Batch Gradient Descent_batch gradient descent原文-优快云博客

本文介绍了批量梯度下降法在优化线性回归中的应用，详细阐述了成本函数、学习曲线和梯度下降公式，并探讨了如何选择合适的步长以确保算法收敛。内容包括不同参数数量的学习曲线和迭代过程的可视化，强调了适当步长选择的重要性。

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Batch Gradient Descent

We use linear regression as example to explain this optimization algorithm.

1. Formula

1.1. Cost Function

We prefer residual sum of squared to evaluate linear regression.

J (θ) = 1 2 m \sum i = 1 n [h θ (x i) - y i] 2

$\begin{align} J(\theta) &= \frac{1}{2m} \sum\limits_{i=1}^{n} \left[ h_{\theta}(x_i) - y_i \right] ^ 2 \end{align}$

1.2. Visualize Cost Function

E.g. 1 :

one parameter only $\theta_1$ –> $h_{\theta}(x) = \theta_1 x_1$

Learning Curve 1

1. Learning Curve 1 ^[1]

E.g. 2 :

two parameters $\theta_0, \theta_1$ –> $h_{\theta}(x) = \theta_0 + \theta_1 x_1$

Learning Curve 2

2. Learning Curve 2 ^[2]

Switch to contour plot

Learning Curve 2 - contour

3. Learning Curve 2 - contour^[2]

1.3. Gradient Descent Formula

For all $\theta_i$

\partial J θ \partial θ i = 1 m \sum i = 1 n [h θ (x i) - y i] \cdot (x i)

$\begin{align} \frac{\partial J_\theta}{\partial \theta_i} = \frac{1}{m} \sum\limits_{i=1}^{n} \left[ h_{\theta}(x_i) - y_i \right] \cdot (x_i) \end{align}$

E.g.,
two parameters $\theta_0, \theta_1$ –> $h_{\theta}(x) = \theta_0 + \theta_1 x_1$

For i = 0 :

\partial J θ \partial θ 0 = 1 m \sum i = 1 n [h θ (x i) - y i] \cdot (x 0)

$\frac{\partial J_\theta}{\partial \theta_0} = \frac{1}{m} \sum\limits_{i=1}^{n} \left[ h_{\theta}(x_i) - y_i \right] \cdot (x_0)$

For i = 1:

\partial J θ \partial θ 1 = 1 m \sum i = 1 n [h θ (x i) - y i] \cdot (x 1)

$\frac{\partial J_\theta}{\partial \theta_1} = \frac{1}{m} \sum\limits_{i=1}^{n} \left[ h_{\theta}(x_i) - y_i \right]\cdot (x_1)$

% Octave
%% =================== Gradient Descent ===================
% Add a column(x0) of ones to X

X = [ones(len, 1), data(:,1)];
theta = zeros(2, 1);
alpha = 0.01;
ITERATION = 1500;
jTheta = zeros(ITERATION, 1);

for iter = 1:ITERATION
    % Perform a single gradient descent on the parameter vector
    % Note: since the theta will be updated, a tempTheta is needed to store the data.
    tempTheta = theta;
    theta(1) = theta(1) - (alpha / len) * (sum(X * tempTheta - Y));  % ignore the X(:,1) since the values are all ones.
    theta(2) = theta(2) - (alpha / len) * (sum((X * tempTheta - Y) .* X(:,2)));

    %% =================== Compute Cost ===================
    jTheta(iter) = sum((X * theta - Y) .^ 2) / (2 * len);
endfor

2. Algorithm

For all $\theta_i$

θ i : = θ i - α \partial \partial θ i J (θ 1, θ 2, \dots, θ n)

$\begin{align} \theta_i := \theta_i - \alpha \frac{\partial}{\partial \theta_i} J(\theta_1, \theta_2, \dots ,\theta_n) \end{align}$

E.g.,
two parameters $\theta_0, \theta_1$ –> $h_{\theta}(x) = \theta_0 + \theta_1 x_1$

For i = 0 :

θ 0 : = θ 0 - α 1 m \sum i = 1 n [h θ (x i) - y i]

$\begin{align} \theta_0 := \theta_0 - \alpha \frac{1}{m} \sum\limits_{i=1}^{n} \left[ h_{\theta}(x_i) - y_i \right] \end{align}$

For i = 1 :

θ 1 : = θ 1 - α 1 m \sum i = 1 n [h θ (x i) - y i] \cdot (x 1)

$\begin{align} \theta_1 := \theta_1 - \alpha \frac{1}{m} \sum\limits_{i=1}^{n} \left[ h_{\theta}(x_i) - y_i \right] \cdot (x_1) \end{align}$

Iterative for multiple times (depends on data content, data size and step size). Finally, we could see the result as below.

Visualize Convergence

3. Analyze

Pros	Cons
Controllable by manuplate stepsize, datasize	Computing effort is large
Easy to program

4. How to Choose Step Size?

Choose an approriate step size is significant. If the step size is too small, it doesn’t hurt the result, but it took even more times to converge. If the step size is too large, it may cause the algorithm diverge (not converge).

The graph below shows that the value is not converge since the step size is too big.

Large Step Size