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87⎛∑ Kj=1f(YE KT⎝i |X i = j, ˆθ⎞K )(1/K)⎠ (5.45)f(Y i |ˆθ K T )This expectation can be found by integrating over Y i with respect to the truedensity of Y i .⎛∫⎝Y i∑ Kj=1f(Y i |X i = j, ˆθ⎞K )(1/K)⎠ f(Y i |θ K T)dY i (5.46)f(Y i |ˆθ K T )As noted previously, f(Y i |θ K T) cancels with the term in the denominator, soequation 5.46 becomes equation 5.47.∫K ∑Y i j=1f(Y i |X i = j, ˆθ K )(1/K)dY i = 1 (5.47)Thus I have shown that the expectation on the left hand side of equation 5.42is equal to 1. The right hand side of the inequality in equation 5.42 is equal to thefollowing.exp( )((DK − D KT ) log(N)/2)Since D K > D KT , we know that the following inequalities hold.N(5.48)((D K − D KT ) log(N)/2)N> 0 (5.49)( )((DK − D KT ) log(N)/2)exp> 1 (5.50)NIn the limit as N → ∞, the inequality in equation 5.50 becomes an equality.Combining equations 5.47 and 5.50, we see the following.∫K ∑Y i j=1f(Y i |X i = j, ˆθ K )(1/K)dY i = 1 < exp( )((DK − D KT ) log(N)/2)N(5.51)