# Super-Resolution Reconstruction of Depth Image Based on Kriging Interpolation
## 2 Depth Image super-resolution Algorithm
### Interpolation Algorithm Based on Kriging Interpolation
That is, for any point \((\mathrm{x},\mathrm{y})\), we have  
\[\mathrm{E}[\mathrm{z}(\mathrm{x},\mathrm{y})]=\mathrm{E}[\mathrm{z}]=\mathrm{c} \tag{4}\]  
\[\mathrm{Var}[\mathrm{z}(\mathrm{x},\mathrm{y})]=\sigma^{2} \tag{5}\]  
In other words, the value at any point \(\mathrm{z}\)(x,y) is composed of the regional mean \(\mathrm{c}\) and the random deviation at that point \(\mathrm{R}\)(x,y):  
\[\mathrm{z}(\mathrm{x},\mathrm{y})=\mathrm{E}[\mathrm{z}(\mathrm{x},\mathrm{y} )]+\mathrm{R}(\mathrm{x},\mathrm{y})=\mathrm{c}+\mathrm{R}(\mathrm{x},\mathrm{ y}) \tag{6}\]  
where \(\mathrm{R}\)(x,y) denotes the deviations at the point (x,y), and their variances are all constants. \[\mathrm{Var}[\mathrm{z}(\mathrm{x},\mathrm{y})]=\sigma^{2} \tag{7}\]  
First, we analyse the unbiased estimation condition:  
\[\mathrm{E}(\mathrm{z}_{\mathrm{o}}-\mathrm{z}_{\mathrm{o}})=0 \tag{8}\]  
Substitute  
\[\mathrm{\hat{z}_{\mathrm{o}}}=\sum\limits_{i=1}^{n}\lambda_{i}\mathrm{z}_{i} \tag{9}\]  
We obtain  
\[\mathrm{E}\left(\sum\limits_{i=1}^{n}\lambda_{i}\mathrm{z}_{i}-\mathrm{z}_{ \mathrm{o}}\right)=0 \tag{10}\]  
Furthermore, because \(\mathrm{E}[\mathrm{z}]=\mathrm{c}\) for any \(\mathrm{z}\), one of the constraints on \(\lambda_{i}\) can be obtained. \[\sum\limits_{i=1}^{n}\lambda_{i}=1 \tag{11}\]  
The analysis estimation error \(\mathrm{Var}(\mathrm{\hat{z}_{\mathrm{o}}}-\mathrm{z}_{\mathrm{o}})\) is denoted here by \(J\):  
\[\mathrm{E}(\mathrm{z}_{\mathrm{o}}-\mathrm{z}_{\mathrm{o}})=0 \tag{3}\]0  
Furthermore, we have  
\[\mathrm{E}(\mathrm{z}_{\mathrm{o}}-\mathrm{z}_{\mathrm{o}})=0 \tag{3}\]1  
\[\mathrm{E}(\mathrm{z}_{\mathrm{o}}-\mathrm{z}_{\mathrm{o}})=0 \tag{3}\]2  
By definition, \(\mathrm{C}_{\mathrm{\hat{y}}}=\mathrm{Cov}\left(\mathrm{z}_{i},\mathrm{z}_{i} \right)=\mathrm{Cov}\left(\mathrm{R}_{i},\mathrm{R}_{j}\right)\), where \(\mathrm{R}_{i}=\mathrm{z}_{i}-\mathrm{c}\) denotes the deviation of the attribute value at point \((\mathrm{x}_{i},\mathrm{y}_{i})\) from the regional average attribute value  
\[\mathrm{E}(\mathrm{z}_{\mathrm{o}}-\mathrm{z}_{\mathrm{o}})=0 \tag{3}\]3Defining the semivariance function \(\mathrm{r_{ij}}=\sigma^{2}-\mathrm{C_{ij}}\), which is substituted into \(\mathrm{J}\), we have the following:  
\[ \begin{array}{c}\mathrm{J}=\sum\limits_{i=1}^{n}\sum\limits_{j=0}^{n}\lambda_ {i}\lambda_{j}\big{(}\sigma^{2}-\mathrm{r_{ij}}\big{)}-2\sum\limits_{i=1}^{n} \lambda_{i}\big{(}\sigma^{2}-\mathrm{r_{io}}\big{)}+\big{(}\sigma^{2}-\mathrm{ r_{oo}}\big{)}\\ =\sum\limits_{i=1}^{n}\sum\limits_{j=0}^{n}\lambda_{i}\lambda_{j}\big{(}\sigma ^{2}\big{)}-\sum\limits_{i=1}^{n}\sum\limits_{j=0}^{n}\lambda_{i}\lambda_{j} \big{(}\mathrm{r_{ij}}\big{)}-2\sum\limits_{i=1}^{n}\lambda_{i}\big{(}\sigma ^{2}\big{)}+2\sum\limits_{i=1}^{n}\lambda_{i}\big{(}\mathrm{r_{io}}\big{)}+ \big{(}\sigma^{2}-\mathrm{r_{oo}}\big{)}\end{array}  \tag{15}\]  
Given that \(\sum_{i=1}^{n}\lambda_{i}=1\), after substitution we have  
\[ \begin{array}{c}\mathrm{J}=\sigma^{2}-\sum\limits_{i=1}^{n}\sum\limits_{j=0 }^{n}\lambda_{i}\lambda_{j}\big{(}\mathrm{r_{ij}}\big{)}-2\sigma^{2}+2\sum \limits_{i=1}^{n}\lambda_{i}\big{(}\mathrm{r_{io}}\big{)}+\big{(}\sigma^{2}- \mathrm{r_{oo}}\big{)}\\ =2\sum\limits_{i=1}^{n}\lambda_{i}\big{(}\mathrm{r_{io}}\big{)}-\sum\limits_{ i=1}^{n}\sum\limits_{j=0}^{n}\lambda_{i}\lambda_{j}\big{(}\mathrm{r_{ij}}\big{)}- \mathrm{r_{oo}}\end{array}  \tag{16}\]  
The goal is to find a set of \(\lambda_{i}\) that minimises \(\mathrm{J}\). As \(\mathrm{J}\) is a function of \(\lambda_{i}\), we can directly calculate the partial derivative of \(\mathrm{J}\) with respect to \(\lambda_{i}\), and find the zero-valued point. That is, we solve the following for \(\lambda_{i}\):  
\[\frac{\partial\mathrm{J}}{\partial\lambda_{i}}=0;\;\mathrm{i}=1,2,3,\ldots\; \mathrm{n} \tag{17}\]  
It is also necessary to satisfy \(\sum_{i=1}^{n}\lambda_{i}=1\). Additionally, the optimisation problem with this constraint is solved using the Lagrange multiplier method by first constructing a new objective function.