Identifiability of Asymmetric Circular and Cylindrical Distributions

Abstract

Identifiability of statistical models is a fundamental and essential condition that is required to prove the consistency of maximum likelihood estimators. The identifiability of the skew families of distributions on the circle and cylinder for estimating model parameters has not been fully investigated in the literature. In this paper, a new method combining the trigonometric moments and the simultaneous Diophantine approximation is proposed to prove the identifiability of asymmetric circular and cylindrical distributions. Using this method, we prove the identifiability of general sine-skewed circular distributions, including the sine-skewed von Mises and sine-skewed wrapped Cauchy distributions, and that of a Möbius transformed cardioid distribution, which can be regarded as asymmetric distributions on the unit circle. In addition, we prove the identifiability of two cylindrical distributions wherein both marginal distributions of a circular random variable are the sine-skewed wrapped Cauchy distribution, and conditional distributions of a random variable on the non-negative real line given the circular random variable are a Weibull distribution and a generalized Pareto-type distribution, respectively.

Appendices

Appendix : Appendix A: Proofs

Appendix A: Proofs

Proof Proof of Theorem 2.1.

We prove this by contradiction. Assume conditions (a) and (b), and that there exist parameters γ₁, γ₂ such that γ₁≠γ₂ and F(x|γ₁) = F(x|γ₂). By these assumptions, it holds that ϕ_i(p|γ₂)/ϕ_i(p|γ₁) = 1 for every i and any p, which contradicts condition (b). □

Lemma A.1.

For any $s\in \mathbb {N}$, any $a_{1},...,a_{s}\in \left [\right .0,2\pi \left .\right )$ and any 𝜖 > 0, there exist infinitely many $p\in \mathbb {N}$ such that |pa_i(mod 2π) | < 𝜖 (i = 1,...,s), where “ mod” indicates the modulo operation.

Proof.

Fix any $s\in \mathbb {N}$ and any 𝜖 > 0. For any a_i ∈ [0,2π) (i = 1,...,s), there exist constants c_i ∈ [0,2) such that a_i = c_iπ (i = 1,...,s). From the simultaneous Diophantine approximation (e.g., see Theorem 1A on page 27 of Schmidt (1996)), for any natural number Q with Q ≥ 2, there exist integers q, p₁,...,p_s such that q ∈{1,...,Q^s − 1}, and

$$ \left| c_{i}-\frac{p_{i}}{q}\right| \leq \frac{1}{qQ},\quad (i=1,...,s) $$

which leads to

$$ p\left| c_{i}\pi-\frac{p_{i}}{q}\pi\right| \leq \frac{p\pi}{qQ}, $$

(A.12)

where p is any natural number. For simplicity, we let $c_{i}^{*}=p_{i}/q$. Then, choosing the number Q with 2πQ^− 1/2 < 𝜖 and $Q^{1/2}\in \mathbb {N}$, and putting p = 2qQ^1/2 in Eq. A.12 yields

$$ p\left| c_{i}\pi-c_{i}^{*}\pi\right| \leq \frac{2\pi}{Q^{1/2}}<\epsilon \qquad (i=1,...,s). $$

(A.13)

In addition, we have $pc_{i}^{*}\pi =2Q^{1/2}p_{i}\pi =0$ (mod 2π). Therefore, because $pc_{i}\pi =(Q^{1/2}p_{i})(2\pi )+(pc_{i}\pi -pc_{i}^{*}\pi )$ and $Q^{1/2}p_{i}\in \mathbb {Z}$, it follows from the inequality (A.13) that

$$ \begin{array}{@{}rcl@{}} |pc_{i}\pi \textrm{(mod 2$\pi$)}| &=&|pc_{i}\pi -pc_{i}^{*}\pi | <\epsilon , \end{array} $$

which completes the proof. Note that we can take infinitely many p = 2qQ^1/2 by choosing Q successfully because there are infinitely many Q with 2πQ^− 1/2 < 𝜖 and $Q^{1/2}\in \mathbb {N}$, and q ≥ 1. □

Proof of Theorem 2.2.

For different parameters γ₁ and γ₂, we consider the following three steps: Step 1. ψ₁≠ψ₂, Step 2. ψ₁ = ψ₂, λ₁≠λ₂, Step 3. ψ₁ = ψ₂, λ₁ = λ₂, and μ₁≠μ₂. Then, under each step, we verify the conditions in Theorem 2.1.

Step 1. We consider two parameter vectors γ₁ and γ₂ with ψ₁≠ψ₂, and set ϕ₁(p|γ) = ρ_p(γ)² where ρ_p(γ) is defined in Eq. 4. Then, the ratio of ϕ₁(p|γ₁) to ϕ₁(p|γ₂) is

$$ \frac{\phi_{1}(p|\boldsymbol{\gamma}_{1})}{\phi_{1}(p|\boldsymbol{\gamma}_{2})}=\frac{\alpha_{0,p}(\boldsymbol{\psi}_{1} )^{2}+({\lambda_{1}^{2}}/4)\left\{ \alpha_{0,p-1}(\boldsymbol{\psi}_{1} )-\alpha_{0,p+1}(\boldsymbol{\psi}_{1} )\right\}^{2}}{\alpha_{0,p}(\boldsymbol{\psi}_{2} )^{2}+({\lambda_{2}^{2}}/4)\left\{ \alpha_{0,p-1}(\boldsymbol{\psi}_{2} )-\alpha_{0,p+1}(\boldsymbol{\psi}_{2} )\right\}^{2}}. $$

(A.14)

By condition (5), there exists a function e_p(ψ) of integer p and positive constant M > 0 such that $\sup _{p\in \mathbb {N}}|e_{p}(\boldsymbol {\psi } )|\leq M$ and

$$ \begin{array}{@{}rcl@{}} \frac{\alpha_{0,p-1}(\boldsymbol{\psi} )-\alpha_{0,p+1}(\boldsymbol{\psi} )}{\alpha_{0,p}(\boldsymbol{\psi} )}=e_{p}(\boldsymbol{\psi} )p^{c}, \end{array} $$

where the constant c is given in condition (iii) of Theorem 2.2.

Hence, Eq. A.14 is expressed as

$$ \begin{array}{@{}rcl@{}} \frac{\alpha_{0,p}(\boldsymbol{\psi}_{1} )^{2}\left\{ 1+({\lambda_{1}^{2}}/4)e_{p}(\boldsymbol{\psi}_{1})^{2}p^{2c} \right\}}{\alpha_{0,p}(\boldsymbol{\psi}_{2} )^{2}\left\{ 1+({\lambda_{2}^{2}}/4)e_{p}(\boldsymbol{\psi}_{2})^{2}p^{2c} \right\}}. \end{array} $$

(A.15)

Because the domain of ϕ₁(p|γ) is $S_{1}(\boldsymbol {\gamma })=\mathbb {Z}$, and the closure of S₁(γ) becomes the extended rational number set $\overline {\mathbb {Z}}$ under the metric d(x,y) described in Section 2. Therefore, $p\in \mathbb {Z}$ is allowed to increase to infinity.

If condition $\{ \alpha _{0,p}(\boldsymbol {\psi }_{1})/\alpha _{0,p}(\boldsymbol {\psi }_{2})\} p^{c}\rightarrow 0$ $(p\to \infty )$ in Eq. 6 holds, then the ratio (A.15) is bounded above by

$$ \begin{array}{@{}rcl@{}} \frac{\alpha_{0,p}(\boldsymbol{\psi}_{1} )^{2}}{\alpha_{0,p}(\boldsymbol{\psi}_{2} )^{2}}\left\{ 1+ \frac{M^{2}}{4}p^{2c}\right\} &=&\left( \frac{\alpha_{0,p}(\boldsymbol{\psi}_{1})}{\alpha_{0,p}(\boldsymbol{\psi}_{2} )}p^{c}\right)^{2}\left\{ p^{-2c}+ \frac{M^{2}}{4}\right\} \\ &\rightarrow& 0\quad (p\to\infty). \end{array} $$

On the other hand, if $\frac {\alpha _{0,p}(\boldsymbol {\psi }_{1})}{\alpha _{0,p}(\boldsymbol {\psi }_{2})}p^{-c}\rightarrow \infty$ $(p\to \infty )$, then the ratio (A.15) is bounded below by

$$ \begin{array}{@{}rcl@{}} \frac{\alpha_{0,p}(\boldsymbol{\psi}_{1} )^{2}}{\alpha_{0,p}(\boldsymbol{\psi}_{2} )^{2}}\frac{1}{ 1+ (M^{2}/4)p^{2c}} &=&\left( \frac{\alpha_{0,p}(\boldsymbol{\psi}_{1})}{\alpha_{0,p}(\boldsymbol{\psi}_{2} )}p^{-c}\right)^{2} \frac{1}{ p^{-2c}+ (M^{2}/4)} \\ &\rightarrow& \infty \quad (p\to\infty). \end{array} $$

Accordingly, under assumption ψ₁≠ψ₂, the ratio $\phi _{1}(p|\boldsymbol {\gamma }_{1})/\phi _{1}(p|\boldsymbol {\gamma }_{2}) \not \rightarrow 1$ $(p\to \infty )$.

Step 2. Next, we consider two parameter vectors γ₁ and γ₂ with ψ₁ = ψ₂ = ψ and λ₁≠λ₂, and set ϕ₂(p|γ) = β_p(γ), which is defined in Eq. 3. Then, by Lemma A.1, for possibly different parameters μ₁ and μ₂ in [0,2π), there exists a sequence $\{ p_{n}\}_{n\in \mathbb {N}}$ with $p_{n}\in \mathbb {N}$ such that $\lim _{n\to \infty }p_{n}=\infty$ and $\lim _{n\to \infty }|p_{n}\mu _{i} \pmod {2\pi }|=0$ (i = 1,2). By using the sequence $\{ p_{n}\}_{n\in \mathbb {N}}$, it follows that

$$ \begin{array}{@{}rcl@{}} &&\left| \sin (p_{n}\mu_{i}) \frac{\alpha_{0,p_{n}}(\boldsymbol{\psi}_{i})}{\alpha_{0,p_{n}-1}(\boldsymbol{\psi}_{i} )-\alpha_{0,p_{n}+1}(\boldsymbol{\psi}_{i} )}\right|\\ &\leq& |\sin (p_{n}\mu_{i})|\frac{1}{\inf_{p\in\mathbb{N}}\left| \frac{\alpha_{0,p-1}(\boldsymbol{\psi}_{i} )-\alpha_{0,p+1}(\boldsymbol{\psi}_{i} )}{\alpha_{0,p}(\boldsymbol{\psi}_{i})}\right|} \\ &\leq& M|\sin (p_{n}\mu_{i})| \\ &\rightarrow& 0\quad (n\to\infty ). \end{array} $$

(A.16)

If λ₂≠ 0, from Eq. A.16, the ratio of ϕ₂(p|γ₁) to ϕ₂(p|γ₂) with p replaced by p_n is

$$ \begin{array}{@{}rcl@{}} \frac{\phi_{2}(p_{n}|\boldsymbol{\gamma}_{1})}{\phi_{2}(p_{n}|\boldsymbol{\gamma}_{2})}\!\!\! &&=\frac{\sin (p_{n}\mu_{1} )\alpha_{0,p_{n}}(\boldsymbol{\psi} )\!+\cos (p_{n}\mu_{1} )\lambda_{1} \{ \alpha_{0,p_{n}-1}(\boldsymbol{\psi} ) - \alpha_{0,p_{n}+1}(\boldsymbol{\psi} )\} /2 }{\sin (p_{n}\mu_{2} )\alpha_{0,p_{n}}(\boldsymbol{\psi} )\!+\cos (p_{n}\mu_{2} )\lambda_{2} \{ \alpha_{0,p_{n}-1}(\boldsymbol{\psi} ) - \alpha_{0,p_{n}+1}(\boldsymbol{\psi} )\} /2 } \\ &&=\frac{\sin (p_{n}\mu_{1})\frac{\alpha_{0,p_{n}}(\boldsymbol{\psi} )}{\alpha_{0,p_{n}-1}(\boldsymbol{\psi} )-\alpha_{0,p_{n}+1}(\boldsymbol{\psi} )}+\frac{\cos (p_{n}\mu_{1} )\lambda_{1}}{2} }{\sin (p_{n}\mu_{2} )\frac{\alpha_{0,p_{n}}(\boldsymbol{\psi} )}{\alpha_{0,p_{n}-1}(\boldsymbol{\psi} )-\alpha_{0,p_{n}+1}(\boldsymbol{\psi} )}+\frac{\cos (p_{n}\mu_{2} )\lambda_{2}}{2} } \\ && \rightarrow\frac{\lambda_{1}}{\lambda_{2}}\ne 1\qquad (n\to\infty ). \end{array} $$

If λ₂ = 0, then $|\phi _{2}(p_{n}|\boldsymbol {\gamma }_{1})/\phi _{2}(p_{n}|\boldsymbol {\gamma }_{2})|\to \infty$ as $n\to \infty$.

Step 3. We consider two parameter vectors γ₁ and γ₂ with ψ₁ = ψ₂ = ψ, λ₁ = λ₂ = λ and μ₁≠μ₂. Let i be the imaginary unit defined as i² = − 1, and set $\phi _{3}(\boldsymbol {\gamma })=\alpha _{1}(\boldsymbol {\gamma })+\mathrm {i}\beta _{1}(\boldsymbol {\gamma })=\exp (\mathrm {i}\mu )\{ \alpha _{0,1}(\boldsymbol {\psi } )+\mathrm {i}\lambda M_{1}(\boldsymbol {\psi } )\}$ which is the polar form of the mean direction where M₁(ψ) = {α_0,0(ψ) − α_0,2(ψ)}/2. Then, we have

$$ \begin{array}{@{}rcl@{}} &\phi_{3}(\boldsymbol{\gamma}_{1})-\phi_{3}(\boldsymbol{\gamma}_{2})=\left\{ \exp (\mathrm{i}\mu_{1})-\exp (\mathrm{i}\mu_{2}) \right\}\{ \alpha_{0,1}(\boldsymbol{\psi} )+\mathrm{i}\lambda M_{1}(\boldsymbol{\psi} )\} . \end{array} $$

Because |α_0,1(ψ) + iλM₁(ψ)|≥|α_0,1(ψ)| > 0 and $\exp (\mathrm {i}\mu _{1})-\exp (\mathrm {i}\mu _{2})\ne 0$, we have ϕ₃(γ₁)≠ϕ₃(γ₂), which completes the proof. □

Proof Proof of Proposition 2.1.

Condition (i) is obvious. It follows from $\alpha _{0,p}(\rho )=E_{0,\rho }\{ {\cos \limits } (p{\varTheta } )\} =\rho ^{p}$ that

$$ \begin{array}{@{}rcl@{}} \frac{\alpha_{0,p-1}(\rho )-\alpha_{0,p+1}(\rho )}{\alpha_{0,p}(\rho )}=\frac{1}{\rho}-\rho . \end{array} $$

Thus, conditions (ii) and (iii) hold with c = 0, which completes the proof. □

To prove Proposition 2.2, we present a lemma on the modified Bessel function.

Lemma A.2.

The following results hold:

$$ \begin{array}{@{}rcl@{}} &&\frac{1}{p!}\left( \frac{\kappa}{2}\right)^{p}\leq I_{p}(\kappa )\leq \frac{1}{p!}\left( \frac{\kappa}{2}\right)^{p}\exp\left( \frac{\kappa^{2}}{4}\right) ,\textrm{ and }\\ &&\frac{A_{p}(\kappa )}{A_{p}(\kappa^{\prime})}=O\left( \left( \frac{\kappa}{\kappa^{\prime}}\right)^{p}\right) \textrm{ for }\kappa <\kappa^{\prime}\quad (p\to\infty ), \end{array} $$

(A.17)

where A_p(κ) = I_p(κ)/I₀(κ).

Proof.

Using the expansion in Mardia and Jupp (2000, p.349), we have

$$ \begin{array}{@{}rcl@{}} I_{p}(\kappa )&=&\sum\limits_{r=0}^{\infty}\frac{1}{(p+r)!r!}\left( \frac{\kappa}{2}\right)^{2r+p} \\ &=&\frac{1}{p!}\left( \frac{\kappa}{2}\right)^{p}\left\{ 1+\frac{1}{(p+1)1!}\left( \frac{\kappa}{2}\right)^{2}+\frac{1}{(p+2)(p+1)2!}\left( \frac{\kappa}{2}\right)^{4}+{\cdots} \right\} \\ &\leq& \frac{1}{p!}\left( \frac{\kappa}{2}\right)^{p}\left\{ 1+\frac{1}{1!}\left( \frac{\kappa^{2}}{4}\right) +\frac{1}{2!}\left( \frac{\kappa^{2}}{4}\right)^{2}+\frac{1}{3!}\left( \frac{\kappa^{2}}{4}\right)^{3}+{\cdots} \right\} \\ &=&\frac{1}{p!}\left( \frac{\kappa}{2}\right)^{p}\exp\left( \frac{\kappa^{2}}{4}\right) , \end{array} $$

(A.18)

where the last equation holds from the Maclaurin expansion for e^x. On the one hand, by Eq. A.18, we have the left side inequality in Eq. A.17. Next, we prove the latter. Let $\kappa <\kappa ^{\prime }$. Because

$$ \frac{A_{p}(\kappa )}{A_{p}(\kappa^{\prime})}=\frac{I_{0}(\kappa^{\prime})I_{p}(\kappa)}{I_{0}(\kappa )I_{p}(\kappa^{\prime})}, $$

applying the inequality (A.17) to I_p(κ) and $I_{p}(\kappa ^{\prime } )$ yields the result. □

Proof Proof of Proposition 2.2.

We verify conditions (i)–(iii) in Theorem 2.2 to demonstrate the identifiability of the SSvM distribution. Because the p th cosine-moment is given by $\alpha _{0,p}:=E_{0,\kappa }\{ {\cos \limits } (p{\varTheta } )\} =I_{p}(\kappa )/I_{0}(\kappa )$, condition (i) holds. In addition, we have

$$ \begin{array}{@{}rcl@{}} \frac{\alpha_{0,p-1}(\kappa )-\alpha_{0,p+1}(\kappa )}{\alpha_{0,p}(\kappa )}&=&\frac{I_{p-1}(\kappa )-I_{p+1}(\kappa )}{I_{p}(\kappa )}. \end{array} $$

(A.19)

By using the result I_ν− 1(z) − I_ν+ 1(z) = (2ν/z)I_ν(z), which is given in Abramowitz and Stegun (1972, p.376, 9.6.26), Eq. A.19 equals (2p)/κ. Therefore, conditions (ii) and Eq. 5 in (iii) hold with c = 1.

Finally, we verify Eq. 6 in condition (iii). It follows from Lemma A.2 that

$$ \begin{array}{@{}rcl@{}} \frac{\alpha_{0,p}(\kappa_{1})}{\alpha_{0,p}(\kappa_{2})}=\frac{I_{p}(\kappa_{1} )/I_{0}(\kappa_{1} )}{I_{p}(\kappa_{2})/I_{0}(\kappa_{2})}=O\left( \left( \frac{\kappa_{1}}{\kappa_{2}}\right)^{p}\right) . \end{array} $$

Thus, if κ₁ > κ₂, then $(\alpha _{0,p}(\kappa _{1})/\alpha _{0,p}(\kappa _{2}))p^{-c}\rightarrow \infty$ $(p\to \infty )$. By contrast, if κ₁ < κ₂, then $(\alpha _{0,p}(\kappa _{1})/\alpha _{0,p}(\kappa _{2}))p^{c}\rightarrow 0$ $(p\to \infty )$, which implies condition (iii). Hence, the result is proved. □

Proof of Proposition 2.3.

For different parameters γ_WS1 = (μ₁,ρ_α1, $\bar {\rho }_{1},\xi _{1})^{T}$ and $\boldsymbol {\gamma }_{\textsc {WS2}}=(\mu _{2},\rho _{\alpha 2},\bar {\rho }_{2},\xi _{2})^{T}$, we consider the following four steps: Step 1: ρ_α1≠ρ_α2, Step 2: ρ_α1 = ρ_α2 and $\bar {\rho }_{1}\ne \bar {\rho }_{2}$, Step 3: ρ_α1 = ρ_α2, $\bar {\rho }_{1}=\bar {\rho }_{2}$, and ξ₁≠ξ₂. Step 4: ρ_α1 = ρ_α2, $\bar {\rho }_{1}=\bar {\rho }_{2}$, ξ₁ = ξ₂, and μ₁≠μ₂.

First, we consider Step 1 with ρ_α1≠ρ_α2, and set ϕ₁(p|γ_WS) = α_p(γ_WS) which is the cosine moment. If ρ_α1 > ρ_α2, then ϕ₁(p|γ_WS1)/ϕ₁(p|γ_WS2) becomes

$$ \begin{array}{@{}rcl@{}} &&\frac{p\bar{\rho}_{1}\rho_{\alpha 1}^{p-1}(1-\rho_{\alpha 1}^{2})\cos (p\mu_{1} +\xi_{1} )+\rho_{\alpha 1}^{p}\cos (p\mu_{1})}{p\bar{\rho}_{2}\rho_{\alpha 2}^{p-1}(1-\rho_{\alpha 2}^{2})\cos (p\mu_{2} +\xi_{2} )+\rho_{\alpha 2}^{p}\cos (p\mu_{2})} \\ &=&\frac{p\bar{\rho}_{1}(\rho_{\alpha 1}/\rho_{\alpha 2})^{p-1}(1-\rho_{\alpha 1}^{2})\cos (p\mu_{1} +\xi_{1} )+(\rho_{\alpha 1}/\rho_{\alpha 2})^{p-1}\rho_{\alpha 1}\cos (p\mu_{1})/p}{\bar{\rho}_{2}(1-\rho_{\alpha 2}^{2})\cos (p\mu_{2} +\xi_{2} )+\rho_{\alpha 2}\cos (p\mu_{2})/p} \\ &\rightarrow& \infty \qquad (p\to \infty ). \end{array} $$

The case when ρ_α1 < ρ_α2 is proved similarly.

Next, we consider Step 2 with ρ_α1 = ρ_α2 = ρ_α and $\bar {\rho }_{1}\ne \bar {\rho }_{2}$, and set ϕ₂(p|γ_WS) to the squared p th mean resultant length, that is ϕ₂(p|γ_WS) = ρ_p(γ_WS)². When $\bar {\rho }_{1}>\bar {\rho }_{2}$, then

$$ \begin{array}{@{}rcl@{}} \frac{\phi_{2} (p|\boldsymbol{\gamma}_{\textsc{WS1}})}{\phi_{2} (p|\boldsymbol{\gamma}_{\textsc{WS2}})}&=&\frac{p^{2}\bar{\rho}_{1}^{2}\rho_{\alpha}^{2(p-1)}(1-\rho_{\alpha}^{2})^{2}+\rho_{\alpha}^{2p}+2p\bar{\rho}_{1}\rho_{\alpha}^{2(p-1)+1}(1-\rho_{\alpha}^{2})\cos \xi_{1}}{p^{2}\bar{\rho}_{2}^{2}\rho_{\alpha}^{2(p-1)}(1-\rho_{\alpha}^{2})^{2}+\rho_{\alpha}^{2p}+2p\bar{\rho}_{2}\rho_{\alpha}^{2(p-1)+1}(1-\rho_{\alpha}^{2})\cos \xi_{2}} \\ &=&\frac{\bar{\rho}_{1}^{2}(1-\rho_{\alpha}^{2})^{2}+\rho_{\alpha}^{2}/p^{2}+2\bar{\rho}_{1}\rho_{\alpha}(1-\rho_{\alpha}^{2})(\cos \xi_{1})/p}{\bar{\rho}_{2}^{2}(1-\rho_{\alpha}^{2})^{2}+\rho_{\alpha}^{2}/p^{2}+2\bar{\rho}_{2}\rho_{\alpha}(1-\rho_{\alpha}^{2}) (\cos \xi_{2})/p} \\ &\rightarrow& \frac{\bar{\rho}_{1}^{2}}{\bar{\rho}_{2}^{2}}>1 \qquad (p\to\infty ). \end{array} $$

The case when $\bar {\rho }_{1}<\bar {\rho }_{2}$ can also be verified similarly.

Next, we consider Step 3 with ρ_α1 = ρ_α2 = ρ_α, $\bar {\rho }_{1}=\bar {\rho }_{2}=\bar {\rho }$, ξ₁≠ξ₂. ϕ₃(p|γ_WS) is set to the characteristic function α_p(γ_WS) + iβ_p(γ_WS). For simplicity of notation, let $B_{p}=p\bar {\rho }\rho _{\alpha }^{p-1}(1-\rho _{\alpha }^{2})$. If ξ₁≠ξ₂, then we have

$$ \begin{array}{@{}rcl@{}} &&\frac{\alpha_{p}(\boldsymbol{\gamma}_{\textsc{WS1}})}{\alpha_{p}(\boldsymbol{\gamma}_{\textsc{WS2}})} \\ &=&\frac{\cos (p\mu_{1} + \xi_{1} ) + (\rho_{\alpha}^{p}/B_{p})\cos (p\mu_{1}) + \mathrm{i}\left\{ \sin (p\mu_{1} +\xi_{1} )+(\rho_{\alpha}^{p}/B_{p})\sin (p\mu_{1})\right\}}{\cos (p\mu_{2} + \xi_{2} ) + (\rho_{\alpha}^{p}/B_{p})\cos (p\mu_{2}) + \mathrm{i}\left\{ \sin (p\mu_{2} +\xi_{2} )+(\rho_{\alpha}^{p}/B_{p})\sin (p\mu_{2})\right\}}.\\ \end{array} $$

(A.20)

By Lemma A.1, there exists a sequence {p_n} of natural numbers such that $p_{n}\to \infty$ and |pμ_i(mod2π)|→ 0(i = 1,2) as $n\to \infty$. Hence, replacing p with p_n in Eq. A.20 and taking the limit leads to

$$ \begin{array}{@{}rcl@{}} \frac{\phi_{3} (p_{n}|\boldsymbol{\gamma}_{\textsc{WS1}})}{\phi_{3} (p_{n}|\boldsymbol{\gamma}_{\textsc{WS2}})} \rightarrow \frac{\cos (\xi_{1})+\mathrm{i}\sin (\xi_{1})}{\cos (\xi_{2})+\mathrm{i}\sin (\xi_{2})} \ne 1 ,\qquad (n\to \infty ), \end{array} $$

for which the condition is verified under Step 3.

Finally, we consider Step 4 with ρ_α1 = ρ_α2 = ρ_α, $\bar {\rho }_{1}=\bar {\rho }_{2}=\bar {\rho }$, ξ₁ = ξ₂ = ξ and μ₁≠μ₂. Here, we set ϕ₄(p|γ_WS) = ϕ₃(p|γ_WS). Then,

$$ \begin{array}{@{}rcl@{}} \phi_{4} (p|\boldsymbol{\gamma}_{\textsc{WS}})\!\!&=&\!B_{p}\cos (p\mu +\xi )+\rho_{\alpha}^{p}\cos (p\mu )+\mathrm{i}\left\{ B_{p}\sin (p\mu +\xi )+\rho_{\alpha}^{p}\sin (p\mu ) \right\} \\ \!&=&\!B_{p}\exp \{ \mathrm{i}(p\mu +\xi )\} +\rho_{\alpha}^{p}\exp (\mathrm{i}p\mu ). \end{array} $$

Hence, we have

$$ \begin{array}{@{}rcl@{}} &&\phi_{4} (p|\boldsymbol{\gamma}_{\textsc{WS1}})-\phi_{4} (p|\boldsymbol{\gamma}_{\textsc{WS2}}) \\ &=&B_{p}\left\{ \exp (\mathrm{i}(p\mu_{1}+\xi ))-\exp (\mathrm{i}(p\mu_{2}+\xi ))\right\} +\rho_{\alpha}^{p}\left( \exp (\mathrm{i}p\mu_{1})-\exp (\mathrm{i}p\mu_{2})\right) \\ &=&\left( B_{p}\exp (\mathrm{i}\xi)+\rho_{\alpha}^{p}\right) \left\{ \exp (\mathrm{i}p\mu_{1})-\exp (\mathrm{i}p\mu_{2})\right\} . \end{array} $$

Because $|B_{p}\exp (\mathrm {i}\xi )+\rho _{\alpha }^{p}|^{2}={B_{p}^{2}}+2B_{p}\rho _{\alpha }^{p}{\cos \limits } \xi +\rho _{\alpha }^{2p}\geq (B_{p}-\rho _{\alpha }^{p})^{2}$ and $B_{p}-\rho _{\alpha }^{p}>0$ for some large p, we have ϕ₄(p|γ_WS1) − ϕ₄(p|γ_WS2)≠ 0 for some large p, which the condition is verified under Step 4. Therefore, the family of distributions (8) is identifiable under Γ_WS. □

Proof of Proposition 2.4.

Because the marginal distribution of Θ is the SSWC, from Proposition 2.1 and Remark 2.2, it is identifiable with respect to the parameter vector (μ,κ,λ)^T. Hence, we begin with the case of (μ₁,κ₁,λ₁)^T = (μ₂,κ₂,λ₂)^T(= (μ,κ,λ)^T). Set $g(\alpha ,\beta )=\beta (1-\tanh (\kappa )\cos \limits (\theta -\mu ))^{1/\alpha }$. It follows from the identifiability of Weibull distribution that f_AL(x|𝜃;α₁,β₁,κ) = f_AL(x|𝜃;α₂,β₂,κ) implies α₁ = α₂ and g(α₁,β₁) = g(α₂,β₂), which leads to β₁ = β₂. This means that for any (α₁,β₁)^T≠(α₂,β₂)^T, f_AL(x|𝜃;α₁,β₁,κ)≠f_AL(x|𝜃;α₂,β₂,κ), which completes the proof. □

Proof of Proposition 2.5.

Because the marginal distribution for Θ in Eq. 9 is the SSWC distribution, the identifiability with respect to the parameter vector (μ,κ,λ)^T holds from Proposition 2.1 and Remark 2.2. The conditional distribution of the linear part of Eq. 9 is given in Eq. 11. We need to check the case with (μ₁,κ₁)^T = (μ₂,κ₂)^T = (μ,κ)^T and δ₁≠δ₂. First, we assume δ₂ > δ₁. In this case, the ratio on the conditional distribution of the linear part becomes

$$ \begin{array}{@{}rcl@{}} &&\frac{f_{X|{\varTheta}}(x | \theta; \delta_{1}, \tau_{1}, \sigma_{1})}{f_{X|{\varTheta}}(x | \theta; \delta_{2}, \tau_{2}, \sigma_{2})}\\ &=& \frac{ \frac{1}{\sigma_{1} \delta_{1}} \!\left( \!\frac{x}{\sigma_{1}}\!\right)^{1/\delta_{1} -1} \!\left\{ 1 - \kappa \cos(\theta - \mu) \right\} \!\!\left[1 \!+ \frac{\tau_{1}}{\delta_{1}} \!\left( \frac{x}{\sigma_{1}}\right)^{1/\delta_{1}} \{1 - \kappa \cos(\theta - \mu) \}\right]^{-(\delta_{1}/\tau_{1} + 1)} }{ \frac{1}{\sigma_{2} \delta_{2}} \!\left( \!\frac{x}{\sigma_{2}}\!\right)^{1/\delta_{2} -1} \!\left\{ 1 - \kappa \cos(\theta - \mu) \right\} \!\!\left[1 \!+ \frac{\tau_{2}}{\delta_{2}} \!\left( \frac{x}{\sigma_{2}}\right)^{1/\delta_{2}} \{1 - \kappa \cos(\theta - \mu) \}\right]^{-(\delta_{2}/\tau_{2} + 1)} } \\ &=& C_{1} x^{1/\delta_{1} - 1/\delta_{2}} \frac{ \left[1 + \frac{\tau_{2}}{\delta_{2}} \left( \frac{x}{\sigma_{2}}\right)^{1/\delta_{2}} \{1 - \kappa \cos(\theta - \mu) \}\right]^{(\delta_{2}/\tau_{2} + 1)} } {\left[1 + \frac{\tau_{1}}{\delta_{1}} \left( \frac{x}{\sigma_{1}}\right)^{1/\delta_{1}} \{1 - \kappa \cos(\theta - \mu) \}\right]^{(\delta_{1}/\tau_{1} + 1)}} =: K_{1}, \end{array} $$

where $C_{1} = \sigma _{2}^{1/\delta _{2}}\delta _{2} / (\sigma _{1}^{1/\delta _{1}}\delta _{1} )$. Since 1/δ₁ − 1/δ₂ > 0, we observe that K₁ → 0 as x → 0. We can similarly observe that for the case with δ₂ < δ₁, $K_{1} \to \infty$ as x → 0. For the next step, we consider the case with (μ₁,κ₁,δ₁)^T = (μ₂,κ₂,δ₂)^T = (μ,κ,δ)^T and τ₁≠τ₂. To simplify expressions, we let 1/τ₂ − 1/τ₁ = Δ_τ and $C_{2}(\theta )=1 - \kappa \cos \limits (\theta - \mu )$. From the expression K₁ and the fact that (δ/τ₂) + 1 = (δ/τ₁) + 1 + Δ_τδ, we see that

$$ \begin{array}{@{}rcl@{}} && \frac{f_{X|{\varTheta}}(x | \theta; \delta_{1}, \tau_{1}, \sigma_{1})}{f_{X|{\varTheta}}(x | \theta; \delta_{2}, \tau_{2}, \sigma_{2} )} = C_{1} \frac{ \left[1 + \frac{\tau_{2}}{\delta} \left( \frac{x}{\sigma_{2}}\right)^{1/\delta} C_{2}(\theta)\right]^{(\delta/\tau_{2} + 1)} } {\left[1 + \frac{\tau_{1}}{\delta} \left( \frac{x}{\sigma_{1}}\right)^{1/\delta} C_{2}(\theta)\right]^{(\delta/\tau_{1} + 1)}} \\ &=& C_{1} \left\{ \frac{ 1 + \frac{\tau_{2}}{\delta} \left( \frac{x}{\sigma_{2}}\right)^{1/\delta}C_{2}(\theta) } { 1 + \frac{\tau_{1}}{\delta} \left( \frac{x}{\sigma_{1}}\right)^{1/\delta}C_{2}(\theta) } \right\}^{\delta/\tau_{1}+1} \left[ 1+ \frac{\tau_{1}}{\delta}\left( \frac{x}{\sigma_{2}} \right)^{1/\delta} C_{2}(\theta) \right]^{{\Delta}_{\tau}}\\ &=:& C_{1}\times K_{2} \times K_{3} . \end{array} $$

We see that K₂ = O(1) as $x\to \infty$. For τ₂ < τ₁ which indicates Δ_τ > 0, we have $K_{3} \to \infty$ as $x \to \infty$. On the contrary, when τ₂ > τ₁ which indicates Δ_τ < 0, we have K₃ → 0 as $x \to \infty$. For the final step, we consider the case with (μ₁,κ₁,δ₁,τ₁)^T = (μ₂,κ₂,δ₂,τ₂)^T = (μ,κ,δ,τ)^T and σ₁≠σ₂. We observe that

$$ \begin{array}{@{}rcl@{}} \frac{f_{X|{\varTheta}}(x | \theta; \delta, \tau, \sigma_{1})}{f_{X|{\varTheta}}(x | \theta; \delta, \tau, \sigma_{2})} = \frac{\sigma_{2}^{1/\delta}\delta }{\sigma_{1}^{1/\delta}\delta } \frac{ \left[1 + \frac{\tau}{\delta} \left( \frac{x}{\sigma_{2}}\right)^{1/\delta} C_{2}(\theta)\right]^{(\delta/\tau + 1)} } {\left[1 + \frac{\tau}{\delta} \left( \frac{x}{\sigma_{1}}\right)^{1/\delta} C_{2}(\theta)\right]^{(\delta/\tau + 1)}} . \end{array} $$

Evaluating the ratio of the conditional distributions at x = 0, we have

$$ \begin{array}{@{}rcl@{}} \frac{f_{X|{\varTheta}}(0 | \theta; \delta, \tau, \sigma_{1})}{f_{X|{\varTheta}}(0| \theta; \delta, \tau, \sigma_{2} ) } = \frac{\sigma_{2}^{1/\delta}}{\sigma_{1}^{1/\delta}} \ne 1, \end{array} $$

which completes the proof of Proposition 2.5. □

Appendix B: Derivation of p th Cosine and Sine Moments

In this section, we describe how to represent p th cosine and sine moments in a circular model with location μ by using p th cosine and sine moments in the circular model with μ = 0. We consider the density Eq. 2 with μ = 0,

$$ \begin{array}{@{}rcl@{}} f^{*}(\theta |\boldsymbol{\gamma}^{*})&:=&f_{0}(\theta |\boldsymbol{\psi} )\left\{ 1+\lambda \sin (\theta )\right\} , \end{array} $$

where γ^∗ = (ψ^T,λ)^T. In addition, we denote its p th cosine and sine moments as $\alpha _{p}^{*}(\boldsymbol {\gamma }^{*}):=E^{*}\{{\cos \limits } (p{\varTheta })\}$ and $\beta _{p}^{*}(\boldsymbol {\gamma }^{*}):=E^{*}\{{\sin \limits } (p{\varTheta })\}$, $(p\in \mathbb {Z})$. Because the density Eq. 2 is written as f(𝜃|γ) = f^∗(𝜃 − μ|γ^∗), the p th cosine moment in the density Eq. 2 is given by

$$ \begin{array}{@{}rcl@{}} \alpha_{p}(\boldsymbol{\gamma})&=&{\int}_{\mu-\pi}^{\mu+\pi}\cos (p\theta )f^{*}(\theta -\mu |\boldsymbol{\gamma}^{*}) d\theta . \end{array} $$

(A.21)

Letting 𝜃 − μ = η, Eq. A.21 becomes

$$ \begin{array}{@{}rcl@{}} &&{\int}_{-\pi}^{\pi}\cos (p(\mu +\eta ))f^{*}(\eta |\boldsymbol{\gamma}^{*})d\eta \\ &=&{\int}_{-\pi}^{\pi}\left\{ \cos (p\mu )\cos(p\eta )-\sin (p\mu )\sin (p\eta )\right\}f^{*}(\eta |\boldsymbol{\gamma}^{*}) d\eta \\ &=&\cos (p\mu )E^{*}\left\{ \cos (p\eta )\right\} -\sin (p\mu)E^{*}\left\{ \sin (p\eta )\right\} \\ &=&\cos (p\mu )\alpha_{p}^{*}(\boldsymbol{\gamma}^{*})-\sin (p\mu )\beta_{p}^{*}(\boldsymbol{\gamma}^{*}). \end{array} $$

Similarly, the p th sine moment is

$$ \begin{array}{@{}rcl@{}} \beta_{p}(\boldsymbol{\gamma})&=&{\int}_{-\pi}^{\pi}\sin (p(\mu +\eta ))f^{*}(\eta |\boldsymbol{\gamma}^{*})d\eta \\ &=&{\int}_{-\pi}^{\pi}\left\{ \sin (p\mu )\cos (p\eta )+\cos (p\mu )\sin (p\eta )\right\} f^{*}(\eta |\boldsymbol{\gamma}^{*})d\eta \\ &=&\sin (p\mu )E^{*}\left\{ \cos (p\eta )\right\} +\cos (p\mu)E^{*}\left\{ \sin (p\eta )\right\} \\ &=&\sin (p\mu )\alpha_{p}^{*}(\boldsymbol{\gamma}^{*}) +\cos (p\mu)\beta_{p}^{*}(\boldsymbol{\gamma}^{*}). \end{array} $$

This result shows that if we know the p th cosine and sine moments when μ = 0, we can automatically derive the p th cosine and sine moments of the circular model with location μ.