# Statistical analysis of mixture distribution - Smith A.F.M

ISBN 0-470-90763-4

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An ‘optimal’ choice of c is that which minimizes var(rt,). Such a c satisfies the stationarity condition

/(c)[l -2F(c)][F1(c)-F2(c)] = 2F(c)[l -F(c)][/|(c)-/2(c)].

Of course, it is not easy to solve this equation and the solution clearly depends on the unknown n, but we follow up this idea with the next example.

Example 4.2.2 Mixture of two known exponentials Suppose

p(x) = nc~x + (l-n)2e~2x, x > 0,0 < 7i < 1.

If

r,(x) = x*/r(s-I-1), 0 0,

Then

ft, = = * + (1-*)/2*.

= (2X - 0/(2* -1)

and

n var Alt = [Hs + 1 )(2S - 1)] " 222s var Xs

= [r(s+ 1)(1 — 2-J)]-2{r(2s + l)F<J.-[r(s+ l)Fj2}- (4-2-5)

The optimal power-moment estimator of n is thus given by the s which minimizes (4.2.5). Also, the relative efficiency of the simple estimator based on the

74 Statistical analysis of finite mixture distributions

sample mean (s= 1) can be obtained as

RE(7t) = var(rt,J/var(tf(1).

Some values of s* and RE(ir) are given below: n 0.1 0.3 0.5 0.7 0.9

s* 1.45 0.90 0.66 0.50 0.36

RE(n) 0.94 0.99 0.95 0.88 0.81

In practice, computation of s* requires a preliminary estimate of n, possibly Htl. Note that these calculations ignore the fact that it may not lie in 0 ^ -ft ^ 1.

Example 4.2.3 Mixture of two unknown exponentials (Rider, 1961, 1962) Suppose

p(x| ij/) = 1 exp( — x/OJ + n20f1 exp( — x/02\ x > 0,

with 7ix, 0, and d2 all unknown. In order to simplify the notation of Example 4.2.2, let us define

n

m, = n~1 Y, xf/H5 + 1), s = 0,1,...,

i = l

where m0 = 1. Given m1, m2, m3, moment estimates are available from

nl6sl+n29s2 = ms, s = 0,1,2,3. (4.2.6)

Eliminating nl and n2 using the first two equations and considering the remaining two, we find that 9X, d2 are two solutions of the quadratic equation

(nij ~ Moni2)02 + 2(m0m3 - m1m2)6 + m\ — m1 w3 = 0.

There is no guarantee that these roots will be real, let alone positive, although they are consistent if the true 9/s are unequal.

Tallis and Light (1968) show that, by using moment equations based on values of s other than 1, 2, 3, spectacular gains in efficiency can sometimes be obtained. (However, they do not attempt the difficult task of calculating an optimal set of power moments in the spirit of Example 4.2.2.) Table 4.2.1 shows the gains in relative efficiency that are possible. Efficiency is measured in terms of the determinant of the asymptotic covariance matrix, given in (4.2.3), relative to that obtained using maximum likelihood. In the table, tj = 0l/02 and the moment equations used were for sx = 1, s2, s3, the latter two having been identified as optimal by searching over a grid of values with increments of \ in each direction. Ei and E2 are relative efficiencies for the ‘natural’ powers (1,2,3) and the suboptimal set (l,s*,s3). The greatest improvements occur if tj is very different from 1 and if fractional moments are used.

Learning about the parameters of a mixture

Table 4.2.1 Relative efficiencies for Example 4.2.3. (Adapted from Tallis and Light, 1968 by permission of the American Statistical Association)

n 1.5 2 5 10

0.942 0.737 0.155 0.041

0.1 e2 0.950 0.829 0.561 0.501

s*>s 3 2.25,2.75 2.00,2.25 0.75,1.50 0.75,1.25

0.784 0.475 0.052 0.008

0.5 e2 0.950 0.872 0.625 0.507

1.75,2.00 1.25,1.50 0.50,0.75 0.50,0.75

0.652 0.343 0.022 0.002

0.9 e2 0.956 0.884 0.665 0.522

S2,S3 1.50,1.75 0.75,1.50 0.25,0.75 0.25,0.50

Example 4.2.4 Mixture of two univariate normals Let

p(x\*J/) = n<t>{x\iil,(Tl) + (\ -n)(f)(x\p2,o2l

0 < 7i < 1, ol, o2 > 0. Estimation using the method of moments in this five-parameter mixture by Pearson (1894) is often thought of as the starting point of the analysis of mixtures. Pearson’s approach was later streamlined by Charlier (1906) and Charlier and Wicksell (1924); perhaps the clearest descriptions of the calculations arc set out in Holgersson and Jorner (1979), Johnson and Kotz (1970a, Section 13.7.2), and Cohen (1967). Equations based on the first five central moments are used and, after much elimination, we are left with the problem of finding a negative root for the famous ‘nonic’ equation. We do not derive the nonic here but rather indicate the ‘reverse’ procedure for calculating parameter

estimates from data.

In what follows, ms and ks denote the sth sample central moment and sample

cumulant, respectively. The nonic equation, to be solved for v, is

with ag = 24, a8 = 0, a7 = 84 k4, a6 = 36 mj, a5 = 90kl + 72fc5m3,

a4 = 444 /c4m3 — 18/c|,

a3 = 288ml - 108 m3Ms + 21kl a2 — — (63mlkl + 12m\ks\

76

Statistical analysis of finite mixture distributions

a, = — 96m3k4, a0 = - 24m63.

Having obtained a negative root v, we calculate

tj = ( - 6m3v3 -I- 2k5v2 + 9m3/c4 + 6m3)/(2v3 + 3fc4v + 4w3)

and

o) = r\ — m3.

Then we compute p = co/v and solve the quadratic equation

s* - pb + V = 0, (4.2.7)

giving roots <5, and b2, with <5j > 0 > S2, say.

We also calculate

{} = (2(o- m3)/3v.

We may now express our estimates in the form

6) = SjP + in 2 - b], j = 1,2,

1x = b 2/(^1 — ^2)» pj = bj + x, j = 1,2.

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