Characteristic Functions Inversion and Continuity — Examples

All displayed masses are nonnegative. The Bernoulli masses sum to one. The binomial sum is $(p+1-p{)}^n=1$, including $n=0$, when there is exactly one term of value one. The Poisson sum is $e^{-\lambda }{\sum }_{k\ge 0}{\lambda }^k\mathrm{/}k!=e^{-\lambda }{e}^{\lambda }=1$. Weighted Dirac sums therefore define Borel probabilities on $\mathbb{R}$. For any such countably supported law with masses $b_k$ at distinct integers, $f_N(x)={\sum }_{k=0}^N{e}^{itk}{1}_{\{k\}}(x)$ tends to $e^{itx}$ almost everywhere for that law and satisfies $\mathrm{\mid }{f}_N\mathrm{\mid }\le 1$. Its integral is the finite sum of values times singleton masses. DCT yields $\phi (t)={\sum }_{k\ge 0}{b}_k{e}^{itk}$, with absolute sum $\sum b_k=1$. Finite supports are the same calculation with zero masses afterwards.

Substitute the Bernoulli masses to get $(1-p)+p{e}^{it}$. For the binomial law the finite sum is ${\sum }_{k=0}^n\left(\genfrac{}{}{0px}{}{n}{k}\right)(p{e}^{it}{)}^k(1-p{)}^{n-k}=(1-p+p{e}^{it}{)}^n$. This is also the product supplied for any already-given family of $n$ independent Bernoulli variables; no existence of an infinite family is needed. For Poisson, absolute convergence allows recognition of the defining series: $e^{-\lambda }{\sum }_{k\ge 0}(\lambda e^{it}{)}^k\mathrm{/}k!=e^{-\lambda }\exp (\lambda e^{it})=\exp (\lambda (e^{it}-1))$. At $p=0$ the first two laws are ${\delta }_0$; at $p=1$ they are ${\delta }_1$ and ${\delta }_n$; $n=0$ and $\lambda =0$ give ${\delta }_0$. The stated formulas give precisely their constant-point transforms, and all values at $t=0$ equal one.

For the finite mixture, the weighted-sum theorem gives a measure of total mass ${\sum }_j{a}_j=1$. For a simple complex function $s={\sum }_{\mathrm{\ell }=1}^q{c}_{\mathrm{\ell }}{1}_{E_{\mathrm{\ell }}}$ on a disjoint measurable partition, the integral definition and finite sums give $\int sd\rho ={\sum }_j{a}_j\int sd{\mu }_j$. Approximate $e^{itx}$ by rounding its real and imaginary parts down to multiples of $2^{-N}$; each approximation is Borel, simple, uniformly bounded by three and converges pointwise. DCT for $\rho$ and for each of the finitely many ${\mu }_j$ passes the simple identity to the limit, giving the mixture formula. Zero weights contribute zero, a one-component mixture returns that component, and no empty mixture has weights summing to one. No AC is used in these explicit sums and approximations.

The nonnegative Borel density defines a measure, and its mass is $(b-a{)}^{-1}{\int }_a^{b}1dx=1$. The primitive $x$ and the integral bridge give this normalization. The density integration identity, applied to positive and negative parts of cosine and sine, yields $\phi (t)=(b-a{)}^{-1}{\int }_a^b(\cos (tx)+i\sin (tx))dx$; all four parts are integrable because the interval is finite and their absolute values are at most one.

For $t\mathrm{\ne }0$, the primitives are $\sin (tx)\mathrm{/}t$ for cosine and $-\cos (tx)\mathrm{/}t$ for sine, by the chain rule. Their derivatives are continuous on $[a,b]$, so FTC and the bridge give $\phi (t)=\frac{\sin (tb)-\sin (ta)-i(\cos (tb)-\cos (ta))}{t(b-a)}=\frac{e^{itb}-e^{ita}}{it(b-a)}\mathrm{.}$ At $t=0$ the integral of the constant one equals one. Continuity of characteristic functions then proves the claimed extension. The endpoints of $[a,b]$ have zero density measure, so using an open or half-open interval gives the same law. The hypothesis $a<b$ prevents division by zero; when $a=b$ this density is not defined, though the distinct Dirac law at $a$ has transform $e^{ita}$. The stated AC is spent on the compact integration bridge and its continuous-integrand prerequisites.

Let $g(x)=e^{-x^2\mathrm{/}2}\mathrm{/}\sqrt{2\pi }$ and let $Z$ have its law on the canonical real probability space. Normalization is supplied by the Gaussian-density lemma. Exponential differentiation and the chain rule give $g^{\mathrm{\prime }}(x)=-xg(x)$. FTC, reflection substitution and the bridge yield ${\int }_{-R}^R\mathrm{\mid }x\mathrm{\mid }g(x)dx=\frac{2}{\sqrt{2\pi }}(1-e^{-R^2\mathrm{/}2})(R>0)\mathrm{.}$ Monotone convergence over positive integer R gives $\mathbb{E}\mathrm{\mid }Z\mathrm{\mid }=2\mathrm{/}\sqrt{2\pi }<\mathrm{\infty }$ by density integration. Apply the moments lemma at order one: ${\phi }^{\mathrm{\prime }}(t)=\int ix{e}^{itx}g(x)dx$, converting real positive/negative parts of the density integral separately.

On $[-R,R]$ both $g$ and the real and imaginary parts of $e^{itx}$ are continuously differentiable. Integration by parts, applied componentwise, gives ${\int }_{-R}^{R}ix{e}^{itx}g(x)dx=-i[e^{itx}g(x){]}_{-R}^R-t{\int }_{-R}^R{e}^{itx}g(x)dx\mathrm{.}$ The boundary term has modulus at most $2g(R)\to 0$. The left integrand is dominated by the integrable $\mathrm{\mid }x\mathrm{\mid }g(x)$, and the last integral by $g(x)$; DCT along integer R therefore gives ${\phi }^{\mathrm{\prime }}(t)=-t\phi (t)$ for every real t. No imaginary displacement of an integration contour is involved.

The real and imaginary components of $H(t)=e^{t^2\mathrm{/}2}\phi (t)$ are differentiable. The product and chain rules and step 2.1 give $H^{\mathrm{\prime }}(t)=e^{t^2\mathrm{/}2}(t\phi (t)+{\phi }^{\mathrm{\prime }}(t))=0$. The zero-derivative theorem applied to each component on $\mathbb{R}$ makes H constant. Its value at zero is $\phi (0)=\int g=1$, so $\phi (t)=e^{-t^2\mathrm{/}2}$. Finally affine scaling gives ${\phi }_{m+\sigma Z}(t)=e^{imt}{\phi }_Z(\sigma t)=e^{imt-{\sigma }^2{t}^2\mathrm{/}2}$. If $\sigma =0$, the random variable is constantly m and its transform is directly $e^{imt}$, agreeing with the formula. The stated AC is inherited from normalization and the compact integration bridge (including their countable-choice prerequisites).

The nonnegative exponential density has integral ${\lim }_{R\to \mathrm{\infty }}{\int }_0^R{e}^{-x}dx={\lim }_R(1-e^{-R})=1$, using the compact primitive, bridge and monotone convergence. For real t the function $e^{(-1+it)x}\mathrm{/}(-1+it)$ is a primitive of $e^{-x}(\cos (tx)+i\sin (tx))$, as direct componentwise differentiation shows; the denominator cannot vanish since its real part is -1. Therefore ${\int }_0^R{e}^{itx}{e}^{-x}dx=\frac{e^{(-1+it)R}-1}{-1+it}⟶\frac{1}{1-it}\mathrm{.}$ The boundary modulus is $e^{-R}$ and the integrand modulus is $e^{-x}$, so DCT justifies the full oscillatory integral. Whenever density integration is used below for a bounded complex test $u+iv$, apply [F16] separately to the nonnegative functions $u^{+},u^{-},v^{+},v^{-}$ and then reassemble their finite integrals componentwise as in [F1]; this gives $\int (u+iv)d(fdx)=\int (u+iv)fdx$ without enlarging [F16]. Reflection substitution on compact intervals and then DCT show that the reflected density has transform $1\mathrm{/}(1+it)$. Splitting the two half-lines gives the probability density $\mathrm{\ell }(x)=e^{-\mathrm{\mid }x\mathrm{\mid }}\mathrm{/}2$ and transform $\frac{1}{2}[(1-it{)}^{-1}+(1+it{)}^{-1}]=(1+t^2{)}^{-1}$.

The increasing arctangent has range $(-\pi \mathrm{/}2,\pi \mathrm{/}2)$, so its limits at the two infinities are the endpoints of that range: any smaller limiting supremum would omit values in the range, and similarly for the infimum. Consequently compact arctangent integration and monotone convergence give ${\int }_{\mathbb{R}}(1+t^2{)}^{-1}dt=\pi$. The Laplace characteristic function is therefore integrable. Density inversion supplies the continuous density $h(x)=(2\pi {)}^{-1}\int e^{-itx}\mathrm{/}(1+t^2)dt$ for the same Laplace law. It equals $\mathrm{\ell }$ at every point: if two continuous densities of the same measure differed at one point, their difference would have a fixed strict sign and magnitude on a small interval, contradicting equal integrals on that interval. Hence $\frac{1}{\pi }{\int }_{\mathbb{R}}\frac{e^{itx}}{1+x^2}dx=2h(-t)=e^{-\mathrm{\mid }t\mathrm{\mid }}\mathrm{.}$ The preceding arctangent integral also normalizes c, and density integration identifies the left side as its characteristic function, including t=0. No first moment of this law has been used.

For $R>0$, logarithmic differentiation with the chain rule and the compact integral bridge give ${\int }_0^R\frac{x}{\pi (1+x^2)}dx=\frac{\log (1+R^2)}{2\pi }\mathrm{.}$ This tends to infinity: log is increasing by its positive derivative, and its inverse relation implies that log of an unbounded positive argument eventually exceeds every real level. Monotone convergence and density integration imply $\int \mathrm{\mid }x\mathrm{\mid }c(x)dx=+\mathrm{\infty }$, already from the positive half-line. The density is finite at x=0 and has total mass one; it is not a zero or point-mass law. AC is retained from density inversion and the compact integration bridge. The auxiliary Laplace density has a corner at zero, but all differentiations above were on individual half-lines and its use in inversion required continuity, not differentiability there.

The Bernoulli transform is $1-p+p{e}^{it}$. Binomial weights $\left(\genfrac{}{}{0px}{}{n}{k}\right)p^k(1-p{)}^{n-k}$ are nonnegative and sum to one by the binomial theorem; they define a weighted Dirac probability, and the same finite expansion gives transform $(1-p+p{e}^{it}{)}^n$. Here every zeroth power means the empty product one, including parameter endpoints. The product rule for the given independent variables yields exactly this transform for their sum, so uniqueness gives its binomial law.

By [F22], $\mathrm{\mid }{e}^{itx}\mathrm{\mid }=1$ for real $t,x$. For $a\ge 0$, the weights $e^{-a}{a}^k\mathrm{/}k!$ sum to $e^{-a}{e}^a=1$ and define a probability. Truncating its exponential integrand to the integers $0,\dots ,N$ gives bounded functions of modulus at most one converging almost everywhere, so DCT identifies the transform with the absolutely convergent series $e^{-a}{\sum }_{k\ge 0}\frac{(a{e}^{it}{)}^k}{k!}=\exp (a(e^{it}-1))\mathrm{.}$ Multiplying the transforms at a=lambda and a=eta gives $\exp ((\lambda +\eta )(e^{it}-1))$, the transform of the constructed Poisson$(\lambda +\eta )$ law. Independence and uniqueness establish the claim, also when either parameter is zero.

For the normal calculation set $g(x)=e^{-x^2\mathrm{/}2}\mathrm{/}\sqrt{2\pi }$, which has mass one. Its derivative is $-xg(x)$. On each half of $[-R,R]$, FTC gives ${\int }_{-R}^R\mathrm{\mid }x\mathrm{\mid }g(x)dx=2(1-e^{-R^2\mathrm{/}2})\mathrm{/}\sqrt{2\pi }$. The bridge and monotone convergence prove the first absolute moment finite. If a complex test $h=u+iv$ satisfies $\int \mathrm{\mid }h\mathrm{\mid }gdx<\mathrm{\infty }$, apply [F18] to the four nonnegative functions $u^{+},u^{-},v^{+},v^{-}$ and reassemble the finite integrals componentwise; hence $\int hd(gdx)=\int hgdx$. Applying this first to $h(x)=e^{itx}$ and then to $h(x)=ix{e}^{itx}$, whose absolute values are $1$ and $\mathrm{\mid }x\mathrm{\mid }$, the moments lemma gives ${\phi }^{\mathrm{\prime }}(t)=\int ix{e}^{itx}g(x)dx$. For fixed real $t$, [F22] writes $e^{itx}=\cos (tx)+i\sin (tx)$; [F14, F15, F23] therefore give $(d\mathrm{/}dx)e^{itx}=-t\sin (tx)+it\cos (tx)=it{e}^{itx}$, also for $t=0$. Compact integration by parts in both real components gives ${\int }_{-R}^{R}ix{e}^{itx}g(x)dx=-i[e^{itx}g(x){]}_{-R}^R-t{\int }_{-R}^R{e}^{itx}g(x)dx\mathrm{.}$ Both differentiated functions have continuous derivatives on the compact interval. The boundary is bounded by $2g(R)\to 0$, while $\mathrm{\mid }x\mathrm{\mid }g$ and g dominate the integrands; DCT yields ${\phi }^{\mathrm{\prime }}=-t\phi$. The real and imaginary derivatives of $e^{t^2\mathrm{/}2}\phi (t)$ are therefore zero by the product and chain rules, so both components are constant. Since $\phi (0)=1$, $\phi (t)=e^{-t^2\mathrm{/}2}$.

The normal definition and affine identity now give transform $e^{i{m}_{j}t-{\sigma }_j^2{t}^2\mathrm{/}2}$ for each input. Its product is $\exp (it{\sum }_j{m}_j-t^2{\sum }_j{\sigma }_j^2\mathrm{/}2)$, exactly the transform of $N({\sum }_j{m}_j,{\sum }_j{\sigma }_j^2)$; the nonnegative square root of the variance sum is the scale in that definition. Uniqueness proves the result. If all variances vanish, each input is constant and the result is the corresponding Dirac law; empty sums give ${\delta }_0$, and one-term sums return the original law. Bernoulli p=0 and p=1 similarly give deterministic zero and n. AC is inherited from Fourier uniqueness and from normal normalization and the compact integral bridge; no companion example is a supplier.

First use $h(u)=(1-\mathrm{\mid }u\mathrm{\mid }{)}_{+}$ as a density in the space variable u. It is nonnegative Borel and $\int h=2{\int }_0^1(1-u)du=1$, so it defines a probability. Its characteristic function q has vanishing imaginary part by the oddness of $h(u)\sin (su)$. For $s\mathrm{\ne }0$, integration by parts with $1-u$ and $\sin (su)\mathrm{/}s$ yields $q(s)=2{\int }_0^1(1-u)\cos (su)du=\frac{2}{s}{\int }_0^1\sin (su)du=\frac{2(1-\cos s)}{s^2}\mathrm{.}$ All factors and derivatives are continuous on $[0,1]$, so FTC and the bridge apply. Density integration is applied to the real and imaginary positive/negative parts. At s=0, q(0)=1, and continuity follows from the characteristic-function lemma. Thus q is nonnegative everywhere and bounded by one, with $q(s)\le 4\mathrm{/}{s}^2$ for $\mathrm{\mid }s\mathrm{\mid }\ge 1$. The primitive $-1\mathrm{/}s$ on $[1,R]$, the bridge and monotone convergence give ${\int }_1^{\mathrm{\infty }}{s}^{-2}ds=1$; reflection gives the other tail. Hence q is integrable.

Apply density inversion to that probability with characteristic function q. It supplies the continuous density $H(y)=(2\pi {)}^{-1}\int e^{-isy}q(s)ds$. This equals h everywhere: if the two continuous densities differed at y, continuity would give a small interval where their difference had one strict sign, contradicting that both densities integrate to the same interval mass. In particular $1=h(0)=H(0)=(2\pi {)}^{-1}\int q$. Therefore $f=q\mathrm{/}(2\pi )$ is nonnegative and integrates to one, so defines a probability. It has exactly the displayed formula and the continuous value $1\mathrm{/}(2\pi )$ at zero.

Density integration and the identity in step 2.1 now give ${\phi }_f(t)=\frac{1}{2\pi }\int e^{itx}q(x)dx=H(-t)=h(-t)=(1-\mathrm{\mid }t\mathrm{\mid }{)}_{+}\mathrm{.}$ This includes t=0 and both endpoints t=±1, where the value is zero; outside the closed interval it is zero as well. The density value at x=0 was fixed by continuity, not division by zero. AC is inherited from the compact bridge and density inversion. The argument applied inversion only to the known law h before establishing that the triangle is a characteristic function.

For each integer $n\ge 1$, the Borel density $1_{[-n,n]}\mathrm{/}(2n)$ integrates to one and hence defines ${\mu }_n$. Apply [F9] separately to the positive and negative parts of the bounded real functions $\cos (tx)$ and $\sin (tx)$ and subtract their finite integrals; reassembling the real and imaginary components shows that its transform is $(2n{)}^{-1}{\int }_{-n}^n{e}^{itx}dx$. When $t\mathrm{\ne }0$, integrating the cosine using $\sin (tx)\mathrm{/}t$ gives $\sin (nt)\mathrm{/}(nt)$, and the sine integral vanishes by oddness (or its primitive $-\cos (tx)\mathrm{/}t$). The compact bridge validates these Lebesgue calculations. At t=0 the integrand is one. For each fixed nonzero t, $\mathrm{\mid }\sin (nt)\mathrm{/}(nt)\mathrm{\mid }\le 1\mathrm{/}(n\mathrm{\mid }t\mathrm{\mid })\to 0$, while at zero the sequence is constantly one.

The function $1_{\{0\}}$ is discontinuous at zero: at $t=1\mathrm{/}k$ its value is zero for every positive integer k, while its value at zero is one. Since every characteristic function is continuous there, it cannot be the characteristic function of any Borel probability. Moreover for every $M\ge 0$, ${\mu }_n([-M,M])=\min (1,M\mathrm{/}n)\to 0$. Every nonempty compact K is contained in such an interval, so eventually ${\mu }_n(K)<1\mathrm{/}2$ and its complement has mass greater than $1\mathrm{/}2$. The empty compact set has complement mass one for every n. No compact set works for error $1\mathrm{/}2$, proving non-tightness. The index n=0 is excluded because the displayed density divides by 2n; a point mass at zero would be a different law. AC is used only through the compact integration bridge and its continuous-integrand prerequisites.

Put $N=m+1\ge 1$ and $\mu =2^{1-N}{\sum }_{\begin{array}{c}0\le k\le N\\ k\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}\end{array}}\left(\genfrac{}{}{0px}{}{N}{k}\right){\delta }_k,\nu =2^{1-N}{\sum }_{\begin{array}{c}0\le k\le N\\ k\mathrm{o}\mathrm{d}\mathrm{d}\end{array}}\left(\genfrac{}{}{0px}{}{N}{k}\right){\delta }_k\mathrm{.}$ All coefficients are nonnegative real images of the natural binomial coefficients. The binomial theorem at (1,1) and (-1,1) says that the sum of the even and odd coefficient totals is $2^N$, while their difference is zero. Each total is $2^{N-1}>0$. Thus these finite weighted measures are Borel probabilities. They are distinct because $\mu (\{0\})=2^{1-N}>0$ and $\nu (\{0\})=0$.

For a polynomial P define $\mathrm{\Delta }P(x)=P(x+1)-P(x)$. The binomial theorem gives $\mathrm{\Delta }(x^j)={\sum }_{r=0}^{j-1}\left(\genfrac{}{}{0px}{}{j}{r}\right)x^r$ for j>=1, while $\mathrm{\Delta }1=0$. By linearity each application lowers a positive degree by at least one and kills a constant, so ${\mathrm{\Delta }}^{N}P=0$ whenever $\mathrm{deg}P<N$. To compute the iterate, induction gives ${\mathrm{\Delta }}^{r}P(x)={\sum }_{k=0}^r(-1{)}^{r-k}\left(\genfrac{}{}{0px}{}{r}{k}\right)P(x+k)\mathrm{.}$ The base r=0 is P(x). Subtract this expression at x from the expression at x+1: the interior coefficient of P(x+k) is $(-1{)}^{r+1-k}[\left(\genfrac{}{}{0px}{}{r}{k-1}\right)+\left(\genfrac{}{}{0px}{}{r}{k}\right)]=(-1{)}^{r+1-k}\left(\genfrac{}{}{0px}{}{r+1}{k}\right)$, and the coefficients at k=0,r+1 are $(-1{)}^{r+1}$ and one. This proves the induction including both endpoints. Taking r=N, x=0 and P(x)=x^j, j<N, yields ${\sum }_{k=0}^N(-1{)}^k\left(\genfrac{}{}{0px}{}{N}{k}\right)k^j=0$. For j=0 the polynomial is constantly one, so its value at zero is one.

Every required moment is finite because the supports are finite. The difference of the jth moments of the two laws equals $2^{1-N}{\sum }_{k=0}^N(-1{)}^k\left(\genfrac{}{}{0px}{}{N}{k}\right)k^j=0$ for $0\le j\le m=N-1$. This proves the promised failure of determination for every m. When m=0 the witnesses are ${\delta }_0$ and ${\delta }_1$ and only total mass is matched. For m=2 the even law has masses 1/4 at 0 and 3/4 at 2, while the odd law has masses 3/4 at 1 and 1/4 at 3. Their means are both 3/2 and their second moments both 3, yet their masses at zero differ. All finite choices are specified by parity, and no AC is used.