View - Statistics - University of Washington

89⎛∑ ∑ KTlog ⎝ j=1 f(Y i |X i = j, ˆθ K T )p(Xi = j|N( ˆX K Ti ), ˆφ⎞K T )⎠ > ((Dif(Y i |ˆθ K KT −D K ) log(N)/2))(5.55)Recall from case 1 that as N → ∞, it follows that ˆθ K T → θK T and ˆφK T →φ K T . Some additional notation is needed at this point. Denote the true densityparameters in θ K Tby µ1 , µ 2 , σ 2 1 , and σ2 2 . Similarly, let the parameters in θK (whichin this case consists of only one component) be denoted by µ K and σ 2 K.We can deduce the asymptotic values of µ K and σ 2 K in terms of the true parameters.Let P 1 be the proportion of pixels for which the true (unobservable)state X i is 1, and similarly let P 2 be the proportion of pixels in state 2. Then µ Kand σ 2 K will be given by equations 5.56 and 5.57.µ K = P 1 µ 1 + P 2 µ 2 (5.56)σK 2 = P 1 (σ1 2 + µ 2 1 − 2P 1 µ 2 1 − 2P 2 µ 1 µ 2 + P1 2 µ 2 1 + P2 2 µ 2 2 + 2P 1 P 2 µ 1 µ 2 )+P 2 (σ2 2 + µ 2 2 − 2P 2 µ 2 2 − 2P 1 µ 1 µ 2 + P1 2 µ 2 1 + P2 2 µ 2 2 + 2P 1 P 2 µ 1 µ 2 )(5.57)The inequality in equation 5.55 holds if the inequality of equation 5.58 holds;this is true for any set of values of m i , since the left hand side of equation 5.58 isless than the left hand side of equation 5.55.(∑ f(Yi |X i = m i , θ K T )p(Xi = m i |N(logˆX K Ti ), φ K )T )> ((Dif(Y i |θ K KT − D K ) log(N)/2))(5.58)Now construct two sets, S 1 and S 2 , as follows. The set S 1 consists of all i suchthat (Y i − µ 1 ) 2 /2σ1 2 < (Y i − µ 2 ) 2 /2σ2, 2 and similarly S 2 is the set of i such that(Y i − µ 1 ) 2 /2σ1 2 > (Y i − µ 2 ) 2 /2σ2. 2 Combining this with equation 5.58, we find that