# CSR-DCF视频目标跟踪论文笔记（2）——关于滤波器Learning的推导（Augmented Lagrangian方法）

2020年08月04日 阅读数：1502

# 2. 滤波器求解目标函数的构建

$\begin{array}{}\text{(1)}& \begin{array}{c}\underset{\mathbf{h}}{\mathrm{arg}min}\underset{d=1}{\overset{{N}_{d}}{\sum }}\left({‖{\mathbf{f}}_{d}\odot {\mathbf{h}}_{d}-\mathbf{g}‖}^{2}+\lambda {‖{\mathbf{h}}_{d}‖}^{2}\right)\\ =\underset{\mathbf{h}}{\mathrm{arg}min}\underset{d=1}{\overset{{N}_{d}}{\sum }}\left({‖{\stackrel{\mathbf{^}}{\mathbf{h}}}_{d}^{H}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left({\stackrel{\mathbf{^}}{\mathbf{f}}}_{d}\right)-{\stackrel{^}{g}}_{d}‖}^{2}+\lambda {‖{\stackrel{\mathbf{^}}{\mathbf{h}}}_{d}‖}^{2}\right)\end{array}\end{array}$

git

$\begin{array}{}\text{(2)}& \begin{array}{c}\underset{\mathbf{h}}{\mathrm{arg}min}{‖\mathbf{f}\odot \mathbf{h}-\mathbf{g}‖}^{2}+\lambda {‖\mathbf{h}‖}^{2}\\ =\underset{\mathbf{h}}{\mathrm{arg}min}{‖{\stackrel{\mathbf{^}}{\mathbf{h}}}^{H}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\stackrel{\mathbf{^}}{\mathbf{f}}\right)-\stackrel{^}{g}‖}^{2}+\lambda {‖\stackrel{\mathbf{^}}{\mathbf{h}}‖}^{2}\end{array}\end{array}$

$\begin{array}{}\text{(3)}& {\mathbf{h}}_{c}-{\mathbf{h}}_{m}=0\end{array}$

github

$\begin{array}{}\text{(4)}& \begin{array}{l}\underset{{\mathbf{h}}_{c},{\mathbf{h}}_{m}}{\mathrm{arg}min}{‖{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}^{H}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\stackrel{\mathbf{^}}{\mathbf{f}}\right)-\stackrel{^}{g}‖}^{2}+\frac{\lambda }{2}{‖{\stackrel{\mathbf{^}}{\mathbf{h}}}_{m}‖}^{2}\\ s.t.\phantom{\rule{thickmathspace}{0ex}}{\mathbf{h}}_{c}-{\mathbf{h}}_{m}=0\end{array}\end{array}$

# 3. 构建Lagrange表达式

$\begin{array}{}\text{(5)}& \mathcal{L}\left({\stackrel{\mathbf{^}}{\mathbf{h}}}_{c},\mathbf{h},\stackrel{\mathbf{^}}{\mathbf{I}}|\mathbf{m}\right)={‖{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}^{H}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\stackrel{\mathbf{^}}{\mathbf{f}}\right)-\stackrel{^}{g}‖}^{2}+\frac{\lambda }{2}{‖{\mathbf{h}}_{m}‖}^{2}+\left[{\stackrel{\mathbf{^}}{\mathbf{I}}}^{H}\left({\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}-{\stackrel{\mathbf{^}}{\mathbf{h}}}_{m}\right)+\stackrel{¯}{{\stackrel{\mathbf{^}}{\mathbf{I}}}^{H}\left({\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}-{\stackrel{\mathbf{^}}{\mathbf{h}}}_{m}\right)}\right]+\mu {‖{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}-{\stackrel{\mathbf{^}}{\mathbf{h}}}_{m}‖}^{2}\end{array}$

$\begin{array}{}\text{(6)}& \mathcal{L}\left({\stackrel{\mathbf{^}}{\mathbf{h}}}_{c},\mathbf{h},\stackrel{\mathbf{^}}{\mathbf{I}}|\mathbf{m}\right)={‖{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}^{H}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\stackrel{\mathbf{^}}{\mathbf{f}}\right)-\stackrel{^}{g}‖}^{2}+\frac{\lambda }{2}{‖{\mathbf{h}}_{m}‖}^{2}+\left[{\stackrel{\mathbf{^}}{\mathbf{I}}}^{H}\left({\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)+\stackrel{¯}{{\stackrel{\mathbf{^}}{\mathbf{I}}}^{H}\left({\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)}\right]+\mu {‖{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}‖}^{2}\end{array}$

$\begin{array}{}\text{(7)}& \mathcal{L}\left({\stackrel{\mathbf{^}}{\mathbf{h}}}_{c},\mathbf{h},\stackrel{\mathbf{^}}{\mathbf{I}}\right)={\mathcal{L}}_{1}+{\mathcal{L}}_{2}+{\mathcal{L}}_{3}+{\mathcal{L}}_{4}\end{array}$

$\begin{array}{}\text{(8)}& \left\{\begin{array}{l}{\mathcal{L}}_{1}={‖{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}^{H}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\stackrel{\mathbf{^}}{\mathbf{f}}\right)-\stackrel{^}{g}‖}^{2}=\left({\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}^{H}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\stackrel{\mathbf{^}}{\mathbf{f}}\right)-\stackrel{\mathbf{^}}{\mathbf{g}}\right){\stackrel{¯}{\left({\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}^{H}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\stackrel{\mathbf{^}}{\mathbf{f}}\right)-\stackrel{\mathbf{^}}{\mathbf{g}}\right)}}^{T}\\ {\mathcal{L}}_{2}=\frac{\lambda }{2}{‖{\mathbf{h}}_{m}‖}^{2}\\ {\mathcal{L}}_{3}={\stackrel{\mathbf{^}}{\mathbf{I}}}^{H}\left({\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)+\stackrel{¯}{{\stackrel{\mathbf{^}}{\mathbf{I}}}^{H}\left({\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)}\\ {\mathcal{L}}_{4}=\mu {‖{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}‖}^{2}=\mu \left({\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right){\stackrel{¯}{\left({\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)}}^{T}\end{array}\end{array}$

# 4. 开始优化，首先对h_c求偏导数

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$\begin{array}{}\text{(9)}& \begin{array}{l}{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}^{\mathrm{o}\mathrm{p}\mathrm{t}}=\underset{{\mathbf{h}}_{c}}{\mathrm{arg}min}L\left({\stackrel{\mathbf{^}}{\mathbf{h}}}_{c},\mathbf{h},\stackrel{\mathbf{^}}{\mathbf{I}}\right)\\ {\mathbf{h}}^{\mathrm{o}\mathrm{p}\mathrm{t}}=\underset{\mathbf{h}}{\mathrm{arg}min}L\left({\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}^{\mathrm{o}\mathrm{p}\mathrm{t}},\mathbf{h},\stackrel{\mathbf{^}}{\mathbf{I}}\right)\end{array}\end{array}$

$\begin{array}{}\text{(10)}& {\mathrm{\nabla }}_{\stackrel{¯}{{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}}}{\mathcal{L}}_{1}+{\mathrm{\nabla }}_{\stackrel{¯}{{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}}}{\mathcal{L}}_{2}+{\mathrm{\nabla }}_{\stackrel{¯}{{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}}}{\mathcal{L}}_{3}+{\mathrm{\nabla }}_{\stackrel{¯}{{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}}}{\mathcal{L}}_{4}\equiv 0\end{array}$

$\begin{array}{}\text{(11)}& \begin{array}{rl}{\mathrm{\nabla }}_{\stackrel{¯}{{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}}}{\mathcal{L}}_{1}& =\frac{\mathrm{\partial }}{\mathrm{\partial }\stackrel{¯}{{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}}}\left[\left({\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}^{H}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\stackrel{\mathbf{^}}{\mathbf{f}}\right)-\stackrel{\mathbf{^}}{\mathbf{g}}\right){\stackrel{¯}{\left({\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}^{H}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\stackrel{\mathbf{^}}{\mathbf{f}}\right)-\stackrel{\mathbf{^}}{\mathbf{g}}\right)}}^{T}\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\stackrel{¯}{{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}}}\left[{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}^{H}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\stackrel{\mathbf{^}}{\mathbf{f}}\right)\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}{\left(\stackrel{\mathbf{^}}{\mathbf{f}}\right)}^{H}{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}-{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}^{H}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\stackrel{\mathbf{^}}{\mathbf{f}}\right){\stackrel{\mathbf{^}}{\mathbf{g}}}^{H}-\stackrel{\mathbf{^}}{\mathbf{g}}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}{\left(\stackrel{\mathbf{^}}{\mathbf{f}}\right)}^{H}{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}+\stackrel{\mathbf{^}}{\mathbf{g}}{\stackrel{\mathbf{^}}{\mathbf{g}}}^{H}\right]\\ & =\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\stackrel{\mathbf{^}}{\mathbf{f}}\right)\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}{\left(\stackrel{\mathbf{^}}{\mathbf{f}}\right)}^{H}{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}-\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\stackrel{\mathbf{^}}{\mathbf{f}}\right){\stackrel{\mathbf{^}}{\mathbf{g}}}^{H}-0+0\\ & =\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\stackrel{\mathbf{^}}{\mathbf{f}}\right)\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}{\left(\stackrel{\mathbf{^}}{\mathbf{f}}\right)}^{H}{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}-\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\stackrel{\mathbf{^}}{\mathbf{f}}\right){\stackrel{\mathbf{^}}{\mathbf{g}}}^{H}\end{array}\end{array}$

$\begin{array}{}\text{(12)}& \begin{array}{rl}{\mathrm{\nabla }}_{\stackrel{¯}{{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}}}{\mathcal{L}}_{2}& ={\mathrm{\nabla }}_{\stackrel{¯}{{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}}}\left[\frac{\lambda }{2}{‖{\mathbf{h}}_{m}‖}^{2}\right]\\ & =0\end{array}\end{array}$

$\begin{array}{}\text{(13)}& \begin{array}{rl}{\mathrm{\nabla }}_{\stackrel{¯}{{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}}}{\mathcal{L}}_{3}& =\frac{\mathrm{\partial }}{\mathrm{\partial }\stackrel{¯}{{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}}}\left[{\stackrel{\mathbf{^}}{\mathbf{I}}}^{H}\left({\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)+\stackrel{¯}{{\stackrel{\mathbf{^}}{\mathbf{I}}}^{H}\left({\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)}\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\stackrel{¯}{{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}}}\left[{\stackrel{\mathbf{^}}{\mathbf{I}}}^{H}{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}-{\stackrel{\mathbf{^}}{\mathbf{I}}}^{H}\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}+{\stackrel{\mathbf{^}}{\mathbf{I}}}^{T}\stackrel{¯}{{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}}-{\stackrel{\mathbf{^}}{\mathbf{I}}}^{T}\stackrel{¯}{\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}}\right]\\ & =0-0+{\stackrel{\mathbf{^}}{\mathbf{I}}}^{T}-0\\ & =\stackrel{\mathbf{^}}{\mathbf{I}}\end{array}\end{array}$

$\begin{array}{}\text{(14)}& \begin{array}{rl}{\mathrm{\nabla }}_{\stackrel{¯}{{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}}}{\mathcal{L}}_{4}& =\frac{\mathrm{\partial }}{\mathrm{\partial }\stackrel{¯}{{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}}}\left[\mu \left({\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right){\stackrel{¯}{\left({\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)}}^{T}\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\stackrel{¯}{{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}}}\left[\mu \left({\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}\cdot {\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}^{H}-{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}\cdot \sqrt{D}{\mathbf{h}}^{H}\mathbf{M}{\mathbf{F}}^{H}-\sqrt{D}{\mathbf{F}\mathbf{M}\mathbf{h}\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}^{H}+\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\cdot \sqrt{D}{\mathbf{h}}^{H}\mathbf{M}{\mathbf{F}}^{H}\right)\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\stackrel{¯}{{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}}}\left[\mu \left({\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}\cdot {\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}^{H}-{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}\cdot \sqrt{D}{\mathbf{h}}^{H}\mathbf{M}{\mathbf{F}}^{H}-\sqrt{D}{\mathbf{F}\mathbf{M}\mathbf{h}\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}^{H}+D\mathbf{F}\mathbf{M}\mathbf{h}{\mathbf{h}}^{H}\mathbf{M}{\mathbf{F}}^{H}\right)\right]\\ & =\mu \left({\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}-0-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}+0\right)\\ & =\mu {\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}-\mu \sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\end{array}\end{array}$

$\begin{array}{}\text{(15)}& \begin{array}{c}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\stackrel{\mathbf{^}}{\mathbf{f}}\right)\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}{\left(\stackrel{\mathbf{^}}{\mathbf{f}}\right)}^{H}{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}-\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\stackrel{\mathbf{^}}{\mathbf{f}}\right){\stackrel{\mathbf{^}}{\mathbf{g}}}^{H}+0+\stackrel{\mathbf{^}}{\mathbf{I}}+\mu \left({\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}-0-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}+0\right)\equiv 0\\ \mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\stackrel{\mathbf{^}}{\mathbf{f}}\right)\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}{\left(\stackrel{\mathbf{^}}{\mathbf{f}}\right)}^{H}{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}-\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\stackrel{\mathbf{^}}{\mathbf{f}}\right){\stackrel{\mathbf{^}}{\mathbf{g}}}^{H}+\stackrel{\mathbf{^}}{\mathbf{I}}+\mu \left({\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}-0-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}+0\right)\equiv 0\end{array}\end{array}$

$\begin{array}{}\text{(16)}& \mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\stackrel{\mathbf{^}}{\mathbf{f}}\right)\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}{\left(\stackrel{\mathbf{^}}{\mathbf{f}}\right)}^{H}{\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}-\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\stackrel{\mathbf{^}}{\mathbf{f}}\right){\stackrel{\mathbf{^}}{\mathbf{g}}}^{H}+\stackrel{\mathbf{^}}{\mathbf{I}}+\mu \left({\stackrel{\mathbf{^}}{\mathbf{h}}}_{c}-0-{\stackrel{\mathbf{^}}{\mathbf{h}}}_{m}+0\right)\equiv 0\end{array}$