res/2.1/quiz.tex

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\title{Quiz: backward propagation intuition}
\author{BME595 DeepLearning}
\date{\today}
\maketitle

Given equations $(1)$ and $(2)$ and that $h_{\bm{\Theta}}(\bm{x}) := \bm{a}^{(L)}$ and $E = E(h_{\bm{\Theta}}(\bm{x}))$, A) \underline{define} $\bm{\delta}^{(\ell)}$ and B) show (or prove) where equations $(3)-(5)$ come from.
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\begin{align}
  \bm{a}^{(1)}&=\bm{x}, \bm{\hat a}^{(1)}=\genfrac{[}{]}{0pt}{1}{+1}{\bm{a}^{(1)}} \\
  \bm{z}^{(\ell+1)}&=\bm{\Theta}^{(\ell)}\bm{\hat a}^{(\ell)}, \bm{a}^{(\ell)}=\sigma(\bm{z}^{(\ell)}), \bm{\hat a}^{(\ell)}=\genfrac{[}{]}{0pt}{1}{+1}{\bm{a}^{(\ell)}} \\
  \bm{\delta}^{(L)}&=\nabla_h E\odot\sigma^\prime(\bm{z}^{(L)}) \\
  \bm{\delta}^{(\ell)}&=[(\bm{\Theta}^{(\ell)})^\top\bm{\delta}^{(\ell+1)}]\odot\sigma^\prime(\bm{z}^{(\ell)}) \\
  \frac{\partial E}{\partial\bm{\Theta}^{(\ell)}}&=\bm{\hat a}^{(\ell)}(\bm{\delta}^{(\ell+1)})^\top
\end{align}
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Finally, C) discuss about $\bm{\delta}^{(1)}$ and $\frac{\partial E}{\partial \bm{x}}$.

As a reminder, $\mathcal{L}(\bm{\Theta}) = \frac{1}{m} \sum_{i=1}^m E(h_{\bm{\Theta}}(\bm{x}^{(i)}))$. Sometimes, $\mathcal{L}(\bm{\Theta})$ is written $J(\bm{\Theta})$ or $J(\bm{\theta})$.

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