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91− ∑ (Y i − µ 2 ) 2i∈S 22σ22(5.62)∑log(f(Y i |θ K )) = −N log( √ 2π) − N log(σ K ) − N/2 (5.63)iCombining equations 5.61 to 5.63, we find that equation 5.60 is equal to equation5.64.N log(σ K ) − |S 1 | log(σ 1 ) − |S 2 | log(σ 2 ) + ∑p(X i = 1) + ∑p(X i = 2)i∈S 1 i∈S 2− ∑ (Y i − µ 1 ) 2− ∑ (Y i − µ 2 ) 2+ N/2 (5.64)i∈S 12σ12 i∈S 22σ22From the construction of the sets S 1 and S 2 , it is clear that the inequality inequation 5.65 holds.∑ (Y i − µ 1 ) 2+ ∑ (Y i − µ 2 ) 2< ∑ (Y i − µ 1 ) 2(5.65)i∈S 12σ12 i∈S 22σ22 i2σ12Thus, the quantity in equation 5.66 is less than the quantity in equation 5.64.N log(σ K ) − |S 1 | log(σ 1 ) − |S 2 | log(σ 2 ) + ∑log(p(X i = 1)) + ∑log(p(X i = 2))i∈S 1 i∈S 2(5.66)Recalling equation 5.60, consistency is implied if the inequality in equation 5.67holds.N log(σ K ) − |S 1 | log(σ 1 ) − |S 2 | log(σ 2 ) + ∑log(p(X i = 1)) + ∑log(p(X i = 2))i∈S 1 i∈S 2> ((D KT − D K ) log(N)/2) (5.67)Note that the minimum value of log(p(X i = 1)) or log(p(X i = 2)) is − log(e 8φ +1), which is approximately equal to −8φ. Suppose, without loss of generality, thatσ 1 > σ 2 . The inequality in equation 5.67 will be assured when equation 5.68 holds.