Fourier Inversion for Lattice Distributions

Why One Period Recovers One Probability

For a continuous distribution, Fourier inversion recovers a density by integrating over a frequency range. For an integer-valued random variable, the key operation is different: integrate its characteristic function over one full period. The oscillating exponential terms cancel for every integer value except the one whose probability we want.

This cancellation is an orthogonality property. If \(X\) takes integer values, its characteristic function is \(2\pi\)-periodic, and the functions \(e^{ikt}\), for integer \(k\), are orthogonal over an interval of length \(2\pi\). The inversion formula is therefore a direct way to extract each probability mass, without assuming that \(X\) has a density or any finite moments.

Definition: A random variable \(X\) is integer-valued if \(P(X\in\mathbb Z)=1\). Its characteristic function is \(\varphi_X(t)=E[e^{itX}]\), for real \(t\). For an integer \(k\), its probability mass at \(k\) is \(P(X=k)\).

The Inversion Formula on the Integers

The central identity extracts the probability at any specified integer \(k\), including values that may have probability zero.

Theorem (Fourier inversion for an integer-valued random variable): If \(P(X\in\mathbb Z)=1\), then for every \(k\in\mathbb Z\),

P(X=k)=\frac{1}{2\pi}\int_{-\pi}^{\pi}\varphi_X(t)e^{-ikt}\,dt.

The same formula holds with any interval of length \(2\pi\) in place of \([-\pi,\pi]\).

Proof. By the definition of the characteristic function, \(\varphi_X(t)=E[e^{itX}]\). The random integrand has absolute value one. On the finite interval \([-\pi,\pi]\), its absolute integral is bounded by \(2\pi\); consequently, Fubini’s theorem permits interchanging expectation and integration:

\frac{1}{2\pi}\int_{-\pi}^{\pi}\varphi_X(t)e^{-ikt}\,dt = E\left[\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{it(X-k)}\,dt\right].

Since \(X\) is integer-valued almost surely, \(X-k\) is an integer almost surely. For any integer \(m\),

\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{imt}\,dt = \begin{cases} 1, & m=0,\\ 0, & m\ne 0. \end{cases}

Indeed, for \(m=0\) the integrand is one, so the integral is \(2\pi\). For a nonzero integer \(m\), direct integration gives

\int_{-\pi}^{\pi}e^{imt}\,dt = \left[\frac{e^{imt}}{im}\right]_{-\pi}^{\pi} = \frac{e^{im\pi}-e^{-im\pi}}{im} =0,

because \(e^{im\pi}=e^{-im\pi}=(-1)^m\). Thus the random quantity inside the expectation is exactly \(1\) when \(X=k\) and \(0\) otherwise. Its expectation is \(P(X=k)\), proving the formula. The characteristic function is \(2\pi\)-periodic because \(e^{i(t+2\pi)X}=e^{itX}\) almost surely. Integrating a periodic function over any interval of one period gives the same integral. \(\square\)

Equivalently, if \(p_j=P(X=j)\), the characteristic function can be written as the absolutely convergent series \(\varphi_X(t)=\sum_{j\in\mathbb Z}p_j e^{ijt}\). Substitution in the inversion formula gives an integral of each term \(e^{i(j-k)t}\). The proof above justifies the extraction directly by expectation, without needing to make a separate interchange of an infinite series and an integral.

Worked Example: Recovering a Binomial Probability

Let \(X\sim\operatorname{Binomial}(4,0.3)\). Its characteristic function is

\varphi_X(t)=(0.7+0.3e^{it})^4.

Use inversion to find \(P(X=2)\). Expand the characteristic function by the binomial theorem:

\varphi_X(t) = \sum_{j=0}^{4}\binom{4}{j}(0.7)^{4-j}(0.3)^j e^{ijt}.

Multiplication by \(e^{-2it}\) makes the term indexed by \(j\) proportional to \(e^{i(j-2)t}\). When integrated from \(-\pi\) to \(\pi\), every such term has integral zero except the term with \(j=2\). Therefore,

\begin{aligned} P(X=2) &=\frac{1}{2\pi}\int_{-\pi}^{\pi}\varphi_X(t)e^{-2it}\,dt\\ &=\binom{4}{2}(0.7)^2(0.3)^2\\ &=6(0.49)(0.09)=0.2646. \end{aligned}

This agrees with the binomial probability mass function. Here inversion recovers one coefficient from the characteristic function’s finite trigonometric expansion.

Recovering Probabilities from Infinite Series

The same orthogonality works when the characteristic function has an infinite expansion. In the next examples, the series can be integrated term by term because the sum of the absolute values of the coefficients is finite. For the Poisson series this sum is \(e^{-2}\sum_{j=0}^{\infty}2^j/j!=1\); for the geometric series it is \((1-q)\sum_{j=0}^{\infty}q^j=1\).

Worked Example: Recovering a Poisson Probability

Let \(X\sim\operatorname{Poisson}(2)\), with characteristic function

\varphi_X(t)=\exp\{2(e^{it}-1)\}.

Rewrite it as a series:

\varphi_X(t) = e^{-2}\exp(2e^{it}) = e^{-2}\sum_{j=0}^{\infty}\frac{2^j}{j!}e^{ijt}.

To find the probability at \(k=3\), multiply by \(e^{-3it}\) and integrate. Absolute summability of the coefficients allows term-by-term integration. Orthogonality removes every term except \(j=3\):

\begin{aligned} P(X=3) &=\frac{1}{2\pi}\int_{-\pi}^{\pi}\varphi_X(t)e^{-3it}\,dt\\ &=e^{-2}\frac{2^3}{3!}\\ &=\frac{4}{3e^2}\approx 0.1804. \end{aligned}

Thus the integral selects exactly the coefficient of \(e^{i3t}\), which is the Poisson mass at 3.

Worked Example: Recovering a Geometric Probability

Suppose \(X\) has the geometric distribution on \(\{0,1,2,\ldots\}\) with \(P(X=j)=(1-q)q^j\), where \(q=0.4\). The characteristic function is the geometric series

\varphi_X(t) = (1-q)\sum_{j=0}^{\infty}q^j e^{ijt} = \frac{1-q}{1-qe^{it}}.

The series representation converges absolutely since \(\sum_{j=0}^{\infty}(1-q)q^j=1\). To recover the mass at 2, use the series form in the inversion integral:

\begin{aligned} P(X=2) &=\frac{1}{2\pi}\int_{-\pi}^{\pi} (1-q)\sum_{j=0}^{\infty}q^j e^{i(j-2)t}\,dt\\ &=(1-q)q^2\\ &=(0.6)(0.16)=0.096. \end{aligned}

As in the finite binomial expansion, all terms except the one indexed by \(j=2\) integrate to zero. The difference is that absolute convergence justifies integrating the infinite series term by term.

Changing the Lattice Spacing

A distribution need not be supported on all integers, or even on consecutive integers. Suppose its values lie on a shifted lattice \(a+h\mathbb Z=\{a+hk:k\in\mathbb Z\}\), where \(a\in\mathbb R\) and \(h>0\). The integration interval must now have length \(2\pi/h\), so that distinct points of this lattice remain orthogonal.

Theorem (inversion on a shifted lattice): If \(P(X\in a+h\mathbb Z)=1\), where \(a\in\mathbb R\) and \(h>0\), then for every \(k\in\mathbb Z\),

P(X=a+hk) = \frac{h}{2\pi} \int_{-\pi/h}^{\pi/h} \varphi_X(t)e^{-it(a+hk)}\,dt.

Proof. Write \(X=a+hJ\) almost surely, where \(J\) is integer-valued, and set \(x=a+hk\). Fubini’s theorem applies because the integrand \(e^{it(X-x)}\) has absolute value one and the integration interval has finite length. Hence

\frac{h}{2\pi}\int_{-\pi/h}^{\pi/h}\varphi_X(t)e^{-itx}\,dt = E\left[\frac{h}{2\pi}\int_{-\pi/h}^{\pi/h}e^{it(X-x)}\,dt\right].

For each outcome, \(X-x=h(J-k)\). If \(J=k\), the expression inside the expectation is 1. Otherwise, \(m=J-k\) is a nonzero integer, and

\frac{h}{2\pi}\int_{-\pi/h}^{\pi/h}e^{ihmt}\,dt = \frac{1}{2\pi}\int_{-\pi}^{\pi}e^{imu}\,du =0,

where \(u=ht\). The random quantity inside the expectation is therefore the indicator of \(X=x\), and its expectation is \(P(X=x)\). This proves the result. \(\square\)

Worked Example: Inversion on the Odd Integers

Let \(N\sim\operatorname{Poisson}(1)\) and set \(X=1+2N\). Then \(X\) is supported on the shifted lattice \(1+2\mathbb Z\), and

\varphi_X(t)=e^{it}\exp(e^{i2t}-1).

The point \(5\) corresponds to \(1+2k=5\), so \(k=2\). The shifted-lattice formula, with \(a=1\) and \(h=2\), gives

\begin{aligned} P(X=5) &=\frac{1}{\pi}\int_{-\pi/2}^{\pi/2} e^{it}\exp(e^{i2t}-1)e^{-5it}\,dt\\ &=\frac{e^{-1}}{\pi}\int_{-\pi/2}^{\pi/2} \sum_{j=0}^{\infty}\frac{e^{i2jt}}{j!}e^{-i4t}\,dt. \end{aligned}

The series is absolutely summable in its coefficients, since \(\sum_{j=0}^{\infty}1/j!=e\), so it may be integrated term by term. The resulting factor is \(e^{i2(j-2)t}\). Its integral over \([-\pi/2,\pi/2]\) is \(\pi\) when \(j=2\), and zero for every other integer \(j\). Thus

P(X=5)=\frac{e^{-1}}{\pi}\frac{\pi}{2!} =\frac{1}{2e}\approx 0.1839.

This is also \(P(N=2)=e^{-1}/2!\), as required by \(X=1+2N\). The shorter interval is essential: the lattice spacing is 2, not 1.

Common Mistake and Practical Use

A common mistake is to use the integer-lattice formula unchanged for a variable supported on a lattice with spacing \(h\ne1\). The factor and interval in the inversion formula are linked: for spacing \(h\), the interval has length \(2\pi/h\) and the multiplier is \(h/(2\pi)\). Using an interval of length \(2\pi\) without checking the spacing does not, in general, make the exponential terms orthogonal.

Another mistake is to treat a characteristic function like a density transform that must be integrated over the entire real line. For an integer-valued random variable, the characteristic function is periodic, and the mass formula uses one period. Integrating across all real \(t\) is not the inversion formula established here.

The practical value of lattice inversion is that it turns a characteristic function into exact point probabilities. This is useful when a distribution is most naturally described through its characteristic function, or when the function has a manageable series or finite expansion. The main steps are to identify the lattice, choose a full period for its spacing, multiply by the exponential corresponding to the target point, and use orthogonality to isolate its mass.

Check Your Understanding

Use the orthogonality argument and lattice formulas to answer the following.

Why does integrating \(e^{i(j-k)t}\) over \([-\pi,\pi]\) give zero when \(j\) and \(k\) are distinct integers?
For \(X\sim\operatorname{Binomial}(3,0.2)\), which term in the expansion of \((0.8+0.2e^{it})^3e^{-it}\) survives integration over one period?
What integration interval and multiplier apply when \(X\) is supported on \(a+3\mathbb Z\)?
Explain why Fubini’s theorem permits exchanging expectation and integration in the integer-lattice proof.
If \(X=2N\) for an integer-valued random variable \(N\), what lattice contains \(X\), and what spacing should be used to recover \(P(X=2k)\)?