import os
os.environ["CUDA_VISIBLE_DEVICES"] = ""
import random
import numpy as np
import pandas as pd
from keras.models import Sequential
from keras.models import Model
from keras.layers import Input
from keras.layers import Dense
from keras.initializers import ConstantLab: Backpropagation
ACTL3143 & ACTL5111 Deep Learning for Actuaries
Backpropagation performs a backward pass to adjust the neural network’s parameters. It’s an algorithm that uses gradient descent to update the neural network weights.
Linear Regression via Batch Gradient Descent
Let \boldsymbol{\theta}^{(t)}=(w^{(t)}, b^{(t)}) be the parameter estimates of the tth iteration. Let \mathcal{D}= \{(x_i, y_i)\}_{i=1}^{N} represents the training batch. Let mean squared error (MSE) be the loss/cost function \mathcal{L}.
Finding the Gradients
- Step 1: Write down \mathcal{L}(\mathcal{D}, \boldsymbol{\theta}^{(t)}) and \hat{y}(x_i; \boldsymbol{\theta}^{(t)}) \begin{align*} \mathcal{L}(\mathcal{D},\boldsymbol{\theta}^{(t)}) &=\frac{1}{N} \sum_{i=1}^{N} \big(\hat{y}(x_i; \boldsymbol{\theta}^{(t)}) - y_i \big)^2 \\ \hat{y}(x_i; \boldsymbol{\theta}^{(t)}) &= w^{(t)}x_i + b^{(t)} \end{align*}
- Step 2: Derive \frac{\partial \mathcal{L}(\hat{y}(x_i; \boldsymbol{\theta}^{(t)}), y_i)}{\partial \hat{y}(x_i; \boldsymbol{\theta}^{(t)})} and \frac{\partial\hat{y}(x_i; \boldsymbol{\theta}^{(t)})}{\partial \boldsymbol{\theta}^{(t)}} \begin{align*} \frac{\partial \mathcal{L}(\hat{y}(x_i; \boldsymbol{\theta}^{(t)}), y_i)}{\partial \hat{y}(x_i; \boldsymbol{\theta}^{(t)})} & = 2 \big(\hat{y}(x_i; \boldsymbol{\theta}^{(t)}) - y_i \big) \\ \frac{\partial\hat{y}(x_i; \boldsymbol{\theta}^{(t)})}{\partial w^{(t)}} & = x_i \\ \frac{\partial\hat{y}(x_i; \boldsymbol{\theta}^{(t)})}{\partial b^{(t)}} & = 1 \end{align*}
- Step 3: Derive \frac{\partial \mathcal{L}(\mathcal{D}, \boldsymbol{\theta}^{(t)})}{\partial \boldsymbol{\theta}^{(t)}} \frac{\partial \mathcal{L}(\mathcal{D}, \boldsymbol{\theta}^{(t)})}{\partial w^{(t)}} = \frac{1}{N}\sum_{i=1}^{N}\frac{\partial \mathcal{L}(\hat{y}(x_i; \boldsymbol{\theta}^{(t)}), y_i)}{\partial \hat{y}(x_i; \boldsymbol{\theta}^{(t)})} \frac{\partial\hat{y}(x_i; \boldsymbol{\theta}^{(t)})}{\partial w^{(t)}} = \frac{2}{N} \sum_{i=1}^{N} \big(\hat{y}(x_i; \boldsymbol{\theta}^{(t)}) - y_i \big) \cdot x_i \tag{1} \frac{\partial \mathcal{L}(\mathcal{D}, \boldsymbol{\theta}^{(t)})}{\partial b^{(t)}} = \frac{1}{N}\sum_{i=1}^{N}\frac{\partial \mathcal{L}(\hat{y}(x_i; \boldsymbol{\theta}^{(t)}), y_i)}{\partial \hat{y}(x_i; \boldsymbol{\theta}^{(t)})} \frac{\partial\hat{y}(x_i; \boldsymbol{\theta}^{(t)})}{\partial b^{(t)}} = \frac{2}{N} \sum_{i=1}^{N} \big(\hat{y}(x_i; \boldsymbol{\theta}^{(t)}) - y_i \big) \cdot 1 \tag{2}
Then, we initialise \boldsymbol{\theta}^{(0)} = (w^{(0)}, b^{(0)}) and then apply gradient descent for t=1, 2, \ldots \begin{align} w^{(t+1)} &= w^{(t)} - \eta \cdot \frac{\partial \mathcal{L}(\mathcal{D}, \boldsymbol{\theta}^{(t)})}{\partial w}\bigg|_{w^{(t)}} \\ b^{(t+1)} &= b^{(t)} - \eta \cdot \frac{\partial \mathcal{L}(\mathcal{D}, \boldsymbol{\theta}^{(t)})}{\partial b}\bigg|_{b^{(t)}} \end{align} using the derivatives derived from Equation 1 and Equation 2. \eta is a chosen learning rate.
Exercise
Use backpropagation algorithm to find \theta^{(3)} with \theta^{(0)}= (w^{(0)} = 1, b^{(0)} = 0). The dataset \mathcal{D} is as follows:
i x_i y_i 1 2 7 2 3 10 3 5 16
That is, the true model would be y_i = 3 x_i + 1, i.e., w = 3, b = 1. Implement batch gradient descent.
Neural Network
For a neural network with H hidden layers:
- L_0 is the input layer (the zeroth hidden layer). L_k represents the kth hidden layer for k\in \{1, 2, \ldots, H\}. L_{H+1} is the output layer (the H+1th hidden layer).
- \phi^{(k)} represents the activation function for the kth hidden layer, with k\in \{1, 2, \ldots, H\}. \phi^{(H+1)} represents the activation function for the output layer.
- \boldsymbol{w}^{(k)}_j represents the weights connecting the activated neurons \boldsymbol{a}^{(k-1)} from the k-1th hidden layer to the jth neuron in the kth hidden layer, where k\in \{1, \ldots, H+1\} and j\in \{1, \ldots, q_{k}\}, i.e., q_{k} denotes the number of neurons in the kth hidden layer. \boldsymbol{a}^{(0)} = \boldsymbol{z}^{(0)} =\boldsymbol{x} by definition.
- b^{(k)}_j represents the bias for the jth neuron in the kth hidden layer.
Gradients For the Output Layer
The gradient for \boldsymbol{w}_1^{(H+1)}, i.e., the weights connecting the neurons in the Hth (last) hidden layer to the first neuron of the output layer, is given by: \frac{\partial \mathcal{L}(\mathcal{D}, \boldsymbol{\theta})}{\partial \boldsymbol{w}^{(H+1)}_1} = \frac{\partial \mathcal{L}(\mathcal{D}, \boldsymbol{\theta})}{\partial \hat{y}_1} \frac{\partial \hat{y}_1}{\partial z^{(H+1)}_1 } \frac{\partial z^{(H+1)}_1}{\partial \boldsymbol{w}^{(H+1)}_1} \tag{3} where
- \hat{y}_1=a^{(H+1)}_1= \phi (z^{(H+1)}_1)
- z^{(H+1)}_1 = \langle \boldsymbol{a}^{(H)}, \boldsymbol{w}_1^{(H+1)} \rangle + b^{(H+1)}_1.
- \langle \cdot, \cdot \rangle represents the inner product.
Example 1
- From input layer L_0 to the first hidden layer L_1: \begin{align*} a^{(1)}_1 &= \phi^{(1)}\big(w^{(1)}_{1, 1}x_1 + w^{(1)}_{2, 1}x_2 + w^{(1)}_{3, 1} x_3 + b^{(1)}_1\big) = \phi^{(1)} (\langle \boldsymbol{w}^{(1)}_{1}, \boldsymbol{x} \rangle + b^{(1)}_1 )\\ a^{(1)}_2 &= \phi^{(1)}\big(w^{(1)}_{1, 2}x_1 + w^{(1)}_{2, 2}x_2 + w^{(1)}_{3, 2} x_3 + b^{(1)}_2\big) = \phi^{(1)} (\langle \boldsymbol{w}^{(1)}_{2}, \boldsymbol{x} \rangle + b^{(1)}_2) \end{align*}
- From the first hidden layer L_1 to the output layer layer L_2: \begin{align*} \hat{y} &= \phi^{(2)}\big(w^{(2)}_{1, 1} a^{(1)}_1 + w^{(2)}_{2, 1} a^{(1)}_2 + b^{(2)}_1\big) = \phi^{(2)}( \langle \boldsymbol{w}^{(2)}_{1}, \boldsymbol{a}^{(1)} \rangle + b^{(2)}_1) \end{align*}
- \phi^{(1)}(z)= S(z) (sigmoid function) and \phi^{(2)}(z) = \exp(z) (exponential function).
Let \boldsymbol{\theta}^{(t)}=(\boldsymbol{w}^{(t)}, \boldsymbol{b}^{(t)})= \Big(\boldsymbol{w}^{(t, 1)}_1, \boldsymbol{w}^{(t, 1)}_2, \boldsymbol{w}^{(t, 2)}_1, b^{(t,1)}_1, b^{(t,1)}_2, b^{(t,2)}_1\Big) be the parameter estimates of the tth iteration. For illustration, we assume the bias terms \big(b^{(t,1)}_1, b^{(t,1)}_2, b^{(t,2)}_1\big) are all zeros.
- For \boldsymbol{w}_1^{(2)}, apply equation Equation 3
- For \boldsymbol{w}^{(1)}_1, apply equation Equation 4
- For \boldsymbol{w}^{(1)}_2, apply equation Equation 4
Implementing Backpropagation in Python
See Week_4_Lab_Notebook.ipynb for more details. The required packages/functions are as follows:
True weights:
w1_1 = np.array([[0.25], [0.5], [0.75]])
w1_2 = np.array([[0.75], [0.5], [0.25]])
w2_1 = np.array([[2.0], [3.0]])Some synthetic data to work with:
# Generate 10000 random observations of 3 numerical features
np.random.seed(0)
X = np.random.randn(10000, 3)
# Sigmoid activation function
def sigmoid(z):
return(1/(1+np.exp(-z)))
# Hidden Layer 1
z1_1 = X @ w1_1 # The first neuron before activation
z1_2 = X @ w1_2 # The second neuron before activation
a1_1 = sigmoid(z1_1) # The first neuron after activation
a1_2 = sigmoid(z1_2) # The second neuron after activation
# Output Layer
z2_1 = np.concatenate((a1_1, a1_2), axis = 1) @ w2_1 # Pre-activation of the ouput
a2_1 = np.exp(z2_1) # Output
# The actual values
y = a2_1From Scratch
# Initialised weights
w1_1_hat = np.array([[0.2], [0.6], [1.0]])
w1_2_hat = np.array([[0.4], [0.8], [1.2]])
w2_1_hat = np.array([[1.0], [2.0]])
losses = []
num_iterations = 5000
for _ in range(num_iterations):
# Compute Forward Passes
# Hidden Layer 1
z1_1_hat = X @ w1_1_hat # The first neuron before activation
z1_2_hat = X @ w1_2_hat # The second neuron before activation
a1_1_hat = sigmoid(z1_1_hat) # The first neuron after activation
a1_2_hat = sigmoid(z1_2_hat) # The second neuron after activation
a1_hat = np.concatenate((a1_1_hat, a1_2_hat), axis = 1)
# Output Layer
z2_1_hat = a1_hat @ w2_1_hat # The output before activation
y_hat = np.exp(z2_1_hat).reshape(len(y), 1) # The ouput
# Track the Losses
loss = (y_hat - y)**2
losses.append(np.mean(loss))
# Compute Deltas
delta2_1 = 2 * (y_hat - y) * np.exp(z2_1_hat)
delta1_1 = w2_1_hat[0] * delta2_1 * sigmoid(z1_1_hat) * (1-sigmoid(z1_1_hat))
delta1_2 = w2_1_hat[1] * delta2_1 * sigmoid(z1_2_hat) * (1-sigmoid(z1_2_hat))
# Compute Gradients
d2_1_hat = delta2_1 * a1_hat
d1_1_hat = delta1_1 * X
d1_2_hat = delta1_2 * X
# Learning Rate
eta = 0.0005
# Apply Batch Gradient Descent
w2_1_hat -= eta * np.mean(d2_1_hat, axis = 0).reshape(2, 1)
w1_1_hat -= eta * np.mean(d1_1_hat, axis = 0).reshape(3, 1)
w1_2_hat -= eta * np.mean(d1_2_hat, axis = 0).reshape(3, 1)
print(w1_1_hat)
print(w1_2_hat)
print(w2_1_hat)[[0.24985576]
[0.5000211 ]
[0.75018656]]
[[0.74987578]
[0.49998626]
[0.25009692]]
[[1.99874327]
[3.00125615]]From Keras
# An initialiser for the weights in the neural network
init1 = Constant([[0.2, 0.4], [0.6, 0.8], [1.0, 1.2]])
init2 = Constant([[1.0], [2.0]])
# Build a neural network
# `use_bias` (whether to include bias terms for the neurons or not) is True by default
# `kernel_initializer` adjusts the initialisations of the weights
x = Input(shape=X.shape[1:], name="Inputs")
a1 = Dense(2, "sigmoid", use_bias=False,
kernel_initializer=init1)(x)
y_hat = Dense(1, "exponential", use_bias=False,
kernel_initializer=init2)(a1)
model = Model(x, y_hat)
# Choosing the optimiser and the loss function
model.compile(optimizer="adam", loss="mse")
# Model Training
# We don't implement early stopping to make the results comparable to the previous section
hist = model.fit(X, y, epochs=5000, verbose=0, batch_size = len(y))
# Print out the weights
print(model.get_weights())[array([[0.25250557, 0.75124925],
[0.49919206, 0.5006289 ],
[0.74791 , 0.2485937 ]], dtype=float32), array([[2.0164008],
[2.983465 ]], dtype=float32)]