# Point Cloud Denoising Algorithm via Geometric Metrics on the Statistical Manifold
## Appendix B Proof of Proposition 4  
Let \(F(U)\) be the objective function in order to minimize \(m\) positive-definite matrices as follows  
\[F(U):=\frac{1}{m}\sum_{i=1}^{m}\text{tr}\bigg{(}U_{i}^{-1}U-\log(U_{i}^{-1}U) -I\bigg{)}. \tag{15}\]  
Suppose \(Z(\tau):=U+\tau A\), then the gradient of \(F(U)\) can be computed as follows by (12)  
\[\langle\
abla F(U),A\rangle = \left.\frac{\mathrm{d}}{\mathrm{d}\tau}\right|_{\tau=0}F(U+\tau A)\] \[= \left.\frac{\mathrm{d}}{\mathrm{d}\tau}\right|_{\tau=0}\frac{1}{ m}\sum_{i=1}^{m}\text{tr}\bigg{(}U_{i}^{-1}Z(\tau)-\log(U_{i}^{-1}Z(\tau))-I \bigg{)}\] \[= \frac{1}{m}\sum_{i=1}^{m}\text{tr}\bigg{(}U_{i}^{-1}A\bigg{)}- \frac{1}{m}\sum_{i=1}^{m}\text{tr}\bigg{(}\frac{\mathrm{d}}{\mathrm{d}\tau} \bigg{|}_{\tau=0}\log(U_{i}^{-1}U)\bigg{)}.\]  
The rest is to calculate the differential term of the above identity. Using Lemmas 1-3, we obtain  
\[\text{tr}(\log Z(\tau)\frac{\mathrm{d}}{\mathrm{d}\tau}\log(U_{i }^{-1}Z(\tau)))\] \[= \int_{0}^{1}\text{tr}\bigg{(}[(U_{i}^{-1}Z(\tau)-I)t+I]^{-1}\frac {d(U_{i}^{-1}Z(\tau))}{d\tau}[(U_{i}^{-1}Z(\tau)-I)t+I]^{-1}\bigg{)}dt\] \[= \int_{0}^{1}\text{tr}\bigg{(}[(U_{i}^{-1}Z(\tau)-I)t+I]^{-2}dtU_{ i}^{-1}A\bigg{)}\] \[= \text{tr}\bigg{(}(U_{i}^{-1}U)^{-1}U_{i}^{-1}A\bigg{)}\] \[= \text{tr}\bigg{(}U^{-1}A\bigg{)}.\]Consequently, we obtain  
\[\
abla F(U)=\text{tr}\bigg{(}\frac{1}{m}\sum_{i=1}^{m}(U_{i}^{-1}-\overline{U}^{- 1})\bigg{)}. \tag{101}\]  
Let \(G(U)\) be the objective function to minimize the \(m+l\) symmetry positive-definite matrices as follows  
\[G(U):= (1-\tau)\frac{1}{m}\sum_{i=1}^{m}\text{tr}\bigg{(}U_{i}^{-1}U- \log(U_{i}^{-1}U)-I\bigg{)}\] \[+\tau\frac{1}{l}\sum_{j=1}^{l}\text{tr}\bigg{(}R_{j}^{-1}U-\log( R_{j}^{-1}U)-I\bigg{)},\]  
then we have  
\[\
abla G(U)=(1-\tau)\frac{1}{m}\sum_{i=1}^{m}\text{tr}(U_{i}^{-1}-U^{-1})+\tau \frac{1}{l}\sum_{j=1}^{l}\text{tr}(R_{j}^{-1}-U^{-1}). \tag{102}\]  
As \(\
abla G(\widehat{U})=0\), we have  
\[(1-\tau)\frac{1}{m}\sum_{i=1}^{m}\text{tr}(U_{i}^{-1}-\widehat{U}^{-1})+\tau \frac{1}{l}\sum_{j=1}^{l}\text{tr}(R_{j}^{-1}-\widehat{U}^{-1})=0. \tag{103}\]  
with the aim of obtaining the linear term about \(\tau\), this study differentiates (103) and sets \(\tau=0\). Thus, we obtain  
\[\frac{1}{l}\sum_{j=1}^{l}\text{tr}(R_{j}^{-1}-\overline{U}^{-1})+\frac{1}{m} \sum_{i=1}^{m}\frac{\text{d}}{\text{d}\tau}\bigg{|}_{\tau=0}\text{tr}(U_{i}^{ -1}-\widehat{U}^{-1})=0. \tag{104}\]  
Note that  
\[0=\frac{\text{d}}{\text{d}\tau}I=\frac{\text{d}}{\text{d}\tau}(\widehat{U} \widehat{U}^{-1})=\bigg{(}\frac{\text{d}}{\text{d}\tau}\widehat{U}\bigg{)} \widehat{U}^{-1}+\widehat{U}\frac{\text{d}}{\text{d}\tau}\widehat{U}^{-1}, \tag{105}\]  
consequently,  
\[\frac{\text{d}}{\text{d}\tau}\bigg{|}_{\tau=0}\widehat{U}^{-1}=-\widehat{U} ^{-1}\frac{\text{d}}{\text{d}\tau}\widehat{U}\bigg{|}_{\tau=0}\widehat{U}^{- 1}=-\widehat{U}^{-1}H\widehat{U}^{-1}. \tag{106}\]  
Next, we substitute (106) into (104) and take the trace of Equation (104) to obtain the influence function of the KLD mean:  
\[H=\overline{U}-\frac{1}{l}\sum_{j=1}^{l}\overline{U}R_{j}^{-1}\overline{U}. \tag{107}\]  
\(\square\)