ISSN 02786419, Moscow University Computational Mathematics and Cybernetics, 2011, Vol. 35, No. 4, pp. 176–183. © Allerton Press, Inc., 2011. Original Russian Text © A.K. Gorshenin, 2011, published in Vestnik Moskovskogo Universiteta. Vychislitel’naya Matematika i Kibernetika, 2011, No. 4, pp. 26–32.
Testing of Statistical Hypotheses in the Splitting Component Model A. K. Gorshenina, b a
Faculty of Computational Mathematics and Cybernetics, Moscow State University, Moscow, 119991 Russia b Institute of Informatics Problems, Russian Academy of Sciences, Moscow, 119333 Russia email:
[email protected] Received April 11, 2011
Abstract—This paper considers the problem of testing the statistical significance of parametrically close components of the mixture of probabilistic distribution. To solve the problem, the author pre sents the asymptotically most powerful test and finds the limiting distributions, power loss, and asymp totic deficiency. The application of a given test for mixtures of normal and gamma distributions is detailed in the present paper. Keywords: mixtures of probabilistic distributions, asymptotically most powerful test, power loss, asymptotic deficiency. DOI: 10.3103/S0278641911040054
1. INTRODUCTION The authors of [1] proposed the asymptotically powerful test for testing the hypotheses of the number of components in a model of the following form (parameter θ ∈ [0, 1]) k
k
p ( x, θ ) = ( 1 – θ )
∑
p i ψ i ( x ) + θψ k + 1 ( x ),
∑p
i
= 1.
(1)
i=1
i=1
This model is aimed at testing the importance of an arbitrary component that is possibly low in weight. However, in solving the problem of reducing the number of components in the fitting model of mixture of probabilistic distributions, we should not exclude practically important components by their mistaken integration into a single component. The presence of components with close values of parameters, includ ing weight, is therefore possible. Note that a majority of bestknown algorithms (EM, SEM, and MCEM algorithms and their var ious modifications) use a prescribed number of components for the statistical decomposition of mix tures and cannot change it during the iterative procedure. In applying such algorithms, however, the problem of distinguishing the parametrically close components of a mixture arises. This is because the EM algorithms do not always distinguish the given components, but integrate them into a single component. This lack of such algorithms appears in testing on samples of mixtures with a known dis tribution [2]. This work suggests an asymptotically optimal test for testing hypotheses of the number of components of a mixture of probabilistic distributions in terms of the maximization of the limit of power test, allowing us to solve the problem of decomposition of “combined” components. The approach extensively investi gated in [1] makes it possible to find the asymptotically most powerful test and to study its properties. The present paper refers to the development of the prior results with allowance for all additional features that emerge due to the substituting the model of component addition (1) for the subsequent splitting compo nent model. 176
TESTING OF STATISTICAL HYPOTHESES
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2. FORMULATION OF THE PROBLEM Assume that each independent observation Xn = (X1, …, Xn) has density in the form of a finite Kcom ponent mixture of densities of some distribution laws K
K
∑
p i ψ i ( x ),
∑p
i
= 1,
p i ≥ 0,
i=1
i=1
where ψi(x), i = 1, …, K, is the density corresponding to the same kernel. It is universally supposed that the mixture is identifiable (if, for the certain distributions, additional conditions are required, they are separately indicated). Let k be some known natural number. We must then test the hypothesis H0 : K = k against the alternative H1 : K = k + 1. In other words, we have to test the significance of (k + 1) component (for example, if the weights pi, i = 1, …, k + 1 are in descending order). This is quite typical when we must be sure of the importance of several components with close parameters, or combine them without signif icant loss of the model’s information value. The answer to this question is of great importance for socalled grid methods of mixture decomposition [3, 4]. To simplify the asymptotic analysis of proposed criteria, let us reduce the above problem of testing the hypotheses of discrete parameter K to the problem of the hypotheses of continuous parameter. To do this, we assume that for some θ ∈ [0, 1], X1 has the following density: k–1
p ( x, θ ) =
∑ p ψ (x) + (p i
i
k
– θ )ψ k ( x ) + θψ ( x ) = f ( x ) + θg ( x ),
i=1
f(x) =
(2)
k
k
∑ p ψ ( x ), ∑ p i
i
i
= 1,
p i ≥ 0, g ( x ) = ψ ( x ) – ψ k ( x ),
0 ≤ θ ≤ pk ;
i=1
i=1
function ψ(x) is the density of the same family of distributions as all ψi(x). Note that, as opposed to the case of component addition, function g(x) is not the density of some distribution as a whole; we therefore cannot apply the results of [1] to the case of component decomposition in an explicit form. We must test the simple hypothesis H0 against the sequence of composite alternatives Hn, 1 in the fol lowing form (since, in general, there is no the uniformly most powerful test for testing the simple hypoth esis against the complicated hypothesis): H 0 : θ = 0,
t H n, 1 : θ = , n
0 < t ≤ C,
C > 0,
where t is an unknown parameter. This in fact tests the hypothesis of whether the given mixture is the kcomponent mixture (at the validity of null hypothesis H0) or the (k + 1)component mixture (at the validity of alternative Hn, 1). 3. THE ASYMPTOTICALLY MOST POWERFUL TEST OF THE VERIFICATION OF HYPOTHESES OF COMPONENT NUMBER OF MIXTURE Let us use the asymptotical approach described, for example, in [5]. According to the Neyman–Pear son lemma, the best test for testing hypothesis H0 against the simple alternative Hn, 1 for any fixed t ∈ (0, C] is based on the logarithm of the likelihood ratio: n
Λn ( t ) =
∑ ( l ( X , tn i
– 1/2
) – l ( X i, 0 ) ),
l ( x, θ ) = log p ( x, θ ).
(3)
i=1
We denote the power of such a test of level α ∈ (0, 1) by β *n ( t ) . Although statistics Λ n ( t ) cannot be used to construct a test for testing the hypothesis H0 against the alternative Hn, 1 due to the unknown parameter t, β *n ( t ) specifies the limit for the power of any test in testing hypothesis H0 against the fixed alternative Hn, 1, t > 0. MOSCOW UNIVERSITY COMPUTATIONAL MATHEMATICS AND CYBERNETICS
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Below, we use the following functions ( j ≥ 1): n
j (j) j+1 1 g ( X i ) ∂ l ( x, θ ) ( x)⎞ j , L ( 1 ) = ⎛ g l ( x ) = = ( – 1 ) ( j – 1 )! . (4) n j ⎝ f(x)⎠ n i = 1 f ( Xi ) ∂θ θ=0 Consider the sufficient conditions at which the logarithm of probability relation is asymptotically nor mal. This is socalled condition of local asymptotic normality, i.e., the possibility of presenting the loga rithm of probability relation Λn(t) in the form
∑
2
(1) t I (5) Λ n ( t ) = tL n – + ξ n ( t ), 2 where the remainder ξn(t) 0 by the probability at hypothesis H0 at n ∞ and Fisher information I = ⺕0(l (1)(X1))2. It should be mentioned that the authors of [1] proved the possibility of writing the logarithm 2
(1) t I of probability relation in the form Λ n ( t ) = tL n – + η n ( t ) for the function ηn(t). 2
∫
Lemma. Let the integral
∫
∞
2
g (x)(
–∞
∑
–1 k–1 p ψ ( x ) ) dx i=1 i i
∞
–1
2
ψ ( x )ψ 1 ( x ) dx be finite at k = 1 in Eq. (2) and the integral
–∞
be finite at k ≥ 2. Then Eq. (5) is fulfilled for the density p(x, θ) from (2).
Proof. We denote the required right neighbourhood of the point θ = 0 by δ. Below, we assume that con dition 0 ≤ θ < δ is fulfilled. A. Clearly, the linear function is absolutely continuous since, if
∑ (b – a ) < δ i
i
i
1,
where (ai, bi) is the
arbitrary system of pairwise crossed intervals, a and b are the finite fixed numbers for the arbitrary linear function y(x) = ax + b. We therefore obtain
∑ y(b ) – y(a ) i
i
= a
i
∑ (b – a ) < a δ i
i
= ε
1
i
at the corresponding selection of δ1. Density p(x, θ) is the linear function by θ at each fixed x ∈ ⺢, and is thus absolutely continuous in terms of θ from the right δ neighbourhood of zero. B. Let us find the derivative ∂p ( x, θ) = ∂ ( f ( x ) + θg ( x )) = g ( x ). ∂θ ∂θ Clearly, the given derivative exists at all x ∈ ⺢ the (Lebesgue measure) for any θ from the right δ neigh bourhood of zero. C. Let us establish the continuity of the function 2 ∂ log p ( x, θ ) 2 g(x) I ( θ ) = ⺕ θ ⎛ ⎞ = ⺕ θ ⎛ ⎞ = ⎝ ⎠ ⎝ f ( x ) + θg ( x )⎠ ∂θ
∞
∫
–∞
2
g (x) dx. f ( x ) + θg ( x )
The function g(x) ≠ 0 almost surely, and by the wellknown property of Lebesgue integral, this means that the function I(θ) > 0 (due to the nonnegativity condition of the integrand the condition I(θ) ≠ 0 is equivalent toI(θ) > 0) almost surely for any θ from the right δneighbourhood of zero (including the value θ = 0). Let us first assume that k = 1 and integration element
∫
∞
2
–1
ψ ( x )ψ 1 ( x ) dx < ∞. Since 0 ≤ θ < δ, we write our estimate for the
–∞
2
g (x) 1 ( ψ ( x ) – 2ψ ( x ) + ψ 2 ( x )ψ –1 ( x ) ). ≤ 1 1 ( 1 – θ )ψ 1 ( x ) + θψ ( x ) ( 1 – δ ) Hence, ∞
⎞ –1 2 1 ⎛ I ( θ ) ≤ ⎜ ψ ( x )ψ 1 ( x ) dx – 1⎟ , (1 – δ)⎝ ⎠
∫
–∞
which implies the finiteness of I(θ) in terms of the lemma. MOSCOW UNIVERSITY COMPUTATIONAL MATHEMATICS AND CYBERNETICS
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We then assume that k ≥ 2 and
∫
∞ –∞
2
g (x)(
∑
–1 k–1 p ψ ( x ) ) dx i=1 i i
179
< ∞. Therefore (with allowance for (2)), –1
2 ⎛k – 1 ⎞ 2 g (x) ≤ g ( x ) ⎜ p i ψ i ( x )⎟ , f ( x ) + θg ( x ) ⎝ ⎠
∑
i=1
which also implies the finiteness of I(θ). Using the Lebesgue theorem of dominated convergence [7], we obtain 2
g (x) lim dx = θ → θ 0 f ( x ) + θg ( x )
∫
∫
2
g (x) dx. f ( x ) + θ0 g ( x )
This relation means the continuity of function I(θ) in the right δ neighbourhood of zero (including the value θ = 0). According to the known results in [6], fulfillment of the given conditions demonstrates the validity of (5). The lemma is proved. To distinguish the hypothesis of the number of components, let us consider a test based on the statistics (1)
L n from (4) as a test with the ultimate power β* ( t ) = Φ ( t I – u α ), where Φ(uα) = 1 – α and Φ(·) is the function of a standard normal distribution. The lemma implies that (1)
(1)
test L n is asymtotically most powerful. According to the central limit theorem, L n at n
∞ has a nor (1)
mal distribution with parameters 0 and I (at the validity of the zero hypothesis). The critical value c n can then be found from the following relations: (1)
(1)
(1)
⺠ n, 0 ( L n > c n ) = α, c n =
Iu α + o ( 1 ),
where ⺠n, 0 denotes the distribution Xn at θ = 0. 3.1. Asymptotic Behavior of the Power Difference It was shown in [5] that the normalized limit of the power difference (the socalled power loss) for a test based on statistics (4) has the form 3
(2) –1 2 ( 1 ) (2) t r ( t ) = ϕ ( u α – t I ) [ ⺔ 0 l ( X 1 ) – I ⺕ 0 l ( X 1 )l ( X 1 ) ], 8 I where ϕ(·) is the density of a standard normal distribution. –1
Let us denote the moments on the order of s of random variable ξ = g ( X 1 )f ( X 1 ) : +∞ s
Ψ s = ⺕ 0 ξ = ⺕ 0 ( g ( X 1 )f ( X 1 ) ) = –1
∫ g ( x )f s
1–s
( x ) dx,
s = 2, 3, 4.
(6)
–∞
It is obvious that I = Ψ2 ,
(1)
(2)
(2)
⺕ 0 l ( X 1 )l ( X 1 ) = – Ψ 3 , ⺔ 0 l ( X 1 ) = Ψ 4 – Ψ 2 , 2
(7)
2
3 Ψ 2 t r ( t ) = ϕ ( u α – t Ψ 2 ) ⎛ Ψ 4 – Ψ 2 – 3 ⎞ . ⎝ Ψ 2⎠ 8 Ψ2
Allowing for quantity r(t) from (7), we can find the asymptotic deficiency (which is similar to the state ments in [1]): 3
2 Ψ 2 t 2r ( t ) d = lim d n ≡ lim ( k n – n ) = = ⎛ Ψ 4 – Ψ 2 – 2 ⎞ . ⎝ n→∞ n→∞ 4Ψ 2 Ψ2 ⎠ t Ψ2 ϕ ( t Ψ2 – uα )
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Here, dn is a deficiency, and kn is the number of observations that the test based on the statistics L n
from (4) requires to attain the same power as the test based on statistics Λn(t) from (3) at alternatives t/ n . The first equality in (8) implies that if the limit exists and is finite, it is an asymptotic deficiency. 3.2. Conditions for the Convergence of Moment Characteristics Ψs Let us consider the conditions that guarantee the finiteness of moment characteristics (6) for some spe cial cases of distribution mixtures. In this section, we use the inequalities (u = 2, 3, a > 0, b ≥ 0) ⎛ ⎜ ⎝
∑ i
⎞ a i⎟ ⎠
–u
⎛ ≤⎜ ⎝
–1
∑ i
u⎞ ai ⎟ , ⎠
1 1 ≤ . a+b a
(9)
Let us write moment characteristics (6) in detail –1
+∞
∫ (ψ (x) + 2
Ψ2 =
2 ψk ( x )
–∞
⎛ k ⎞ – 2ψ ( x )ψ k ( x ) ) ⎜ p i ψ i ( x )⎟ dx, ⎝i = 1 ⎠
∑
+∞
Ψ3 =
∫ (ψ (x) – 3
3 ψk ( x )
2
– 3ψ ( x )ψ k ( x ) +
⎛ 2 3ψ ( x )ψ k ( x ) ) ⎜
–∞ +∞
Ψ4 =
∫ (ψ (x) + 4
4 ψk ( x )
+
2 2 6ψ k ( x )ψ ( x )
–2
⎞ p i ψ i ( x )⎟ dx, ⎝i = 1 ⎠
3
– 4ψ ( x )ψ k ( x ) –
k
∑
⎛ 3 4ψ ( x )ψ k ( x ) ) ⎜
–3
⎞ p i ψ i ( x )⎟ dx. ⎝i = 1 ⎠
–∞
k
∑
Note that we must fulfill the following conditions in terms of the lemma: ∞
∫ψ
∞ 2
–1 ( x )ψ 1 ( x ) dx
< ∞,
∫
k = 1,
–∞
–∞
–1
⎛k – 1 ⎞ g ( x ) ⎜ p i ψ i ( x )⎟ dx < ∞, ⎝i = 1 ⎠
∑
2
k ≥ 2,
which obviously guarantee the finiteness of the moment characteristic Ψ2. We therefore take into account the conditions of the convergence of the given integrals in considering the certain distributions. In the case of k ≥ 2, these conditions can be found with allowance for the inequality ⎛k – 1 ⎞ g ( x ) ⎜ p i ψ i ( x )⎟ ⎝i = 1 ⎠ 2
–1
∑
2
≤ (ψ +
2 ⎛ ψk ) ⎜
–1
⎞ p i ψ i ( x )⎟ . ⎝i = 1 ⎠ k–1
∑
The convergence of the integral from the right member of the inequality results in the convergence of the left member. Futhermore, it is clear from (9) that ⎛ k ⎞ ⎜ p i ψ i ( x )⎟ ⎝i = 1 ⎠
∑
–u –u
≤ Cψ k ( x ),
C > 0.
The integrals from ψ(x) and ψk(x) are finite since the given functions are the densities of some distri butions. The problem of the finiteness of moment characteristics (6) is thus solved by using the conditions of the convergence of the following integrals: ∞
∫ ψ ( x )ψ u
–u + 1 ( x ) dx, k
2 ≤ u ≤ 4,
–∞
which are considered in [1] for the normal and gamma distributions. MOSCOW UNIVERSITY COMPUTATIONAL MATHEMATICS AND CYBERNETICS
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Normal Distribution From the above, we can derive the following sufficient conditions for the convergence of the moment characteristics in model (2) for the mixture of gamma distributions: 2 2 2 2 2 2 (10) σ < 4 σ k , k ≥ 1, σ < 2 max σ j , σ k < 2 max σ j , k ≥ 2. 3 1≤j≤k–1 1≤j≤k–1 Gamma Distribution In this case, we can derive the following sufficient conditions for convergence of the moment charac teristics in the model (2) for mixtures of gammadistributors: ⎧1 ⎫ 1 β ≥ max ⎨ ( 3β 1 + 1 ), ( β 1 + 1 ) ⎬, 2 ⎩4 ⎭
3 α > α1 , 4
k = 1,
⎧ ⎫ β ≥ max ⎨ 1 ( 3β k + 1 ), 1 ( β k + 1 ), 1 min ( β i + 1 ) ⎬, 2 21 ≤ i ≤ k – 1 ⎩4 ⎭
(11)
⎧1 3 ⎫ α > max ⎨ min α j, α k ⎬, 2 ⎩ 1≤j≤k–1 4 ⎭ 1 1 β k ≥ min ( β i + 1 ), α k > min α j , k ≥ 2. 21 ≤ i ≤ k – 1 21 ≤ j ≤ k – 1 Our next theorem thus follows from the above: Theorem. Let the sufficient conditions of the finiteness of moment characteristics Ψs, where s = 2, 3, 4, be fulfilled from (6). The mixture in (2) is identifiable. For the splitting component model, the test for testing the hypothesis of a k component mixture against the alternative of (k + 1) component in terms of statistics (1) Ln
n
1 g ( X i ) = n i = 1 f ( Xi )
∑
then has the following properties: (i) At the validity of the zero hypothesis, these statistics have a normal distribution with parameters 0 and Ψ2 at n ∞ (1)
ᑦ ( Ln H0 )
N ( 0, Ψ 2 ).
(ii) At the validity of the alternative, these statistics have a normal distribution with parameters tΨ2 and Ψ2 at n ∞ (1)
ᑦ ( L n H n, 1 )
N ( tΨ 2, Ψ 2 ).
(iii) The given test is asymptotically more powerful with the ultimate power (for the given level α ∈ (0, 1)) in the form β* ( t ) = Φ ( t Ψ 2 – u α ). (iv) The power loss for the given test is 2
3 Ψ3 ⎞ t ϕ ( u – t Ψ ) ⎛ Ψ – Ψ 2 – r ( t ) = . α 2 ⎝ 4 2 Ψ2 ⎠ 8 Ψ2 (v) The asymptotic deficiency for the given test is 2
2 Ψ 2 t d = ⎛ Ψ 4 – Ψ 2 – 3 ⎞ . ⎝ 4Ψ 2 Ψ 2⎠ Note 1. At the finite mixture of normal laws for the finiteness of moment characteristics Ψs, it is suffi cient to require the fulfillment of conditions (10) and (11) in considering the finite mixture of normal laws for the finiteness of moment characteristics Ψs and of gamma distribution, respectively.
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Note 2. The theorem implies the fulfillment of condition Ψ2 > 0. Its validity was stated in point C of the lemma proof. 4. EXAMPLES OF CERTAIN MIXTURES OF PROBABILISTIC DISTRIBUTIONS This section considers the special cases of mixtures for which the expressions of integrals (6) can be written in an explicit form. It considers identifiable mixtures (for normal and gamma distributions, this is the condition of mixture finiteness, due to theorems proved in [8]). To reduce the sizes of the formulae, we restrict our consideration to testing the hypothesis of a single component mixture against the alternative of a twocomponent mixture for all types of distributions; i.e., parameter k from the formula of density p(x, θ) in (2) is taken to be unity. Let us write the type of moment characteristics Ψs, s = 2, 3, 4 from (6) for the given case (assuming that the conditions of finiteness of corresponding moment characteristics are fulfilled for each type of distri bution): +∞
Ψ2 =
∫
–∞
2
ψ (x) dx – 1, ψ1 ( x ) +∞
Ψ4 =
4
ψ (x)
+∞
+∞
3
2
ψ ( x ) – 3 ψ ( x ) + 2, Ψ3 = dx 2 ψ1 ( x ) ψ ( x ) 1 –∞ –∞
∫
+∞
3
ψ (x)
∫
+∞
(12)
2
ψ (x)
dx – 4 dx + 6 – 3. ∫ ∫ ψ (x) ∫ ψ (x) ψ (x)
–∞
3 1
2 1
–∞
–∞
1
The integrals in (12) are some moment characteristics determined for different types of distributions in [1]. Note that all these quantities can be used in finding the distributions, asymptotic power, power loss, and asymptotic deficiency in the theorem. Normal Distribution Let conditions (10) be fulfilled. We test the hypothesis that the density of each observation is the normal law against the alternative that the density is a mixture of two normal laws. Thus, x – a1 ) ⎫ 1 exp ⎧ – ( ψ 1 ( x ) = ⎬, ⎨ 2 σ 1 2π 2σ 1 ⎭ ⎩ 2
2 x – a) ⎫ 1 exp ⎧ – ( ψ ( x ) = . ⎨ 2 ⎬ σ 2π ⎩ 2σ ⎭
In this case, formulae (12) can be obtained by using the following result of [1] (for s = 2, 3, 4): σ 2 ⎧ ⎛ sa 2 σ ⎫ 1 – ( s – 1 )a 1 ⎞ 2 2 ⎪⎝ σ 1⎠ σ ( s – 1 )a 1 sa 2 ⎪ ψ (x) – 2 ⎬. = exp ⎨ + 2 s–1 2 2 2 s–1 2 ψ ( x ) 2 ( sσ – ( s – 1 )σ ) 2σ 2σ ⎪ ⎪ 1 1 σ sσ 1 – ( s – 1 )σ –∞ 1 ⎩ ⎭
+∞
∫
s
s σ1
Gamma Distribution Let conditions (11) be fulfilled. We test the hypothesis that the density of each observation is defined by the gamma distribution against the alternative that the density is a mixture of two gamma distributions. Thus, β1
α 1 β1 – 1 –α1 x ψ 1 ( x ) = e , x Γ ( β1 )
β
α β – 1 –αx ψ ( x ) = x e , Γ(β)
x≥0
In this case, formulae (12) can be obtained by using the following result of [1] (for s = 2, 3, 4): +∞
∫ 0
s–1
s sβ ψ (x) α 1 Γ ( β 1 )Γ ( sβ – ( s – 1 )β 1 ) = . ( s – 1 )β 1 sβ – ( s – 1 )β 1 s s–1 ψ1 ( x ) Γ (β) α1 ( sα – ( s – 1 )α 1 )
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ACKNOWLEDGMENTS This work was supported by the Russian Foundation for Basic Research, project no. 110112026 ofim. REFERENCES 1. V. E. Bening, A. K. Gorshenin, and V. Yu. Korolev, “Asymptotically Optimal Criterion for Checking the Hypotheses of the Number of Components of a Mixture of Probabilistic Distributions,” Inform. Primen. 5 (3), 4–16 (2011). 2. A. K. Gorshenin, V. Yu. Korolev, and A. M. Tursunbaev, “Median Modifications of EM and SEMAlgorithms for Division of Mixtures of Probabilistic Distributions and Their Application to the Decomposition of Volatility of Financial Time Series,” Inform. Primen. 2 (4), 12–47 (2008). 3. V. Yu. Korolev, ProbabilisticStatistical Analysis of Chaotic Processes with the Help of Mixed Gaussian Models. Decomposition of Volatility of Financial Indexes and Turbulent Plasma (Inst. Probl. Inf. RAN, Moscow, 2007) [in Russian]. 4. V. Yu. Korolev, ProbabilisticStatistical Methods of the Decomposition of Volatility of Chaotic Processes (Mosk. Gos. Univ., Moscow, 2011) [in Russian]. 5. V. E. Bening, Asymptotic Theory of Testing Statistical Hypothesis: Efficient Statistics, Optimally, Power Loss and Deficiency (VSP, Utrecht, 2000). 6. J. Hájek, “Asymptotically Most Powerful RankOrder Tests,” Ann. Math. Stat. 33, 1124–1147 (1962). 7. A. H. Kolmogorov and S. V. Fomin, Elements of the Theory of Functions and Functional Analysis (Nauka, Moscow, 1976; Graylock Press, Albany, New York, 1961). 8. H. Teicher, “Identifiability of Finite Mixtures,” Ann. Math. Stat. 34, 1265–1269 (1963).
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