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80⎛∑ ⎞Kj=1f(Yg i (Y i ) = log ⎝i |X i = j, θ 1 )p(X i = j|N 1 (X i ), φ 1 )∑ ⎠KT(5.21)j=1 f(Y i |X i = j, θ 2 )p(X i = j|N 2 (X i ), φ 2 )Here, θ 1 , θ 2 , φ 1 , and φ 2 are fixed parameter values; N 1 (X i ) and N 2 (X i ) arefixed neighborhoods of X i . Let us further assume that f(Y i ) denotes a Gaussianor Gaussian mixture density. Then g i ∈ L 1 , that is, ∫ ∞−∞ |g i (Y i )|f(Y i )dY i < ∞.Proof of Lemma 1.Since g i is a function of Y i , I need to show that ∫ ∞−∞ |g i(Y i )|f(Y i )dY i < ∞. NowN 1 (X i ) and N 2 (X i ) are fixed, so p(X i = j|N 1 (X i ), φ 1 ) and p(X i = j|N 2 (X i ), φ 2 )are also fixed for a given i and I denote them by p 1 j and p 2 j. I can now write g i asfollows.g i (Y i ) = log⎛⎝∑ Kj=1f(Y i |X i = j, θ 1 )p 1 j∑ KTj=1 f(Y i |X i = j, θ 2 )p 2 j⎞⎠ (5.22)Note that both the numerator and denominator inside the logarithm in equation5.22 are Gaussian mixtures. I assume that all variances are nonzero. Thusg i (Y i ) is finite over any finite interval, and its integral is finite when integratedover any finite interval.Let S 1 denote the state with largest variance in the numerator of equation 5.22,and similarly S 2 for the denominator. These terms will dominate in the tails; inother words, there exists a value Y A such that for all Y i > Y A (and also for allY i < −Y A ), the following two inequalities will hold.K∑f(Y i |X i = j, θ 1 )p 1 j < f(Y A |X i = S 1 , θ 1 ) (5.23)j=1K T∑f(Y i |X i = j, θ 2 )p 2 j < f(Y A|X i = S 2 , θ 2 ) (5.24)j=1We can now write the following, in which C denotes a constant.