The below is some part of this thesis.
https://kaiminghe.github.io/publications/pami12guidedfilter.pdf.
I could not understand how equation (4) is calculated. I know that I need to do “partial differentiation $=0$” of (2). I tried to get equation (4). But the result is not same. Please let me know how the equation (4) is calculated from (2) and (3).
Now we define the guided filter. The key assumption of the guided filter is a local linear model between the guidance $I$ and the filtering output $q$. We assume that $q$ is a linear transform of $I$ in a window $\omega_k$ centered at the pixel $k$:
$$q_i=a_k I_i+b_k, \forall i \in \omega_k \tag1$$
where $a_k$, $b_k$ are some linear coefficients assumed to be constant in $\omega_k$.
We seek a solution that minimizes the difference between $q$ and $p$ while maintaining the linear model. Specifically, we minimize the following cost function in the window $\omega_k$:
$$E\left(a_k, b_k\right)=\sum_{i \in \omega_k}\left(\left(a_k I_i+b_k-p_i\right)^2+\epsilon a_k^2\right) \tag2$$
Here, $\epsilon$ is a regularization parameter penalizing large $a_k$.
This equation is the linear ridge regression model and its solution is given by
$$b_k=\bar{p}_k-a_k \mu_k \tag3$$
$$a_k=\frac{\frac{1}{|\omega|} \sum_{i \in \omega_k} I_i p_i-\mu_k \bar{p}_k}{\sigma_k^2+\epsilon} \tag4$$
Here, $\mu_k$ and $\sigma_k^2$ are the mean and variance of $I$ in $\omega_k,|\omega|$ is the number of pixels in $\omega_k$, and $\bar{p}_k=\frac{1}{|\omega|} \sum_{i \in \omega_k} p_i$ is the mean of $p$ in $\omega_k$. We obtained the linear coefficients $\left(a_k, b_k\right)$.
I solved it by myself. By partial differentiation of equation 2,
$\begin{aligned} & \sum_{i \in \omega_k}\left\{2 I_i\left(a_k I_i+b_k-p_i\right)+2 \varepsilon a_k\right\}=0 \\ & \sum_{i \in \omega_k}\left\{I_i\left(a_k I_i+b_k-p_i\right)+\varepsilon a_k\right\}=0 \\ & \sum_{i \in \omega_k}\left\{I_i\left(a_k I_i+b_k\right)+\varepsilon a_k\right\}=\sum_{i \in \omega_k} I_i p_i \\ & b_k=\bar{p}_k-a_k \mu_k \\ & \sum_{i \in \omega_k}\left\{I_i\left(a_k I_i+\bar{p}_k-a_k \mu_k\right)+\varepsilon a_k\right\}=\sum_{i \in \omega_k} I_i p_i \\ & \sum_{i \in \omega_k}\left\{I_i\left(a_k I_i-a_k \mu_k\right)+\varepsilon a_k\right\}=\sum_{i \in \omega_k} I_i p_i-\sum_{i \in \omega_k} I_i \bar{p}_k \\ & a_k \sum_{i \in \omega_k}\left\{I_i\left(I_i-\mu_k\right)+\varepsilon\right\}=\sum_{i \in \omega_k} I_i p_i-\sum_{i \in \omega_k} I_i \bar{p}_k \\ & a_k=\frac{\frac{1}{|\omega|}\left(\sum_{i \in \omega_k} I_i p_i-\sum_{i \in \omega_k} I_i \bar{p}_k\right)}{\frac{1}{|\omega|} \sum_{i \in \omega_k}\left\{I_i\left(I_i-\mu_k\right)+\varepsilon\right\}} \\ & a_k=\frac{\frac{1}{|\omega|}\left(\sum_{i \in \omega_k} I_i p_i-\sum_{i \in \omega_k} I_i \bar{p}_k\right)}{\frac{1}{|\omega|} \sum_{i \in \omega_k}\left\{I_i\left(I_i-\mu_k\right)+\varepsilon\right\}}=\frac{\frac{\sum_{i \in \omega_k} I_i p_i}{|\omega|}-\bar{p}_k \frac{\sum_{i \in \omega_k} I_i}{|\omega|}}{\frac{\sum_{i \in \omega_k} I_i^2}{|\omega|}-\mu_k \frac{\sum_{i \in \omega_k} I_i}{|\omega|}+\frac{\Sigma_{i \in \omega_k} \varepsilon}{|\omega|}} \\ & a_k=\frac{\frac{1}{|\omega|} \sum_{i \in \omega_k} I_i p_i-\bar{p}_k \mu_k}{\frac{\sum_{i \in \omega_k} I_i^2}{|\omega|}-\mu_k^2+\varepsilon}=\frac{\frac{1}{|\omega|} \sum_{i \in \omega_k} I_i p_i-\mu_k \bar{p}_k}{\sigma_k^2+\varepsilon} \\ & \sigma_k^2=\frac{\sum_{i \in \omega_k} I_i^2}{|\omega|}-\mu_k^2 \\ & \end{aligned}$
$\sigma_k^2=\ \frac{\sum_{i\in\omega_k}\left(I_i-\mu_k\right)^2}{\left|\omega\right|} \\ = \frac{\sum_{i\in\omega_k}{(I_i^2-{2\mu}_kI_i+\mu_k^2)}}{\left|\omega\right|} \\ = \frac{\sum_{i\in\omega_k}\ I_i^2}{\left|\omega\right|}\ -\frac{{2\mu}_k\sum_{i\in\omega_k}\ I_i}{\left|\omega\right|} + \frac{\sum_{i\in\omega_k}\mu_k^2}{\left|\omega\right|} \\ = \frac{\sum_{i\in\omega_k}\ I_i^2}{\left|\omega\right|}\ -{2\mu}_k^2+ \mu_k^2 \\ = \frac{\sum_{i\in\omega_k}\ I_i^2}{\left|\omega\right|}\ -\mu_k^2 \\ \sigma_k^2=\frac{\sum_{i\in\omega_k}{(I_i-\mu_k)}^2}{\left|\omega\right|}=\frac{\sum_{i\in\omega_k}\ I_i^2}{\left|\omega\right|}-\mu_k^2 \\$
It was proved.