Polya's Criterion

Last updated: 2026-04-12

Polya’s criterion, in the context of probability theory, is a condition used to determine whether a function is a characteristic function.
Combined with the property that characteristic functions are non-negative definite, Pólya's criterion ensures that the function is non-negative definite.
Theorem (Polya's criterion)
Let φ(t)\varphi(t) be real nonnegative and have φ(0)=\varphi(0)= 1,φ(t)=φ(−t)1, \varphi(t)=\varphi(-t), and φ\varphi is decreasing and convex on (0,∞)(0, \infty) with
lim⁡t↓0φ(t)=1,lim⁡t↑∞φ(t)=0\begin{equation*}\lim _{t \downarrow 0} \varphi(t)=1, \quad \lim _{t \uparrow \infty} \varphi(t)=0\end{equation*}
Then there is a probability measure ν\nu on (0,∞)(0, \infty), so that
φ(t)=∫0∞(1−∣ts∣)+ν(ds)\begin{equation*}\varphi(t)=\int_0^{\infty}\left(1-\left|\frac{t}{s}\right|\right)^{+} \nu(d s)\end{equation*}
and hence φ\varphi is a characteristic function.

Proof of Polya's criterion

The foolowing proof is due to [Durrett2019] pages 119

The assumption that lim⁡t→0φ(t)=1\lim _{t \rightarrow 0} \varphi(t)=1 is necessary because the function φ(t)=\varphi(t)= 1{0}(t)1_{\{0\}}(t) which is 1 at 0 and 0 otherwise satisfies all the other hypotheses. We could allow lim⁡t→∞φ(t)=c>0\lim _{t \rightarrow \infty} \varphi(t)=c>0 by having a point mass of size cc at 0, but we leave this extension to the reader.

Radon-Nikodym measure

\quad Proof. Let φ′\varphi^{\prime} be the right derivative of ϕ\phi, i.e.,
φ′(t)=lim⁡h\0φ(t+h)−φ(t)h\begin{equation*}\varphi^{\prime}(t)=\lim _{h \backslash 0} \frac{\varphi(t+h)-\varphi(t)}{h}\end{equation*}
Since φ\varphi is convex this exists and is right continuous and increasing. So we can let μ\mu be the measure on (0,∞)(0, \infty) with μ(a,b]=φ′(b)−φ′(a)\mu(a, b]=\varphi^{\prime}(b)-\varphi^{\prime}(a) for all 0≤a<b<∞0 \leq a<b<\infty, and let ν\nu be the measure on (0,∞)(0, \infty) with dν/dμ=sd \nu / d \mu=s.

Integrating over new measure

Now φ′(t)→0\varphi^{\prime}(t) \rightarrow 0 as t→∞t \rightarrow \infty (for if φ′(t)↓−ϵ\varphi^{\prime}(t) \downarrow-\epsilon we would have φ(t)≤1−ϵt\varphi(t) \leq 1-\epsilon t for all tt ), so
−φ′(s)=∫s∞r−1ν(dr)\begin{equation*}-\varphi^{\prime}(s)=\int_s^{\infty} r^{-1} \nu(d r)\end{equation*}
Integrating again and using Fubini's theorem we have for t≥0t \geq 0
φ(t)=∫t∞∫s∞r−1ν(dr)ds=∫t∞r−1∫trdsν(dr)=∫t∞(1−tr)ν(dr)=∫0∞(1−tr)+ν(dr)\begin{align*}\varphi(t) & =\int_t^{\infty} \int_s^{\infty} r^{-1} \nu(d r) d s=\int_t^{\infty} r^{-1} \int_t^r d s \nu(d r) \\& =\int_t^{\infty}\left(1-\frac{t}{r}\right) \nu(d r)=\int_0^{\infty}\left(1-\frac{t}{r}\right)^{+} \nu(d r)\end{align*}
Using φ(−t)=φ(t)\varphi(-t)=\varphi(t) to extend the formula to t≤0t \leq 0 we have (*). Setting t=0t=0 in (∗)(*) shows ν\nu has total mass 1.

Non smooth case

If φ\varphi is piece-wise linear, ν\nu has a finite number of atoms and the result follows from fact that weighted average of characteristic function is again characteristic function. To prove the general result, let νn\nu_n be a sequence of measures on (0,∞)(0, \infty) with a finite number of atoms that converges weakly to ν\nu and let
φn(t)=∫0∞(1−∣ts∣)+νn(ds)\begin{equation*}\varphi_n(t)=\int_0^{\infty}\left(1-\left|\frac{t}{s}\right|\right)^{+} \nu_n(d s)\end{equation*}
Since s→(1−∣t/s∣)+s \rightarrow(1-|t / s|)^{+}is bounded and continuous, φn(t)→φ(t)\varphi_n(t) \rightarrow \varphi(t) and the desired result follows from Levy's Continuity Theorem. □\Box

Applications of Polya's criterion

We can apply Polya's criterion to verify whether certain functions are characteristic. Simple examples which can be deduced to characteristic from the plot are
φ1(t)=exp⁡(−∣t∣)φ2(t)={1−∣t∣,∣t∣≤1,0,∣t∣>1.\begin{align*}\varphi_1(t) &= \exp(-|t|) \\\varphi_2(t) &= \begin{cases} 1 - |t|, & |t| \leq 1 , \\ 0, & |t|>1 .\end{cases}\end{align*}

Exponent with power less two

We need helper lemma to prove the main theorem.
Lemma
If cj→0c_j \to 0, aj→∞a_j \to \infty, and ajcj→λa_j c_j \to \lambda, then (1+cj)aj→eλ(1 + c_j)^{a_j} \to e^{\lambda}.
\quad Proof. As x→0x \to 0, log⁡(1+x)x→1\frac{\log(1 + x)}{x} \to 1, so ajlog⁡(1+cj)→λa_j \log(1 + c_j) \to \lambda, and the desired result follows. □\Box
The case α=1\alpha = 1 correspond to the Cauchy distribution and the case α=2\alpha = 2 correspond to a standard distribution.
Theorem
exp⁡(−∣t∣α)\exp(-|t|^\alpha) is a characteristic function for 0<α<20 < \alpha < 2
The case α=1\alpha = 1 correspond to the Cauchy distribution and the case α=2\alpha = 2 correspond to a standard distribution.
\quad Proof. The key idea of the proof is to approximate the function exp⁡(−∣t∣α)\exp(-|t|^\alpha) as a sequence of characteristic functions. By applying Lévy's Continuity Theorem, it follows that exp⁡(−∣t∣α)\exp(-|t|^\alpha) is characteristic function as well.
The function which help us to make approximation is
ψ(t)=1−(1−cos⁡t)α/2.\begin{equation*}\psi(t) = 1 - (1 - \cos t)^{\alpha/2}.\end{equation*}
Then for any for any β\beta and ∣x∣<1|x| < 1 we have the formula
(1−x)β=∑n=0∞(βn)(−x)n\begin{equation*}(1 - x)^\beta = \sum_{n=0}^{\infty} \binom{\beta}{n} (-x)^n\end{equation*}
where
(βn)=β(β−1)⋯(β−n+1)1⋅2⋅…⋅n.\begin{equation*}\binom{\beta}{n} = \frac{\beta (\beta - 1) \cdots (\beta - n + 1)}{1 \cdot 2 \cdot \ldots \cdot n}.\end{equation*}
Now we can represent ψ(t)\psi(t) as a infinite series
ψ(t)=1−(1−cos⁡t)α/2=∑n=1∞cn(cos⁡t)n,\begin{equation*}\psi(t) = 1 - (1 - \cos t)^{\alpha/2} = \sum_{n=1}^{\infty} c_n (\cos t)^n,\end{equation*}
where
cn=(α/2n)(−1)n+1,\begin{equation*}c_n = \binom{\alpha/2}{n} (-1)^{n+1},\end{equation*}
cn≥0c_n \geq 0(we used α<2\alpha < 2), and ∑ncn=1\sum_n c_n = 1 (take t=0t=0 in the definition of ψ(t)\psi(t)). \\
To confirm that ψ(t)\psi(t) is a characteristic function, we utilize the fact that cos⁡(t)\cos(t) is the characteristic function of a coin flip, i.e., P(X=1)=P(X=−1)=1/2P(X = 1) = P(X = -1) = 1/2. By using the binomial coefficients, we can show that cos⁡n(t)\cos^n(t) is also a characteristic function. Since a weighted average of characteristic functions remains a characteristic function, it follows that ψ(t)\psi(t) is indeed a characteristic function. \\
From analysis 1−cos⁡t∼t2/21 - \cos t \sim t^2/2 as t→0t \to 0, so
1−cos⁡(2tn1/α)∼t2n2/α.\begin{equation}1 - \cos( \frac{\sqrt{2} t}{n^{1/\alpha}}) \sim \frac{t^2}{n^{2/\alpha}}.\end{equation}
The representation of exp⁡(−∣t∣α)\exp(-|t|^\alpha) as limit of characteristic functions is due to Lévy's Continuity Theorem and above-mentioned lemma,
lim⁡n→∞{ψ(2tn1/α)}n=lim⁡n→∞{1−(1−cos⁡(2tn1/α))α/2}n=exp⁡(−∣t∣α)□\begin{equation*}\lim_{n \to \infty} \{\psi(\frac{\sqrt{2} t}{n^{1/\alpha}})\}^n = \lim_{n \to \infty} \left\{ 1 - (1 - \cos( \frac{\sqrt{2} t}{n^{1/\alpha}}))^{\alpha/2} \right\}^n = \exp(-|t|^\alpha) \Box\end{equation*}

References