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**Sample text**

1 1 1 ß +ß z. e 0 . 1 I ß +ß z. 1/(1+e 0 I I). 1 These equations do not have explicit solutions and must be solved by numerical methods. Hence it is of interest to determine for which observed values of t l and t 2 there are solutions. 6 the likelihood equations have a unique set of solu- 50 tions, if (tl't 2) is an interior point of the convex extension of the support. In this case it is easy to derive the support. If, namely, t1=i, it follows that exactly i of the binary variables have the value 1.

Hence if z(1)Sz(2)S ... Sz(n) are the z's in order of magnitude, the minimum and maximum values of t 2 are It is further easy to see that the set obtained by connecting the 2n points {O,O}, and {i, z(n_i+l)+ ... +z(n)}' i=1, ... ,n-1 is a convex set. Hence the likelihood equations have solutions if (t 1,t 2) does not coincide i=1, ... ,10 the convex with any of the 2n boundary points. =i, 1 extension of the support is shown in fig. 2. 30 20 10 0~~~~ o __________~______________- - T 5 10 Fig. 2.

M}, and nK(T1, ... Jl J m llij ( °l,.. ·,Ok)· A discussion of estimation problems in exponential families with non-identically distributed random variables was given by Nordberg (1980). 2. 7) = tT + h1(t) - nK( T). p( 0) is the canonical parameter and t=t(xl""'xn) the sufficient statistic for T. 7) is equivalent to E[TI Tl = t, and to nK'( T) = t. ln{Eexp(tT+h1(t))}. 7) dlnHt Ir) = t - nK'( r), the theorem follows. D. In case the ML-estimator is a solution to the likelihood equation, it can thus be found by simply equating the observed value of the sufficient statistic and its mean value.