# Dynamic Indoor Localization Using Maximum Likelihood Particle Filtering
## Appendix C Proofs  
**Proof of Lemma 1. ** We have  
\[p(x_{t}|Z_{t})=p(\theta_{t}|y_{t},Z_{t})p(y_{t}|Z_{t}).\]  
Now  
\[p(y_{t}|Z_{t}) =\frac{1}{p(z_{t}|Z_{t-1})}p(y_{t},z_{t}|Z_{t-1})\] \[\propto p(z_{t}|y_{t})p(y_{t}|Z_{t-1}),\]and  
\[p(\theta_{t}|y_{t},Z_{t}) \propto p(\theta_{t},y_{t}|Z_{t})\] \[\propto p(\theta_{t},y_{t},z_{t}|Z_{t-1})\] \[\propto p(z_{t}|\theta_{t},y_{t})p(\theta_{t},y_{t}|Z_{t-1})\] \[\propto p(z_{t}|y_{t})p(\theta_{t},y_{t}|Z_{t-1})\] \[\propto p(\theta_{t},y_{t}|Z_{t-1})\] \[\propto p(\theta_{t}|y_{t},Z_{t-1}).\]  
Proof of Lemma 2.: We split the proof in steps:  
1. It follows straightforwardly from [34] (SS 5.4) that \[p\Big{(}x_{t}|x_{t-1|t-1}^{i}\Big{)}=\sum_{w:x_{t}=f_{t-1}(w)}\Big{|}\text{det} \mathbf{J}_{t-1}^{i}(w)\Big{|}^{-1}p(w_{t-1}=w).\] Since the variance of \(n_{t-1}\) is small, we can do the following approximation, \[p(y_{t}|Z_{t}) =\frac{1}{p(z_{t}|Z_{t-1})}p(y_{t},z_{t}|Z_{t-1})\]0 \[p(y_{t}|Z_{t}) =\frac{1}{p(z_{t}|Z_{t-1})}p(y_{t},z_{t}|Z_{t-1})\]1 (A1) where to simplify the notation we used \(f_{t-1}^{-i}=\big{(}f_{t-1}^{i}\big{)}^{-1}\). 2. We have \[f_{t-1}^{-i}(x_{t})= \left[ \begin{array}{c}\delta_{t-1}^{-i}(y_{t})\\ \frac{\theta_{t-1}-\delta_{t-1}^{i}(y_{t})-\delta_{t-1}^{-i}\delta_{t-1}^{-i}( y_{t})\theta_{t-1}}{\theta_{t-1}^{i}\delta_{t-1}^{-i}(y_{t})\theta_{t-1}}\\ \end{array} \right]\] (A2) \[= v_{t}^{i}(y_{t})+\xi_{t}^{i}(y_{t})\theta_{t}.\] (A3) Then, from (A2), \[\mathbf{J}_{t-1}^{-i}(x_{t})=\left[ \begin{array}{c}\mathbf{J}_{t-1}^{-i}(y _{t})\\ \ast\end{array} \right.\frac{1}{b_{t-1}^{i}\delta_{t-1}^{-i}(y_{t})}\left. \right],\] where \(\ast\) denotes a non-zero term whose value is irrelevant. Then \[\Big{|}\text{det}\mathbf{J}_{t-1}^{-i}(x_{t})\Big{|}=\left|\frac{\text{det} \mathbf{J}_{t-1}^{-i}(y_{t})}{b_{t-1}^{i}\circ\delta_{t-1}^{-i}(y_{t})}\right|.\] (A4)
3. Now, using (10) and (A1), we get \[p(y_{t},\theta_{t}|Z_{t-1})= p(x_{t}|Z_{t-1})\] \[= \int p(x_{t}|x_{t-1})p(x_{t-1}|Z_{t-1})dy_{t-1}\theta_{t-1}\] \[\simeq \sum_{i=1}^{I}\alpha_{t-1|t-1}^{i}p\Big{(}x_{t}|x_{t-1|t-1}^{i} \Big{)}\] \[\simeq \sum_{i=1}^{I}\alpha_{t-1|t-1}^{i}\Big{|}\text{det}\mathbf{J}_{t -1}^{-i}(x_{t})\Big{|}\mathcal{N}\Big{(}f_{t-1}^{-i}(x_{t});0,Q\Big{)}.\] (A5) But, from (A3), \[\mathcal{N}\Big{(}f_{t-1}^{-i}(x_{t});0,Q\Big{)}=\mathcal{N}\Big{(}v_{t}^{i}(y _{t})+\xi_{t}^{i}(y_{t})\theta_{t};0,Q\Big{)}.\]Now  
\[\Big{(}v_{t}^{i}(y_{t})+c_{t}^{i}(y_{t})\theta_{t}\Big{)}^{\top}Q^{-1} \Big{(}v_{t}^{i}(y_{t})+c_{t}^{i}(y_{t})\theta_{t}\Big{)}\] \[= \xi_{t}^{i}(y_{t})^{\top}Q^{-1}\xi_{t}^{i}(y_{t})\theta_{t}^{2}+2v_ {t}^{i}(y_{t})^{\top}Q^{-1}\xi_{t}^{i}(y_{t})\theta_{t}+v_{t}^{i}(y_{t})^{\top }Q^{-1}v_{t}^{i}(y_{t})\] \[= \xi_{t}^{i}(y_{t})^{\top}Q^{-1}\xi_{t}^{i}(y_{t})\theta_{t}^{2}-2c _{t}^{i}(y_{t})^{\top}Q^{-1}\xi_{t}^{i}(y_{t})\mu_{t}^{i}(y_{t})\theta_{t}\] \[+\xi_{t}^{i}(y_{t})^{\top}Q^{-1}\xi_{t}^{i}(y_{t})\mu_{t}^{i}(y_{ t})^{2}+v_{t}^{i}(y_{t})^{\top}Q^{-1}v_{t}^{i}(y_{t})-\xi_{t}^{i}(y_{t})^{ \top}Q^{-1}\xi_{t}^{i}(y_{t})\mu_{t}^{i}(y_{t})^{2}\] \[p(y_{t},\theta_{t}|Z_{t-1})= p(x_{t}|Z_{t-1})\]0 \[p(y_{t},\theta_{t}|Z_{t-1})= p(x_{t}|Z_{t-1})\]1  
Hence  
\[p(y_{t},\theta_{t}|Z_{t-1})= p(x_{t}|Z_{t-1})\]2 \[p(y_{t},\theta_{t}|Z_{t-1})= p(x_{t}|Z_{t-1})\]3  
4. Putting (A4) and (A6) into (A5) we obtain \[p(y_{t},\theta_{t}|Z_{t-1})= p(x_{t}|Z_{t-1})\]4 and the result follows. Proof of Lemma 3.: If \(x_{t-1}=x_{t-1|t-1}^{i}\), doing a first order Taylor expansion we obtain  
\[p(y_{t},\theta_{t}|Z_{t-1})= p(x_{t}|Z_{t-1})\]5 \[p(y_{t},\theta_{t}|Z_{t-1})= p(x_{t}|Z_{t-1})\]6  
So  
\[p(y_{t},\theta_{t}|Z_{t-1})= p(x_{t}|Z_{t-1})\]7  
The result then follows since  
\[p(y_{t},\theta_{t}|Z_{t-1})= p(x_{t}|Z_{t-1})\]8 \[p(y_{t},\theta_{t}|Z_{t-1})= p(x_{t}|Z_{t-1})\]9  
Proof of Lemma 4.: Let  
\[= \int p(x_{t}|x_{t-1})p(x_{t-1}|Z_{t-1})dy_{t-1}\theta_{t-1}\]0  
Recall that \(N\) is the dimension of \(z_{t}\).