), what is the limiting distribution?
- If \(\phi_n(t)=e^{-2t^2/n}\), what is the limiting distribution?
- Why must the pointwise limit be checked at every real \(t\), rather than only at zero?
- For the uniform example, what are the pointwise limit's values at zero and at a fixed nonzero \(t\)?
- What continuity condition does the continuity theorem require, and why does the uniform example fail it?