1.9 Multilayered Network of Neurons
Your first Deep Neural Network
Learning Algorithm
Forward Propogation
W_2 = \begin{bmatrix}
0.4 & 0.5 & -0.3 \\
-0.1 & -0.4 & -0.5\\
0.8 & 0.2 & 0.9\\
\end{bmatrix}
b = [\ \ 0\ \ \ 0\ \ 0\ \ ]
y = [\ \ 1 \ \ \ 0\ \ ]
x = [\ \ 2\ \ \ 5 \ \ \ 3 \ \ ]
\(a_1 = W_1*x + b_1 = [ -2.2 -0.1 5.3 ]\)
\(h_1 = tanh(a_1) = [ -0.97 -0.1 0.99 ]\)
\(a_2 = W_2*h_1 + b_2 = [ 0.41 -0.24 1.24 ]\)
Output :
\(L(\Theta) = -\text{log}\hat{y_l} = 0.43 \)
W_1 = \begin{bmatrix}
0.1 & 0.3 & 0.8\\
-0.3 & -0.2 & 0.5 \\
-0.3 & 0.1 & 0.4 \\
\end{bmatrix}
W_3 = \begin{bmatrix}
0.3 & -0.5\\
0.1 & 0.2\\
-0.1 & -0.4\\
\end{bmatrix}
\(h_2 = tanh(a_2) = [ 0.39 -0.24 0.84 ]\)
\(a_3 = W_3 * h_2 + b_3 = [ 0.01 -0.58 ]\)
\(\hat{y} = softmax(a_3) = [ 0.64 0.36 ]\)
Learning Algorithm
Backward Propogation
W_2 = \begin{bmatrix}
0.4 & 0.5 & -0.3 \\
-0.1 & -0.4 & -0.5\\
0.8 & 0.2 & 0.9\\
\end{bmatrix}
b = [\ \ 0\ \ \ 0\ \ 0\ \ ]
y = [\ \ 1 \ \ \ 0\ \ ]
x = [\ \ 2\ \ \ 5 \ \ \ 3 \ \ ]
\( \nabla_{a_L} \mathscr{L}(\theta)\) :
W_1 = \begin{bmatrix}
0.1 & 0.3 & 0.8\\
-0.3 & -0.2 & 0.5 \\
-0.3 & 0.1 & 0.4 \\
\end{bmatrix}
W_3 = \begin{bmatrix}
0.3 & -0.5\\
0.1 & 0.2\\
-0.1 & -0.4\\
\end{bmatrix}
=
\begin{bmatrix}
\frac{\partial \mathscr{L}{\theta}}{\partial a_{L1}} \\
\\
\frac{\partial \mathscr{L}{\theta}}{\partial a_{L2
}}\\
\end{bmatrix}
=
\begin{bmatrix}
- (\mathbb{1}_{\ell = 1} - \hat{y}_1)\\
\\
- (\mathbb{1}_{\ell = 2} - \hat{y}_2)\\
\end{bmatrix}
=
\begin{bmatrix}
- (1 - 0.64)\\
\\
- (0 - 0.36)\\
\end{bmatrix}
=
\begin{bmatrix}
- 0.36\\
\\
0.36\\
\end{bmatrix}
Learning Algorithm
Backward Propogation
W_2 = \begin{bmatrix}
0.4 & 0.5 & -0.3 \\
-0.1 & -0.4 & -0.5\\
0.8 & 0.2 & 0.9\\
\end{bmatrix}
b = [\ \ 0\ \ \ 0\ \ 0\ \ ]
y = [\ \ 1 \ \ \ 0\ \ ]
x = [\ \ 2\ \ \ 5 \ \ \ 3 \ \ ]
\( \nabla_{h_2} \mathscr{L}(\theta)\) :
W_1 = \begin{bmatrix}
0.1 & 0.3 & 0.8\\
-0.3 & -0.2 & 0.5 \\
-0.3 & 0.1 & 0.4 \\
\end{bmatrix}
W_3 = \begin{bmatrix}
0.3 & -0.5\\
0.1 & 0.2\\
-0.1 & -0.4\\
\end{bmatrix}
=
\begin{bmatrix}
\frac{\partial \mathscr{L}{\theta}}{\partial h_{21}} \\
\\
\frac{\partial \mathscr{L}{\theta}}{\partial h_{22
}}\\
\\
\frac{\partial \mathscr{L}{\theta}}{\partial h_{23
}}\\
\end{bmatrix}
=
\begin{bmatrix}
(W_{2,., 1})^{T} \textcolor{red}{\nabla_{a_{3}} \mathscr{L}({\theta})}\\
\\
(W_{2,., 2})^{T} \textcolor{red}{\nabla_{a_{3}} \mathscr{L}({\theta})}\\
\\
(W_{2,., 3})^{T} \textcolor{red}{\nabla_{a_{3}} \mathscr{L}({\theta})}\\
\end{bmatrix}
=
\begin{bmatrix}
0.23\\
\\
- 0.03\\
\\
0.09
\end{bmatrix}
\( \nabla_{a_3} \mathscr{L}(\theta)\) :
=
\begin{bmatrix}
\frac{\partial \mathscr{L}{\theta}}{\partial a_{L1}} \\
\\
\frac{\partial \mathscr{L}{\theta}}{\partial a_{L2
}}\\
\end{bmatrix}
=
\begin{bmatrix}
- 0.36\\
\\
0.36\\
\end{bmatrix}
Learning Algorithm
Backward Propogation
W_2 = \begin{bmatrix}
0.4 & 0.5 & -0.3 \\
-0.1 & -0.4 & -0.5\\
0.8 & 0.2 & 0.9\\
\end{bmatrix}
b = [\ \ 0\ \ \ 0\ \ 0\ \ ]
y = [\ \ 1 \ \ \ 0\ \ ]
x = [\ \ 2\ \ \ 5 \ \ \ 3 \ \ ]
\( \nabla_{a_2} \mathscr{L}(\theta)\) :
W_1 = \begin{bmatrix}
0.1 & 0.3 & 0.8\\
-0.3 & -0.2 & 0.5 \\
-0.3 & 0.1 & 0.4 \\
\end{bmatrix}
W_3 = \begin{bmatrix}
0.3 & -0.5\\
0.1 & 0.2\\
-0.1 & -0.4\\
\end{bmatrix}
=
\begin{bmatrix}
\frac{\partial \mathscr{L}{\theta}}{\partial a_{21}} \\
\\
\frac{\partial \mathscr{L}{\theta}}{\partial a_{22
}}\\
\\
\frac{\partial \mathscr{L}{\theta}}{\partial a_{23
}}\\
\end{bmatrix}
=
\begin{bmatrix}
\textcolor{red}{\frac{\partial \mathscr{L}({\theta})}{\partial h_{21}}} g'(a_{21})\\
\\
\textcolor{red}{\frac{\partial \mathscr{L}({\theta})}{\partial h_{22}}} g'(a_{22})\\
\\
\textcolor{red}{\frac{\partial \mathscr{L}({\theta})}{\partial h_{23}}} g'(a_{23})\\
\end{bmatrix}
=
\begin{bmatrix}
0.23 * 0.84\\
\\
- 0.03 * 0.96\\
\\
0.09 * 0.28
\end{bmatrix}
\( \nabla_{h_2} \mathscr{L}(\theta)\) :
=
\begin{bmatrix}
\frac{\partial \mathscr{L}{\theta}}{\partial h_{21}} \\
\\
\frac{\partial \mathscr{L}{\theta}}{\partial h_{22
}}\\
\\
\frac{\partial \mathscr{L}{\theta}}{\partial h_{23
}}\\
\end{bmatrix}
=
\begin{bmatrix}
0.23\\
\\
- 0.03\\
\\
0.09
\end{bmatrix}
=
\begin{bmatrix}
0.19\\
\\
- 0.03\\
\\
0.02
\end{bmatrix}
Learning Algorithm
Backward Propogation
W_2 = \begin{bmatrix}
0.4 & 0.5 & -0.3 \\
-0.1 & -0.4 & -0.5\\
0.8 & 0.2 & 0.9\\
\end{bmatrix}
b = [\ \ 0\ \ \ 0\ \ 0\ \ ]
y = [\ \ 1 \ \ \ 0\ \ ]
x = [\ \ 2\ \ \ 5 \ \ \ 3 \ \ ]
\( \nabla_{W_2} \mathscr{L}(\theta)\) :
W_1 = \begin{bmatrix}
0.1 & 0.3 & 0.8\\
-0.3 & -0.2 & 0.5 \\
-0.3 & 0.1 & 0.4 \\
\end{bmatrix}
W_3 = \begin{bmatrix}
0.3 & -0.5\\
0.1 & 0.2\\
-0.1 & -0.4\\
\end{bmatrix}
=
\begin{bmatrix}
\textcolor{red}{\frac{\partial \mathscr{L}{\theta}}{\partial a_{21}}}h_{11} & \textcolor{red}{\frac{\partial \mathscr{L}{\theta}}{\partial a_{21}}}h_{12} & \textcolor{red}{\frac{\partial \mathscr{L}{\theta}}{\partial a_{21}}}h_{13}\\
\\
\textcolor{red}{\frac{\partial \mathscr{L}{\theta}}{\partial a_{22}}}h_{11} & \textcolor{red}{\frac{\partial \mathscr{L}{\theta}}{\partial a_{22}}}h_{12} & \textcolor{red}{\frac{\partial \mathscr{L}{\theta}}{\partial a_{22}}}h_{13}\\
\\
\textcolor{red}{\frac{\partial \mathscr{L}{\theta}}{\partial a_{23}}}h_{11} & \textcolor{red}{\frac{\partial \mathscr{L}{\theta}}{\partial a_{23}}}h_{12} & \textcolor{red}{\frac{\partial \mathscr{L}{\theta}}{\partial a_{23}}}h_{13}\\\\
\end{bmatrix}
\( \nabla_{a_2} \mathscr{L}(\theta)\)
=
\begin{bmatrix}
-0.19 & 0.03 & -0.02\\
\\
- 0.02 & 0.01 & -0.01\\
\\
0.2 & -0.03 & 0.02 \\
\end{bmatrix}
=
\begin{bmatrix}
0.19\\
\\
- 0.03\\
\\
0.02
\end{bmatrix}
=
\begin{bmatrix}
-0.97\\
\\
-0.1\\
\\
0.99
\end{bmatrix}
h_{1}
Learning Algorithm
Backward Propogation
W_2 = \begin{bmatrix}
0.4 & 0.5 & -0.3 \\
-0.1 & -0.4 & -0.5\\
0.8 & 0.2 & 0.9\\
\end{bmatrix}
b = [\ \ 0\ \ \ 0\ \ 0\ \ ]
y = [\ \ 1 \ \ \ 0\ \ ]
x = [\ \ 2\ \ \ 5 \ \ \ 3 \ \ ]
\( \nabla_{b_2} \mathscr{L}(\theta)\) :
W_1 = \begin{bmatrix}
0.1 & 0.3 & 0.8\\
-0.3 & -0.2 & 0.5 \\
-0.3 & 0.1 & 0.4 \\
\end{bmatrix}
W_3 = \begin{bmatrix}
0.3 & -0.5\\
0.1 & 0.2\\
-0.1 & -0.4\\
\end{bmatrix}
= \textcolor{red}{\nabla_{a_2} \mathscr{L}(\theta)}
\( \nabla_{a_2} \mathscr{L}(\theta)\)
=
\begin{bmatrix}
0.19\\
\\
- 0.03\\
\\
0.02
\end{bmatrix}
=
\begin{bmatrix}
0.19\\
\\
- 0.03\\
\\
0.02
\end{bmatrix}
Evaluation
How do you check the performance of a deep neural network?
(c) One Fourth Labs
Test Data
Accuracy=\frac{\text{Number of correct predictions}}{\text{Total number of predictions}}
= \frac{2}{4} = 50\%
Indian Liver Patient Records \(^{*}\)
- whether person needs to be diagnosed or not ?
| Age |
| 65 |
| 62 |
| 20 |
| 84 |
| Albumin |
| 3.3 |
| 3.2 |
| 4 |
| 3.2 |
| T_Bilirubin |
| 0.7 |
| 10.9 |
| 1.1 |
| 0.7 |
| y |
| 0 |
| 0 |
| 1 |
| 1 |
.
.
.
| Predicted |
| 0 |
| 1 |
| 1 |
| 0 |
Take-aways
What are the new things that we learned in this module ?
(c) One Fourth Labs
\( x_i \in \mathbb{R} \)
Accuracy=\frac{\text{Number of correct predictions}}{\text{Total number of predictions}}
Loss
Model
Data
Task
Evaluation
Learning
Real inputs
Tasks with Real Inputs and Real Outputs
Back-propagation
Squared Error Loss :
L(\Theta) = \frac{1}{N} \displaystyle\sum_{i=1}^N \displaystyle\sum_{i=1}^d (\hat{y}_{ij} - y_{ij})^2
Cross Entropy Loss:
L(\Theta) = -\frac{1}{N} \displaystyle\sum_{i=1}^N \displaystyle\sum_{i=1}^d y_{ij}\log{(\hat{y}_{ij})}
\hat{y} = \frac{1}{1+e^{-(w_{21}*(\frac{1}{1+e^{- (w_{11}*x_1 + w_{12}*x_2 + b_1)}}) + w_{22}*(\frac{1}{1+e^{- (w_{13}*x_1 + w_{14}*x_2 + b_1)}}) + b_2)}}
Copy of Working Example
By preksha nema
