Logistic Regression – How Is the Cost Function from Logistic Regression Differentiated?

derivativegradient descentlogisticregression

I am doing the Machine Learning Stanford course on Coursera.

In the chapter on Logistic Regression, the cost function is this:

Then, it is differentiated here:

I tried getting the derivative of the cost function, but I got something completely different.

How is the derivative obtained?

Which are the intermediary steps?

Best Answer

Adapted from the notes in the course, which I don't see available (including this derivation) outside the notes contributed by students within the page of Andrew Ng's Coursera Machine Learning course.

In what follows, the superscript $(i)$ denotes individual measurements or training "examples."

$\small \frac{\partial J(\theta)}{\partial \theta_j} = \frac{\partial}{\partial \theta_j} \,\frac{-1}{m}\sum_{i=1}^m \left[ y^{(i)}\log\left(h_\theta \left(x^{(i)}\right)\right) + (1 -y^{(i)})\log\left(1-h_\theta \left(x^{(i)}\right)\right)\right] \\[2ex]\small\underset{\text{linearity}}= \,\frac{-1}{m}\,\sum_{i=1}^m \left[ y^{(i)}\frac{\partial}{\partial \theta_j}\log\left(h_\theta \left(x^{(i)}\right)\right) + (1 -y^{(i)})\frac{\partial}{\partial \theta_j}\log\left(1-h_\theta \left(x^{(i)}\right)\right) \right] \\[2ex]\Tiny\underset{\text{chain rule}}= \,\frac{-1}{m}\,\sum_{i=1}^m \left[ y^{(i)}\frac{\frac{\partial}{\partial \theta_j}h_\theta \left(x^{(i)}\right)}{h_\theta\left(x^{(i)}\right)} + (1 -y^{(i)})\frac{\frac{\partial}{\partial \theta_j}\left(1-h_\theta \left(x^{(i)}\right)\right)}{1-h_\theta\left(x^{(i)}\right)} \right] \\[2ex]\small\underset{h_\theta(x)=\sigma\left(\theta^\top x\right)}=\,\frac{-1}{m}\,\sum_{i=1}^m \left[ y^{(i)}\frac{\frac{\partial}{\partial \theta_j}\sigma\left(\theta^\top x^{(i)}\right)}{h_\theta\left(x^{(i)}\right)} + (1 -y^{(i)})\frac{\frac{\partial}{\partial \theta_j}\left(1-\sigma\left(\theta^\top x^{(i)}\right)\right)}{1-h_\theta\left(x^{(i)}\right)} \right] \\[2ex]\Tiny\underset{\sigma'}=\frac{-1}{m}\,\sum_{i=1}^m \left[ y^{(i)}\, \frac{\sigma\left(\theta^\top x^{(i)}\right)\left(1-\sigma\left(\theta^\top x^{(i)}\right)\right)\frac{\partial}{\partial \theta_j}\left(\theta^\top x^{(i)}\right)}{h_\theta\left(x^{(i)}\right)} - (1 -y^{(i)})\,\frac{\sigma\left(\theta^\top x^{(i)}\right)\left(1-\sigma\left(\theta^\top x^{(i)}\right)\right)\frac{\partial}{\partial \theta_j}\left(\theta^\top x^{(i)}\right)}{1-h_\theta\left(x^{(i)}\right)} \right] \\[2ex]\small\underset{\sigma\left(\theta^\top x\right)=h_\theta(x)}= \,\frac{-1}{m}\,\sum_{i=1}^m \left[ y^{(i)}\frac{h_\theta\left( x^{(i)}\right)\left(1-h_\theta\left( x^{(i)}\right)\right)\frac{\partial}{\partial \theta_j}\left(\theta^\top x^{(i)}\right)}{h_\theta\left(x^{(i)}\right)} - (1 -y^{(i)})\frac{h_\theta\left( x^{(i)}\right)\left(1-h_\theta\left(x^{(i)}\right)\right)\frac{\partial}{\partial \theta_j}\left( \theta^\top x^{(i)}\right)}{1-h_\theta\left(x^{(i)}\right)} \right] \\[2ex]\small\underset{\frac{\partial}{\partial \theta_j}\left(\theta^\top x^{(i)}\right)=x_j^{(i)}}=\,\frac{-1}{m}\,\sum_{i=1}^m \left[y^{(i)}\left(1-h_\theta\left(x^{(i)}\right)\right)x_j^{(i)}- \left(1-y^{i}\right)\,h_\theta\left(x^{(i)}\right)x_j^{(i)} \right] \\[2ex]\small\underset{\text{distribute}}=\,\frac{-1}{m}\,\sum_{i=1}^m \left[y^{i}-y^{i}h_\theta\left(x^{(i)}\right)- h_\theta\left(x^{(i)}\right)+y^{(i)}h_\theta\left(x^{(i)}\right) \right]\,x_j^{(i)} \\[2ex]\small\underset{\text{cancel}}=\,\frac{-1}{m}\,\sum_{i=1}^m \left[y^{(i)}-h_\theta\left(x^{(i)}\right)\right]\,x_j^{(i)} \\[2ex]\small=\frac{1}{m}\sum_{i=1}^m\left[h_\theta\left(x^{(i)}\right)-y^{(i)}\right]\,x_j^{(i)} $

The derivative of the sigmoid function is

$\Tiny\begin{align}\frac{d}{dx}\sigma(x)&=\frac{d}{dx}\left(\frac{1}{1+e^{-x}}\right)\\[2ex] &=\frac{-(1+e^{-x})'}{(1+e^{-x})^2}\\[2ex] &=\frac{e^{-x}}{(1+e^{-x})^2}\\[2ex] &=\left(\frac{1}{1+e^{-x}}\right)\left(\frac{e^{-x}}{1+e^{-x}}\right)\\[2ex] &=\left(\frac{1}{1+e^{-x}}\right)\,\left(\frac{1+e^{-x}}{1+e^{-x}}-\frac{1}{1+e^{-x}}\right)\\[2ex] &=\sigma(x)\,\left(\frac{1+e^{-x}}{1+e^{-x}}-\sigma(x)\right)\\[2ex] &=\sigma(x)\,(1-\sigma(x)) \end{align}$

Related Solutions

Solved – Logistic Regression Cost Function issue in Matlab

In matlab or R I would suggest to use vectorized implementation, which logistic loss is just

$$ \mathbf 1^T (-b \log p -(1-b) \log(1-p)) $$

Where

$$p=\frac 1 {1+\exp{(-Ax)}}$$

In above notation, $A$ is data matrix and $b$ is the binary response.

Here is the matlab code (with gradient calculation)

function [J grad] = lg_loss(A,b,x)
    sigmoid = @(z) 1.0 ./ (1.0 + exp(-z));
    p=sigmoid(A*x);
    J=sum(-b.*log(p)-(1-b).*log(1-p));
    grad=A'*(p-b);
end

Regression – Understanding Cost Function in OLS Linear Regression

As you seem to realize, we certainly don't need the $1/m$ factor to get linear regression. The minimizers will of course be exactly the same, with or without it. One typical reason to normalize by $m$ is so that we can view the cost function as an approximation to the "generalization error", which is the expected square loss on a randomly chosen new example (not in the training set):

Suppose $(X,Y),(X^{(1)},Y^{(1)}),\ldots,(X^{(m)},Y^{(m)})$ are sampled i.i.d. from some distribution. Then for large $m$ we expect that $$ \frac{1}{m} \sum _{i=1}^m \left(h_\theta(X^{(i)})-Y^{(i)}\right)^2 \approx \mathbb{E}\left(h_\theta(X)-Y\right)^2. $$

More precisely, by the Strong Law of Large Numbers, we have $$ \lim_{m\to\infty} \frac{1}{m} \sum _{i=1}^m \left(h_\theta(X^{(i)})-Y^{(i)}\right)^2 = \mathbb{E}\left(h_\theta(X)-Y\right)^2 $$ with probability 1.

Note: Each of the statements above are for any particular $\theta$, chosen without looking at the training set. For machine learning, we want these statements to hold for some $\hat{\theta}$ chosen based on its good performance on the training set. These claims can still hold in this case, though we need to make some assumptions on the set of functions $\{h_\theta \,|\, \theta \in \Theta\}$, and we'll need something stronger than the Law of Large Numbers.

Best Answer

Related Solutions

Solved – Logistic Regression Cost Function issue in Matlab

Regression – Understanding Cost Function in OLS Linear Regression

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