One needs to compute the value of the integral
Z dyt+1P(yt+1|{x}t1+1,{y}t1)E[trV|x1t+1,yt1+1]tr Kt+xt+1xTt+1 −1 +xTt+1K−t+11xt+1 . However, Z dyt+1P(yt+1|{x}t1+1,{y} t 1)E[trV|xt1+1,y t+1 1 ] = Z dyt+1 d
∏
i=1T(
yt+1)i (βt)i,2(αt)i,(Mtxt+1)i, γt+1 2 d∑
i=1 (βt+1)i (αt+1)i−1 ,and depends only on the values ofαt+1andβtas a result of (19) and Lemma 4.4, and is independent
of the value of xt+1. Thus, we arrive at the minimization of the following expression:
tr " K−t 1−K −1 t xt+1xTt+1K−t 1 1+xT t+1K−t 1xt+1 ! +xTt+1 K−t 1−K −1 t xt+1xTt+1K−t 1 1+xT t+1K−t 1xt+1 ! xt+1 # = " tr(K−t 1)−trK −1 t xt+1xtT+1K−t 1 1+xT t+1Kt−1xt+1 + x T t+1Kt−1xt+1 1+xT t+1K−t 1xt+1 # , = " tr(K−t 1)−x T t+1Kt−1K−t 1xt+1 1+xT t+1Kt−1xt+1 + x T t+1Kt−1xt+1 1+xT t+1K−t 1xt+1 # , =1+x T t+1Kt−1K−t 1xt+1 1+xTt+1K−t 1xt+1 . References
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