Aleshin-Guendel 2020 - Mixed-Moments Theorem

Summary

Cell probabilities of any $M_{th}$ model depend on the mixing distribution only through its mixed moments (Lemma 1), and equality of conditional cell probabilities is equivalent to the moment vectors being proportional (Theorem 1). So identifiability can be checked by asking whether proportional moments force equal missing-cell probabilities.

Overview

The theorem converts Definition 1 (a statement about probabilities) into an algebraic statement about moment vectors, which is easier to attack by linear algebra.

Main Content

Lemma 1 (cell probabilities are mixtures of moments)

For $h\in H^{\ast }$,

$${\pi }_{Q,h}=\sum _{h^{\prime }\in H^{\ast }}{c}_{h,h^{\prime }}{m}_{Q,h^{\prime }},c_{h,h^{\prime }}=(-1{)}^{{\sum }_k{h}_k^{\prime }-h_k}\prod _{k=1}^{K}I(h_k\le h_k^{\prime }),m_{Q,h^{\prime }}=E_Q(\prod _{k=1}^K{\lambda }_k^{h_k^{\prime }}).$$

Proof. By the multibinomial theorem ${\prod }_k{\lambda }_k^{h_k}(1-{\lambda }_k{)}^{1-h_k}={\sum }_{h^{\prime }\in H^{\ast }}{c}_{h,h^{\prime }}{\prod }_k{\lambda }_k^{h_k^{\prime }}$ (the empty-product term cancels for $h\ne 0$). Take expectations under $Q$.

In matrix form ${\mathbf{\pi }}_Q^{\ast }=C{\mathbf{m}}_Q$, where ${\mathbf{\pi }}_Q^{\ast }=({\pi }_{Q,h}{)}_{h\in H^{\ast }}$, ${\mathbf{m}}_Q=(m_{Q,h}{)}_{h\in H^{\ast }}$ and $C=(c_{h,h^{\prime }})$. $C$ is upper triangular (patterns ordered as binary digits) with non-zero diagonal, hence invertible.

Theorem 1

For any two distributions $Q,R$ on $(0,1{]}^K$, ${\tilde{\pi }}_Q={\tilde{\pi }}_R$ is equivalent to ${\mathbf{m}}_Q=A{\mathbf{m}}_R$ for some $A>0$. Proof. ${\tilde{\pi }}_Q={\tilde{\pi }}_R⟺{\mathbf{\pi }}_Q^{\ast }/(1-{\pi }_{Q,0})={\mathbf{\pi }}_R^{\ast }/(1-{\pi }_{R,0})⟺{\mathbf{\pi }}_Q^{\ast }=A{\mathbf{\pi }}_R^{\ast }$ with $A=(1-{\pi }_{Q,0})/(1-{\pi }_{R,0})>0$. By Lemma 1 this is $C{\mathbf{m}}_Q=AC{\mathbf{m}}_R$, and invertibility of $C$ gives ${\mathbf{m}}_Q=A{\mathbf{m}}_R$.

Consequence

To show a family $\mathcal{Q}$ is identifiable: show that ${\mathbf{m}}_Q=A{\mathbf{m}}_R$ (some $Q,R\in \mathcal{Q}$, $A>0$) implies ${\pi }_{Q,0}={\pi }_{R,0}$. Equivalent to $A=1$, because $A=(1-{\pi }_{Q,0})/(1-{\pi }_{R,0})$. To show non-identifiability: construct $Q,R$ with ${\mathbf{m}}_Q=A{\mathbf{m}}_R$ and $A\ne 1$.

Examples

$K=2$ check of Lemma 1

$H^{\ast }=\{01,10,11\}$. ${\pi }_{Q,11}=E[{\lambda }_1{\lambda }_2]=m_{11}$; ${\pi }_{Q,10}=E[{\lambda }_1(1-{\lambda }_2)]=m_{10}-m_{11}$; ${\pi }_{Q,01}=m_{01}-m_{11}$. The sign pattern $(-1{)}^{\sum h^{\prime }-\sum h}$ matches. Then ${\pi }_{Q,0}=1-m_{10}-m_{01}+m_{11}$.

import numpy as np, itertools
K = 2
H = [h for h in itertools.product([0,1], repeat=K) if any(h)]
C = np.array([[ (-1)**(sum(hp)-sum(h)) * all(h[k] <= hp[k] for k in range(K)) for hp in H] for h in H], float)
print(C, np.linalg.det(C))   # triangular, det = +-1

Connections

• The multiplicative factor $A$ is exactly the ratio of observed-mass $(1-{\pi }_0)$ between the two models; it is the quantity a conditional likelihood cannot pin down.
• Same logic as the Poisson non-identifiability of $N$ in Barker 2018 - Poisson Alternative and Model Discrimination: only a product (here $A\cdot$moments) is estimable.

See Also

• Aleshin-Guendel 2020 - Latent Class Identifiability Theorem and LCMCR
• Aleshin-Guendel 2020 - Mth Model and Identifiability Definition