Appendix
Proof of Lemma
3.1
Proof
Note that in the proof of Lemma 2.2, the terms \(p(\varTheta _{k-1}^i|\mathcal {I}_{k-1})\) and \(p(\varTheta _{k-1}^i|\mathcal {I}_{k})\), that is, \(\alpha ^i_{k|k-1}\) and \(\alpha ^i_{k|k}\), are not involved. It means that, the derivation of \(p(x_k|\varTheta _{k-1}^i,\mathcal {I}_{k})\) is independent of \(p(\varTheta _{k-1}^i|\mathcal {I}_{k})\). Similarly, the derivation of \( p(\breve{x}_k|\varTheta _{k-1}^i,\mathcal {I}_{k})\) is independent of \(\breve{p}(\varTheta _{k-1}^i|\mathcal {I}_{k})\). Since \(p(\breve{x}_k|\varTheta ^i_{k-1},\mathcal {I}_{k})\) is the same function as \(p(x_k|\varTheta ^i_{k-1},\mathcal {I}_{k})\). It follows from Lemma 2.2, we have
$$\begin{aligned} p(\breve{x}_k|\varTheta _{k-1}^i,\mathcal {I}_{k}) = \mathcal {N}_{\breve{x}_k}(m^i_{k|k},S_{k|k}). \end{aligned}$$
Likewise,
$$\begin{aligned} p(\breve{x}_k|\varTheta _{k-1}^i,\mathcal {I}_{k-1}) = \mathcal {N}_{\breve{x}_k}(m^i_{k|k-1},S_{k|k-1}), \end{aligned}$$
where \(m^i_{k|k-1}\) and \(m^i_{k|k}\) evolve in the same way as (2.7).
Define \(\breve{p}(\varTheta ^i_{k-1}|\mathcal {I}_{k-1})\) in (3.6a) as \(\breve{\alpha }^i_{k|k-1}\), and \(\breve{p}(\varTheta ^i_{k-1}|\mathcal {I}_{k})\) in (3.6b) as \(\breve{\alpha }^i_{k|k}\).
Firstly, we check (3.8a). Let \(1 \le i\le 2^{k-1}\),
$$\begin{aligned} \breve{\alpha }^i_{k|k-1} \triangleq {}&\breve{p}(\varTheta _{k-1}^i|\mathcal {I}_{k-1})\\ = {}&p(\nu _{k-1}=0)\breve{p}(\varTheta _{k-2}^i|\mathcal {I}_{k-1})\\ = {}&\bar{\nu }\breve{\alpha }^i_{k-1|k-1}. \end{aligned}$$
Similarly, we will have \(\breve{\alpha }^i_{k|k-1} = \nu \breve{\alpha }^{i-2^{k-1}}_{k-1|k-1}\), for \(2^{k-1}+1 \le i\le 2^k\). Hence (3.8a) holds.
Then we check (3.8b). The proof is by mathematical induction. For \(k=1\), \(\breve{\alpha }^i_{1|1} \triangleq p(\varTheta ^i_0|y_1)\). Thus \(\breve{\alpha }^i_{1|1}=\alpha ^i_{1|1}\) at time \(k=1\). Take \(i=1\), from (2.30),
$$\begin{aligned} \breve{\alpha }^1_{1|1}={}&\alpha ^1_{1|1}\\ ={}&\frac{ \alpha ^1_{1|0}p(y_1|\nu _0=0)}{\alpha ^1_{1|0}p(y_1|\nu _0=0)+\alpha ^2_{1|0}p(y_1|\nu _0=1)}. \end{aligned}$$
By the definition of \(\lambda ^i_{k-1}\) and \(\psi _{k-1}^i(y_k)\) in (3.8b), \(\psi _0^i(y_1) = p(y_1|\nu _0=i)\), for \(i=0\) or 1. Note that \(\breve{\alpha }^1_{1|0}=p(\varTheta ^1_0)=\alpha ^1_{1|0}\), then \(\breve{\alpha }^1_{1|1}=\lambda _0^0\breve{\alpha }^1_{1|0}\) is obtained. By the same derivations, we have \(\breve{\alpha }^2_{1|1}=\lambda _0^1\breve{\alpha }^2_{1|0}\). Then (3.8b) holds for \(k=1\).
Suppose that (3.8b) is true for \(1,\ldots , k\). The condition for \(k+1\) is examined as follows. Let \(1 \le i\le 2^k\), by (2.2), \(\varTheta _{k}^i = \{\nu _k=0 ,\varTheta _{k-1}^i \}\).
$$\begin{aligned} {}&\breve{\alpha }^i_{k+1|k+1}\\ \triangleq {}&\breve{p}(\varTheta _{k}^i|\mathcal {I}_{k+1})\\ \triangleq {}&p(\nu _k=0|y_{k+1}) \breve{p}(\varTheta _{k-1}^i|\mathcal {I}_{k})\\ ={}&\frac{ p(y_{k+1}|\nu _k=0)p(\nu _k=0) \breve{p}(\varTheta _{k-1}^i|\mathcal {I}_{k})}{p(y_{k+1}|\nu _k=0)p(\nu _k=0)+p(y_{k+1}|\nu _k=1)p(\nu _k=1)}\\ ={}&\frac{\psi _k^0(y_{k+1}) }{\bar{\nu }\psi _k^0(y_{k+1}) + \nu \psi _k^1(y_{k+1})} p(\nu _k=0)\breve{p}(\varTheta _{k-1}^i|\mathcal {I}_{k})\\ ={}&\frac{\psi _k^0(y_{k+1}) }{\bar{\nu }\psi _k^0(y_{k+1}) + \nu \psi _k^1(y_{k+1})} \breve{\alpha }^i_{k+1|k}\\ ={}&\lambda ^0_k\breve{\alpha }^i_{k+1|k}. \end{aligned}$$
For \(2^k+1 \le i \le 2^{k+1}\), by the same derivations,
$$\begin{aligned} \breve{\alpha }^i_{k+1|k+1} = {}&\frac{\psi _k^1(y_{k+1})p(\nu _k=1) }{\bar{\nu }\psi _k^0(y_{k+1}) + \nu \psi _k^1(y_{k+1})}\breve{\alpha }^{i-2^k}_{k|k}\\ ={}&\lambda ^1_k\breve{\alpha }^i_{k+1|k}. \end{aligned}$$
Thus (3.8b) holds for \(k+1\). The proof is completed.
Proof of Lemma
3.3
Proof
Note that \(\sum _{i=1}^{2^k}\breve{\alpha }^i_{k|k}=1\). From (3.12c), by using (2.7) and (3.8),
$$\begin{aligned} \breve{x}_{k+1|k}&= \sum _{i=1}^{2^{k+1}}\breve{\alpha }^i_{k+1|k}m^i_{k+1|k} \nonumber \\&= \sum _{i=1}^{2^k}\breve{\alpha }^i_{k+1|k}m^i_{k+1|k} + \sum _{i={2^k+1}}^{2^{k+1}}\breve{\alpha }^i_{k+1|k}m^i_{k+1|k} \\&= \sum _{i=1}^{2^k}\bar{\nu }\breve{\alpha }^i_{k|k}Am^i_{k|k} + \sum _{i=1}^{2^k}\nu \breve{\alpha }^i_{k|k}(Am^i_{k|k}+Bu_k) \\&= A\breve{x}_{k|k}+ \nu Bu_k. \end{aligned}$$
Then (3.14a) is proved. (2.12a) can be proved by following the same derivations as above. Then consider \(\breve{\varPsi }_{k+1|k}\) in (3.13b), by using (2.7) and (3.8),
$$\begin{aligned} \breve{\varPsi }_{k+1|k} \triangleq&\sum _{i=1}^{2^{k+1}}\breve{\alpha }^i_{k+1|k}(m^i_{k+1|k}-\breve{x}_{k+1|k})^2_I \nonumber \\ =&\, \bar{\nu }\sum _{i=1}^{2^k}\breve{\alpha }^i_{k|k}(Am^i_{k|k}-A\breve{x}_{k|k}-\nu B u_k)^2_I\end{aligned}$$
(3.20)
$$\begin{aligned}&+\nu \sum _{i=1}^{2^k}\breve{\alpha }^i_{k|k}(Am^i_{k|k}-A\breve{x}_{k|k}+\bar{\nu }B u_k)^2_I. \end{aligned}$$
(3.21)
Expanding the summation part in (3.20) as follows:
$$\begin{aligned}&\sum _{i=1}^{2^k}\breve{\alpha }^i_{k|k}(Am^i_{k|k}-A\breve{x}_{k|k}-\nu B u_k)^2_I \nonumber \\ =&\, \sum _{i=1}^{2^k}\breve{\alpha }^i_{k|k} \big ((Am^i_{k|k}-A\breve{x}_{k|k})^2_I +(\nu B u_k)^2_I -2\nu B u_k(Am^i_{k|k}-A\breve{x}_{k|k})^{\prime } \big ) \nonumber \\ =&\, A\breve{\varPsi }_{k|k}A^{\prime } +(\nu B u_k)^2_I -2\nu B u_k\sum _{i=1}^{2^k}\breve{\alpha }^i_{k|k}( Am^i_{k|k}-A\breve{x}_{k|k})^{\prime }. \end{aligned}$$
(3.22)
Similarly, for the summation part in (3.21),
$$\begin{aligned}&\sum _{i=1}^{2^k}\breve{\alpha }^i_{k|k}(Am^i_{k|k}-A\breve{x}_{k|k}+\bar{\nu }B u_k)^2_I\nonumber \\ =&\, A\breve{\varPsi }_{k|k}A^{\prime } +(\bar{\nu }B u_k)^2_I +2\bar{\nu }B u_k\sum _{i=1}^{2^k}\breve{\alpha }^i_{k|k}( Am^i_{k|k}-A\breve{x}_{k|k})^{\prime }. \end{aligned}$$
(3.23)
Substituting (3.22) and (3.23) into (3.20) and (3.21), respectively, by some simple algebraic computations, we get
$$\begin{aligned} \breve{\varPsi }_{k+1|k}&= \bar{\nu }\left[ A\breve{\varPsi }_{k|k}A^{\prime }-2\sum _{i=1}^{2^k}\breve{\alpha }^i_{k|k}\big ( Am^i_{k|k}-A\breve{x}_{k|k})(\nu B u_{k+1})^{\prime } \big )\right. \\&+\left. (\nu B u_{k+1})(\nu B u_{k+1})^{\prime } \right] \\&+\nu \left[ A\breve{\varPsi }_{k|k}A^{\prime }+2\sum _{i=1}^{2^k}\breve{\alpha }^i_{k|k}\Big ( Am^i_{k|k}-A\breve{x}_{k|k})(\bar{\nu }B u_{k+1})^{\prime } \Big )\right. \\&+\left. (\bar{\nu }B u_{k+1})(\bar{\nu }B u_{k+1})^{\prime }\right] \\&= A\breve{\varPsi }_{k|k}A^{\prime } + \bar{\nu }(\nu B u_{k+1})(\nu B u_{k+1})^{\prime } + \nu (\bar{\nu }B u_{k+1})(\bar{\nu }B u_{k+1})^{\prime }\\&= A\breve{\varPsi }_{k|k}A^{\prime } + \bar{U}_k \end{aligned}$$
where \(\bar{U}_k=\bar{\nu }\nu B u_ku_k^{\prime }B^{\prime }\). From (3.12d),
$$\begin{aligned} \breve{P}_{k+1|k}&= S_{k+1|k} + \breve{\varPsi }_{k+1|k}\\&= A(S_{k|k} + \breve{\varPsi }_{k|k})A^{\prime } +\bar{U}_k +Q\\&= A\breve{P}_{k|k}A^{\prime } +\bar{U}_k +Q. \end{aligned}$$
Hence (3.14b) is proved. By the same derivations, (2.12b) can be obtained as well.
Next, we derive \(\breve{x}_{k+1|k+1}\) and \(\breve{P}_{k+1|k+1}\). Note that \(\bar{\nu }\lambda _k^0+\nu \lambda _k^1=1\). From (3.12a), by using (2.7) and (3.8),
$$\begin{aligned} \breve{x}_{k+1|k+1}={}&\sum _{i=1}^{2^{k+1}}\breve{\alpha }^i_{k+1|k+1}m^i_{k+1|k+1} \\ ={}&\lambda _k^0 \sum _{i=1}^{2^{k}}\breve{\alpha }^i_{k+1|k} \Big (m^i_{k+1|k} + K_{k+1}\left( y_{k+1} - Cm^i_{k+1|k}\right) \Big ) \\ {}&+ \lambda _k^1 \sum _{i=2^k}^{2^{k+1}}\breve{\alpha }^i_{k+1|k} \Big (m^i_{k+1|k} + K_{k+1}\left( y_{k+1} - Cm^i_{k+1|k}\right) \Big ) \\ ={}&\bar{\nu }\lambda _k^0 \sum _{i=1}^{2^{k}}\breve{\alpha }^i_{k|k}(I-K_{k+1}C)Am^i_{k|k} \\ {}&+ \nu \lambda _k^1 \sum _{i=1}^{2^k}\breve{\alpha }^i_{k|k}(I-K_{k+1}C)(Am^i_{k|k}+Bu_k) \\ {}&+ K_{k+1}y_{k+1} \\ ={}&(I-K_{k+1}C)(A\breve{x}_{k|k} + \nu \lambda _k^1 Bu_k) + K_{k+1}y_{k+1}. \end{aligned}$$
Equation (3.14c) is proved. Consider \(\breve{\varPsi }_{k+1|k+1}\) in (3.13a), by (2.7) and (3.8),
$$\begin{aligned} {}&\breve{\varPsi }_{k+1|k+1} \nonumber \\ \triangleq {}&\sum _{i=1}^{2^{k+1}}\breve{\alpha }^i_{k+1|k+1}(m^i_{k+1|k+1}-\breve{x}_{k+1|k+1})^2_I \nonumber \\ ={}&\bar{\nu }\lambda _{k}^0\sum _{i=1}^{2^{k}}\breve{\alpha }^i_{k|k} \big ( (I - K_{k+1}C)(Am^i_{k|k} - A\breve{x}_{k|k} - \nu \lambda _k^1 Bu_k) \big )^2_I \end{aligned}$$
(3.24)
$$\begin{aligned}&+ \nu \lambda _{k}^1\sum _{i=1}^{2^{k}}\breve{\alpha }^i_{k|k}\big ( (I - K_{k+1}C)(Am^i_{k|k} + Bu_k - A\breve{x}_{k|k} - \nu \lambda _k^1 Bu_k)\big )^2_I. \end{aligned}$$
(3.25)
Expanding (3.24) and (3.25) in the same way as (3.20) and (3.21) and collecting the common terms, then we will readily have
$$\begin{aligned} \breve{\varPsi }_{k+1|k+1} = (I - K_{k+1}C)( A\breve{\varPsi }_{k|k}A^{\prime } + U_k) (I - K_{k+1}C)^{\prime } \end{aligned}$$
where \(U_k=\bar{\nu }\nu \lambda _{k}^0\lambda _{k}^1Bu_ku_k^{\prime }Bu_k^{\prime }\). From (3.12b),
$$\begin{aligned} \breve{P}_{k+1|k+1} = {}&S_{k+1|k+1} + \breve{\varPsi }_{k+1|k+1}\\ ={}&(I - K_{k+1}C)\Big ( A(S_{k|k}+\breve{\varPsi }_{k|k})A^{\prime } + U_k +Q\Big ) (I - K_{k+1}C)^{\prime }\\ {}&+ K_{k+1} R K_{k+1}^{\prime }\\ ={}&\mathbb {K}_{k+1} ( A\breve{P}_{k|k}A^{\prime } + U_k +Q )\mathbb {K}_{k+1}^{\prime } + K_{k+1} R K_{k+1}^{\prime }. \end{aligned}$$
Hence (3.14d) is proved. The proof is completed.
Proof of Lemma
3.6
Proof
(i): Proof of \(\mathbb {E}_{\mathcal {I}}[P] \le \mathbb {E}_{\mathcal {I}}[\tilde{P}]\).
In the following proof, for consistent presentation in the computation of the mathematical expectation of the random variables, we use the notation \(\int _\theta (\cdot )\text{ d }\theta \), instead of \(\sum _{\theta }\), for discrete random variable. Such treatment does not affect the result.
Denote the mean of x and \(\tilde{x}\) as \(\mu \triangleq \mathrm{{e}}(x)\) and \(\tilde{\mu } \triangleq \mathrm{{e}}(\tilde{x})\), respectively.
$$\begin{aligned} \mu ={}&\int _\theta \left( \textstyle \int _x x p(x|\theta ,\mathcal {I})\text{ d }x \right) p(\theta |\mathcal {I}) \text{ d }\theta \nonumber \\ ={}&\mathbb {E}_{\theta }\Big [ \mathbb {E}_x[x|\theta ,\mathcal {I}] \big |\mathcal {I}\Big ] \end{aligned}$$
(3.26)
$$\begin{aligned} \tilde{\mu } ={}&\int _\theta \left( \int _x x p(x|\theta ,\mathcal {I})\text{ d }x \right) p(\theta ) \text{ d }\theta \nonumber \\ = {}&\mathbb {E}_{\theta }\Big [ \mathbb {E}_x[\tilde{x}|\theta ,\mathcal {I}]\Big ]. \end{aligned}$$
(3.27)
As mentioned in the motivation of constructing the auxiliary estimator, we intend to present P and \(\tilde{P}\) in form like (3.4). (That is, (3.34) and (3.35)).
$$\begin{aligned} P&= \int _\theta \left( \int _x (x-\mu )^2p(x|\theta ,\mathcal {I})\text{ d }x \right) p(\theta |\mathcal {I}) \text{ d }\theta \end{aligned}$$
(3.28)
$$\begin{aligned} \tilde{P}&= \int _\theta \left( \int _x (x-\tilde{\mu })^2p(x|\theta ,\mathcal {I})\text{ d }x \right) p(\theta ) \text{ d }\theta . \end{aligned}$$
(3.29)
Note that in (3.27) and (3.29), the replacement of \(\tilde{x}\) with x will not affect the value of the integration. By expanding the term \((x-\mu )^2\) in (3.28) into \(x^2-2\mu x+\mu ^2\), and substituting \(\mu \) by \(\textstyle \int _\theta \left( \textstyle \int _x x p(x|\theta ,\mathcal {I})\text{ d }x \right) p(\theta |\mathcal {I}) \text{ d }\theta \) in (3.26), (or immediately using (3.2)), P can be rewritten as follows.
$$\begin{aligned} P&= \int _\theta \left( \int _x x^2 p(x|\theta ,\mathcal {I})\text{ d }x \right) p(\theta |\mathcal {I}) \text{ d }\theta \end{aligned}$$
(3.30)
$$\begin{aligned}&-\left( \int _\theta \left( \int _x x p(x|\theta ,\mathcal {I})\text{ d }x \right) p(\theta |\mathcal {I}) \text{ d }\theta \right) ^2. \end{aligned}$$
(3.31)
Let
$$\begin{aligned} \varLambda \triangleq \int _\theta \left( \int _x xp(x|\theta ,\mathcal {I})\text{ d }x \right) ^2p(\theta |\mathcal {I}) \text{ d }\theta , \end{aligned}$$
then \(P = (3.30) -\varLambda +\varLambda -(3.31)\). Firstly collect (3.30) \(-\varLambda \),
$$\begin{aligned}&\int _\theta \left\{ \left( \int _x x^2 p(x|\theta ,\mathcal {I})\text{ d }x\right) - \left( \int _x xp(x|\theta ,\mathcal {I})\text{ d }x\right) ^2 \right\} p(\theta |\mathcal {I}) \text{ d }\theta \nonumber \\ =&\,\int _\theta \left\{ (\mathbb {E}_x[x^2|\theta ,\mathcal {I}] - (\mathbb {E}_x[x|\theta ,\mathcal {I}])^2 \right\} p(\theta |\mathcal {I}) \text{ d }\theta \nonumber \\ =&\,\int _\theta \mathrm {cov}_x(x|\theta ,\mathcal {I}) p(\theta |\mathcal {I}) \text{ d }\theta \nonumber \\ =&\,\mathbb {E}_{\theta }\big [\mathrm {cov}_x(x|\theta ,\mathcal {I}) \big |\mathcal {I}\big ]. \end{aligned}$$
(3.32)
where it follows from (3.2) that
$$\begin{aligned} \mathrm {cov}_x(x|\theta ,\mathcal {I})=\mathbb {E}_x[x^2|\theta ,\mathcal {I}] - (\mathbb {E}_x[x|\theta ,\mathcal {I}])^2. \end{aligned}$$
By the same derivation, collect \(\varLambda -(3.31)\) as follows:
$$\begin{aligned}&\int _\theta \left( \int _x xp(x|\theta ,\mathcal {I})\text{ d }x \right) ^2p(\theta |\mathcal {I}) \text{ d }\theta - \left( \int _\theta \left( \int _x xp(x|\theta ,\mathcal {I})\text{ d }x \right) p(\theta |\mathcal {I}) \text{ d }\theta \right) ^2 \nonumber \\ =&\, \int _\theta \left( \mathbb {E}_x[x|\theta ,\mathcal {I}] \right) ^2p(\theta |\mathcal {I}) \text{ d }\theta - \left( \int _\theta \mathbb {E}_x[x|\theta ,\mathcal {I}] p(\theta |\mathcal {I}) \text{ d }\theta \right) ^2 \nonumber \\ =&\, \mathbb {E}_{\theta }\big [\left( \mathbb {E}_x[x|\theta ,\mathcal {I}] \right) ^2\big |\mathcal {I}\big ] - \left( \mathbb {E}_{\theta }\big [\left( \mathbb {E}_x[x|\theta ,\mathcal {I}] \right) \big |\mathcal {I}\big ]\right) ^2 \nonumber \\ =&\, \mathrm {cov}_{\theta }\big (\mathbb {E}_x[x|\theta ,\mathcal {I}]\big |\mathcal {I}\big ). \end{aligned}$$
(3.33)
where (3.33) follows from (3.2) by viewing \(\mathbb {E}_x[x|\theta ,\mathcal {I}]\) as a function of the random variable \(\theta \).
By combining (3.32) and (3.33),
$$\begin{aligned} P&= \mathbb {E}_{\theta }\big [\mathrm {cov}_x(x|\theta ,\mathcal {I}) \big |\mathcal {I}\big ]+ \mathrm {cov}_{\theta }\big (\mathbb {E}_x[x|\theta ,\mathcal {I}]\big |\mathcal {I}\big ), \end{aligned}$$
(3.34)
and by the same way \(\tilde{P}\) can be presented as follows.
$$\begin{aligned} \tilde{P}&= \mathbb {E}_{\theta }\big [\mathrm {cov}_x(x|\theta ,\mathcal {I})\big ]+ \mathrm {cov}_{\theta }\big (\mathbb {E}_x[x|\theta ,\mathcal {I}]\big ) . \end{aligned}$$
(3.35)
Then we compare \(\mathbb {E}_{\mathcal {I}}[P]\) with \(\mathbb {E}_{\mathcal {I}}[\tilde{P}]\). Applying the equality (3.3) to the first part of (3.35), we have
$$\begin{aligned} \mathbb {E}_{\mathcal {I}}\Big [ \mathbb {E}_{\theta }\big [\mathrm {cov}_x(x|\theta ,\mathcal {I})\big ] \Big ] = \mathbb {E}_{\mathcal {I}}\Big [ \mathbb {E}_{\theta }\big [\mathrm {cov}_x(x|\theta ,\mathcal {I})\big |\mathcal {I}\big ] \Big ] , \end{aligned}$$
(3.36)
which is equal to the expectation of the first part of (3.34).
We go on to compare the expectation of their second parts. From (3.4), due to \(\mathrm {cov}( \mathrm{{e}}[X|Y] ) \ge 0\), we have \(\mathrm {cov}(X) \ge \mathrm{{e}}[\mathrm {cov}(X|Y)]\). By substituting X with \(\mathbb {E}_x[x|\theta ,\mathcal {I}]\) and Y with \(\mathcal {I}\), we have
$$\begin{aligned} \mathrm {cov}_{\theta }\big ( \mathbb {E}_x[x|\theta ,\mathcal {I}] \big ) \ge \mathbb {E}_{\mathcal {I}}\Big [ \mathrm {cov}_{\theta }\big ( \mathbb {E}_x[x|\theta ,\mathcal {I}] \big |\mathcal {I}\big ) \Big ]. \end{aligned}$$
In the preceding inequity, the left part is a function of random variable \(\mathcal {I}\), and the right part is a constant value. Taking the expectation to the both sides, the sign of the inequity remains.
$$\begin{aligned} \mathbb {E}_{\mathcal {I}}\Big [ \mathrm {cov}_{\theta }\big ( \mathbb {E}_x[x|\theta ,\mathcal {I}] \big ) \Big ]&\ge \mathbb {E}_{\mathcal {I}}\Big [ \mathrm {cov}_{\theta }\big ( \mathbb {E}_x[x|\theta ,\mathcal {I}] \big |\mathcal {I}\big ) \Big ]. \end{aligned}$$
(3.37)
In (3.37) the left part equates to the expectation of the second term of (3.35), and the right part equates to the expectation of the second term of (3.34). Combining (3.34)–(3.37), \(\mathbb {E}_{\mathcal {I}}[\tilde{P}] \ge \mathbb {E}_{\mathcal {I}}[P]\) is proved.
(ii): Proof of \(\mathbb {E}_{\mathcal {I}, y} [P_1] \le \mathbb {E}_{\mathcal {I}, y} [\breve{P}]\)
We firstly prove that \(P = \mathbb {E}_{y}[P_1]\) and \(\tilde{P}= \mathbb {E}_{y}[\breve{P}]\). Since \(P=\mathbb {E}_{y}[P|y]\) always holds, by (3.30) and (3.31)
$$\begin{aligned} P ={}&\mathbb {E}_{y}[P|y] \nonumber \\ ={}&\mathbb {E}_{y}\Big [ \int _\theta \left( \int _x x^2 p(x|\theta ,\mathcal {I})\text{ d }x \right) p(\theta |\mathcal {I}) \text{ d }\theta \nonumber \\ {}&-\left( \int _\theta \left( \int _x x p(x|\theta ,\mathcal {I})\text{ d }x \right) p(\theta |\mathcal {I}) \text{ d }\theta \right) ^2 \big |y\Big ]\nonumber \\ ={}&\mathbb {E}_{y}\Big [ \int _\theta \left( \int _x x^2 p(x|\theta ,\mathcal {I},y)\text{ d }x \right) p(\theta |\mathcal {I},y) \text{ d }\theta \end{aligned}$$
(3.38)
$$\begin{aligned} {}&-\left( \int _\theta \left( \int _x x p(x|\theta ,\mathcal {I},y)\text{ d }x \right) p(\theta |\mathcal {I},y) \text{ d }\theta \right) ^2 \Big ] \\ ={}&\mathbb {E}_{y}[P_1]. \nonumber \end{aligned}$$
(3.39)
By (3.2), it is easy to check that in (3.38) and (3.39) the terms within the brackets of \(\mathbb {E}_{y}[\cdot ]\) equates \(P_1\). Following the same derivations above,
$$\begin{aligned} \tilde{P}={}&\mathbb {E}_{y}[\tilde{P}|y] \nonumber \\ ={}&\mathbb {E}_{y}\Big [ \int _\theta \left( \int _{\tilde{x}} \tilde{x}^2 p(\tilde{x}|\theta ,\mathcal {I})\text{ d }\tilde{x}\right) p(\theta ) \text{ d }\theta \nonumber \\ {}&-\left( \int _\theta \left( \int _{\tilde{x}} \tilde{x}p(\tilde{x}|\theta ,\mathcal {I})\text{ d }\tilde{x}\right) p(\theta ) \text{ d }\theta \right) ^2 \big |y\Big ]\nonumber \\ ={}&\mathbb {E}_{y}\Big [ \int _\theta \left( \int _{\tilde{x}} \tilde{x}^2 p(\tilde{x}|\theta ,\mathcal {I},y)\text{ d }\tilde{x}\right) p(\theta |y) \text{ d }\theta \end{aligned}$$
(3.40)
$$\begin{aligned} {}&-\left( \int _\theta \left( \int _{\tilde{x}} \tilde{x}p(\tilde{x}|\theta ,\mathcal {I},y)\text{ d }\tilde{x}\right) p(\theta |y) \text{ d }\theta \right) ^2 \Big ] \\ ={}&\mathbb {E}_{y}[\breve{P}]. \nonumber \end{aligned}$$
(3.41)
By viewing \(\tilde{x}\) in (3.40) and (3.41) as \(\breve{x}\), it still follows (3.2) that \(\breve{P}\) equates the terms within the brackets of \(\mathbb {E}_{y}[\cdot ]\) in (3.40) and (3.41). Due to the result \(\mathbb {E}_{\mathcal {I}}[P] \le \mathbb {E}_{\mathcal {I}}[\tilde{P}]\) as proved in (i),
$$\begin{aligned} \mathbb {E}_{\mathcal {I},y} [P_1] = \mathbb {E}_{\mathcal {I}}\Big [ \mathbb {E}_{y}[P_1] \Big ] = \mathbb {E}_{\mathcal {I}}[P] \le \mathbb {E}_{\mathcal {I}}[\tilde{P}] = \mathbb {E}_{\mathcal {I}}\Big [ \mathbb {E}_{y}[\breve{P}] \Big ] = \mathbb {E}_{\mathcal {I},y} [\breve{P}]. \end{aligned}$$
Thus, \(\mathbb {E}_{\mathcal {I},y} [P_1] \le \mathbb {E}_{\mathcal {I},y} [\breve{P}]\) is proved.
