Show the package imports
import random
import numpy as np
import pandas as pdOptimisation
ACTL3143 & ACTL5111 Deep Learning for Actuaries
Dense Layers in Matrices
Logistic regression
Observations: \mathbf{x}_{i,\bullet} \in \mathbb{R}^{2}.
Target: y_i \in \{0, 1\}.
Predict: \hat{y}_i = \mathbb{P}(Y_i = 1).
The model
For \mathbf{x}_{i,\bullet} = (x_{i,1}, x_{i,2}): z_i = x_{i,1} w_1 + x_{i,2} w_2 + b
\hat{y}_i = \sigma(z_i) = \frac{1}{1 + \mathrm{e}^{-z_i}} .
import sympy
sympy.plot("1/(1 + exp(-z))");
Multiple observations
data = pd.DataFrame({"x_1": [1, 3, 5], "x_2": [2, 4, 6], "y": [0, 1, 1]})
data| x_1 | x_2 | y | |
|---|---|---|---|
| 0 | 1 | 2 | 0 |
| 1 | 3 | 4 | 1 |
| 2 | 5 | 6 | 1 |
Let w_1 = 1, w_2 = 2 and b = -10.
w_1 = 1; w_2 = 2; b = -10
data["x_1"] * w_1 + data["x_2"] * w_2 + b 0 -5
1 1
2 7
dtype: int64Matrix notation
Have \mathbf{X} \in \mathbb{R}^{3 \times 2}.
X_df = data[["x_1", "x_2"]]
X = X_df.to_numpy()
Xarray([[1, 2],
[3, 4],
[5, 6]])Let \mathbf{w} = (w_1, w_2)^\top \in \mathbb{R}^{2 \times 1}.
w = np.array([[1], [2]])
warray([[1],
[2]])\mathbf{z} = \mathbf{X} \mathbf{w} + b , \quad \mathbf{a} = \sigma(\mathbf{z})
z = X.dot(w) + b
zarray([[-5],
[ 1],
[ 7]])1 / (1 + np.exp(-z))array([[0.01],
[0.73],
[1. ]])Using a softmax output
Observations: \mathbf{x}_{i,\bullet} \in \mathbb{R}^{2}. Predict: \hat{y}_{i,j} = \mathbb{P}(Y_i = j).
Target: \mathbf{y}_{i,\bullet} \in \{(1, 0), (0, 1)\}.
The model: For \mathbf{x}_{i,\bullet} = (x_{i,1}, x_{i,2}) \begin{aligned} z_{i,1} &= x_{i,1} w_{1,1} + x_{i,2} w_{2,1} + b_1 , \\ z_{i,2} &= x_{i,1} w_{1,2} + x_{i,2} w_{2,2} + b_2 . \end{aligned}
\begin{aligned} \hat{y}_{i,1} &= \text{Softmax}_1(\mathbf{z}_i) = \frac{\mathrm{e}^{z_{i,1}}}{\mathrm{e}^{z_{i,1}} + \mathrm{e}^{z_{i,2}}} , \\ \hat{y}_{i,2} &= \text{Softmax}_2(\mathbf{z}_i) = \frac{\mathrm{e}^{z_{i,2}}}{\mathrm{e}^{z_{i,1}} + \mathrm{e}^{z_{i,2}}} . \end{aligned}
Multiple observations
data| x_1 | x_2 | y_1 | y_2 | |
|---|---|---|---|---|
| 0 | 1 | 2 | 1 | 0 |
| 1 | 3 | 4 | 0 | 1 |
| 2 | 5 | 6 | 0 | 1 |
Choose:
w_{1,1} = 1, w_{2,1} = 2,
w_{1,2} = 3, w_{2,2} = 4, and
b_1 = -10, b_2 = -20.
w_11 = 1; w_21 = 2; b_1 = -10
w_12 = 3; w_22 = 4; b_2 = -20
data["x_1"] * w_11 + data["x_2"] * w_21 + b_10 -5
1 1
2 7
dtype: int64Matrix notation
Have \mathbf{X} \in \mathbb{R}^{3 \times 2}.
Xarray([[1, 2],
[3, 4],
[5, 6]])\mathbf{W}\in \mathbb{R}^{2\times2}, \mathbf{b}\in \mathbb{R}^{2}
W = np.array([[1, 3], [2, 4]])
b = np.array([-10, -20])
display(W); barray([[1, 3],
[2, 4]])array([-10, -20])\mathbf{Z} = \mathbf{X} \mathbf{W} + \mathbf{b} , \quad \mathbf{A} = \text{Softmax}(\mathbf{Z}) .
Z = X @ W + b
Zarray([[-5, -9],
[ 1, 5],
[ 7, 19]])np.exp(Z) / np.sum(np.exp(Z),
axis=1, keepdims=True)array([[9.82e-01, 1.80e-02],
[1.80e-02, 9.82e-01],
[6.14e-06, 1.00e+00]])Optimisation
Gradient-based learning
Make a guess:
Show derivatives: Reveal function:
Show derivatives: Reveal function:
Gradient descent pitfalls
Go over all the training data
Called batch gradient descent.
for i in range(num_epochs):
gradient = evaluate_gradient(loss_function, data, weights)
weights = weights - learning_rate * gradientPick a random training example
Called stochastic gradient descent.
for i in range(num_epochs):
rnd.shuffle(data)
for example in data:
gradient = evaluate_gradient(loss_function, example, weights)
weights = weights - learning_rate * gradientTake a group of training examples
Called mini-batch gradient descent.
for i in range(num_epochs):
rnd.shuffle(data)
for b in range(num_batches):
batch = data[b * batch_size : (b + 1) * batch_size]
gradient = evaluate_gradient(loss_function, batch, weights)
weights = weights - learning_rate * gradientMini-batch gradient descent
Why?
- Because we have to (data is too big)
- Because it is faster (lots of quick noisy steps > a few slow super accurate steps)
- The noise helps us jump out of local minima
Learning rates
Learning rates #2
“a nice way to see how the learning rate affects Stochastic Gradient Descent. we can use SGD to control a robot arm - minimizing the distance to the target as a function of the angles θᵢ. Too low a learning rate gives slow inefficient learning, too high and we see instability”
Learning rate schedule
In training the learning rate may be tweaked manually.
We need non-zero derivatives
This is why can’t use accuracy as the loss function for classification.
Also why we can have the dead ReLU problem.
Loss and derivatives
Example: linear regression
\hat{y}(x) = w x + b
For some observation \{ x_i, y_i \}, the (MSE) loss is
\text{Loss}_i = (\hat{y}(x_i) - y_i)^2
For a batch of the first n observations the loss is
\text{Loss}_{1:n} = \frac{1}{n} \sum_{i=1}^n (\hat{y}(x_i) - y_i)^2
Derivatives
Since \hat{y}(x) = w x + b,
\frac{\partial \hat{y}(x)}{\partial w} = x \text{ and } \frac{\partial \hat{y}(x)}{\partial b} = 1 .
As \text{Loss}_i = (\hat{y}(x_i) - y_i)^2, we know \frac{\partial \text{Loss}_i}{\partial \hat{y}(x_i) } = 2 (\hat{y}(x_i) - y_i) .
Chain rule
\frac{\partial \text{Loss}_i}{\partial \hat{y}(x_i) } = 2 (\hat{y}(x_i) - y_i), \,\, \frac{\partial \hat{y}(x)}{\partial w} = x , \, \text{ and } \, \frac{\partial \hat{y}(x)}{\partial b} = 1 .
Putting this together, we have
\frac{\partial \text{Loss}_i}{\partial w} = \frac{\partial \text{Loss}_i}{\partial \hat{y}(x_i) } \times \frac{\partial \hat{y}(x_i)}{\partial w} = 2 (\hat{y}(x_i) - y_i) \, x_i
and \frac{\partial \text{Loss}_i}{\partial b} = \frac{\partial \text{Loss}_i}{\partial \hat{y}(x_i) } \times \frac{\partial \hat{y}(x_i)}{\partial b} = 2 (\hat{y}(x_i) - y_i) .
Stochastic gradient descent (SGD)
Start with \boldsymbol{\theta}_0 = (w, b)^\top = (0, 0)^\top.
Randomly pick i=5, say x_i = 5 and y_i = 5.
\hat{y}(x_i) = 0 \times 5 + 0 = 0 \Rightarrow \text{Loss}_i = (0 - 5)^2 = 25.
The partial derivatives are \begin{aligned} \frac{\partial \text{Loss}_i}{\partial w} &= 2 (\hat{y}(x_i) - y_i) \, x_i = 2 \cdot (0 - 5) \cdot 5 = -50, \text{ and} \\ \frac{\partial \text{Loss}_i}{\partial b} &= 2 (0 - 5) = - 10. \end{aligned} The gradient is \nabla \text{Loss}_i = (-50, -10)^\top.
SGD, first iteration
Start with \boldsymbol{\theta}_0 = (w, b)^\top = (0, 0)^\top.
Randomly pick i=5, say x_i = 5 and y_i = 5.
The gradient is \nabla \text{Loss}_i = (-50, -10)^\top.
Use learning rate \eta = 0.01 to update \begin{aligned} \boldsymbol{\theta}_1 &= \boldsymbol{\theta}_0 - \eta \nabla \text{Loss}_i \\ &= \begin{pmatrix} 0 \\ 0 \end{pmatrix} - 0.01 \begin{pmatrix} -50 \\ -10 \end{pmatrix} \\ &= \begin{pmatrix} 0 \\ 0 \end{pmatrix} + \begin{pmatrix} 0.5 \\ 0.1 \end{pmatrix} = \begin{pmatrix} 0.5 \\ 0.1 \end{pmatrix}. \end{aligned}
SGD, second iteration
Start with \boldsymbol{\theta}_1 = (w, b)^\top = (0.5, 0.1)^\top.
Randomly pick i=9, say x_i = 9 and y_i = 17.
The gradient is \nabla \text{Loss}_i = (-223.2, -24.8)^\top.
Use learning rate \eta = 0.01 to update \begin{aligned} \boldsymbol{\theta}_2 &= \boldsymbol{\theta}_1 - \eta \nabla \text{Loss}_i \\ &= \begin{pmatrix} 0.5 \\ 0.1 \end{pmatrix} - 0.01 \begin{pmatrix} -223.2 \\ -24.8 \end{pmatrix} \\ &= \begin{pmatrix} 0.5 \\ 0.1 \end{pmatrix} + \begin{pmatrix} 2.232 \\ 0.248 \end{pmatrix} = \begin{pmatrix} 2.732 \\ 0.348 \end{pmatrix}. \end{aligned}
Batch gradient descent (BGD)
For the first n observations \text{Loss}_{1:n} = \frac{1}{n} \sum_{i=1}^n \text{Loss}_i so
\begin{aligned} \frac{\partial \text{Loss}_{1:n}}{\partial w} &= \frac{1}{n} \sum_{i=1}^n \frac{\partial \text{Loss}_{i}}{\partial w} = \frac{1}{n} \sum_{i=1}^n \frac{\partial \text{Loss}_{i}}{\hat{y}(x_i)} \frac{\partial \hat{y}(x_i)}{\partial w} \\ &= \frac{1}{n} \sum_{i=1}^n 2 (\hat{y}(x_i) - y_i) \, x_i . \end{aligned}
\begin{aligned} \frac{\partial \text{Loss}_{1:n}}{\partial b} &= \frac{1}{n} \sum_{i=1}^n \frac{\partial \text{Loss}_{i}}{\partial b} = \frac{1}{n} \sum_{i=1}^n \frac{\partial \text{Loss}_{i}}{\hat{y}(x_i)} \frac{\partial \hat{y}(x_i)}{\partial b} \\ &= \frac{1}{n} \sum_{i=1}^n 2 (\hat{y}(x_i) - y_i) . \end{aligned}
BGD, first iteration (\boldsymbol{\theta}_0 = \boldsymbol{0})
| x | y | y_hat | loss | dL/dw | dL/db | |
|---|---|---|---|---|---|---|
| 0 | 1 | 0.99 | 0 | 0.98 | -1.98 | -1.98 |
| 1 | 2 | 3.00 | 0 | 9.02 | -12.02 | -6.01 |
| 2 | 3 | 5.01 | 0 | 25.15 | -30.09 | -10.03 |
So \nabla \text{Loss}_{1:3} is
nabla = np.array([df["dL/dw"].mean(), df["dL/db"].mean()])
nabla array([-14.69, -6. ])so with \eta = 0.1 then \boldsymbol{\theta}_1 becomes
theta_1 = theta_0 - 0.1 * nabla
theta_1array([1.47, 0.6 ])BGD, second iteration
| x | y | y_hat | loss | dL/dw | dL/db | |
|---|---|---|---|---|---|---|
| 0 | 1 | 0.99 | 2.07 | 1.17 | 2.16 | 2.16 |
| 1 | 2 | 3.00 | 3.54 | 0.29 | 2.14 | 1.07 |
| 2 | 3 | 5.01 | 5.01 | 0.00 | -0.04 | -0.01 |
So \nabla \text{Loss}_{1:3} is
nabla = np.array([df["dL/dw"].mean(), df["dL/db"].mean()])
nabla array([1.42, 1.07])so with \eta = 0.1 then \boldsymbol{\theta}_2 becomes
theta_2 = theta_1 - 0.1 * nabla
theta_2array([1.33, 0.49])Glossary
- batches, batch size
- gradient-based learning, hill-climbing
- stochastic (mini-batch) gradient descent