I am learning machine learning from Andrew Ng's open-class notes and coursera.org. I am trying to understand how the cost function for the logistic regression is derived. I will start with the cost function for linear regression and then get to my question with logistic regression.

(Btw a similar question was asked here, which answers the question how the derivative of cost function was derived but not the cost function itself.)

1) Linear regression uses the following hypothesis: $$ h_\theta(x) = \theta_0 + \theta_1 x$$

Accordingly, the cost function is defined as:

$$J(\theta) = \dfrac {1}{2m} \displaystyle \sum_{i=1}^m \left (h_\theta (x^{(i)}) - y^{(i)} \right)^2$$


2) The logistic regression uses a sigmoid/logistic function which is $ 0 \leq h_\theta (x) \leq 1 $.

Defined as :

$$ \begin{align*} & h_\theta (x) = \dfrac{1}{1 + e^{-(\theta^T x)}} \newline \newline \end{align*} $$

Accordingly, our cost function has also changed. However, instead of plugging-in the new h(x) equation directly, we used logarithm.

$$ \begin{align*} & J(\theta) = \dfrac{1}{m} \sum_{i=1}^m Cost(h_\theta(x^{(i)}),y^{(i)}) \newline & Cost(h_\theta(x),y) = -log(h_\theta(x)) \; & \text{if y = 1} \newline & Cost(h_\theta(x),y) = -log(1-h_\theta(x)) \; & \text{if y = 0} \end{align*} $$

And the new cost function is defined as:

$$ J(\theta) = - \frac{1}{m} \sum_{i=1}^m [y^{(i)}\log (h_\theta (x^{(i)})) + (1 - y^{(i)})\log (1 - h_\theta(x^{(i)}))]$$

From class notes ".. the more hypothesis is off from y, the larger the cost function output. If our hypothesis is equal to y, then our cost is 0."

It's also mentioned in the class notes that MLE (maximum-likelihood estimation) is used to derive the logs in the cost function. I can see how logs function and set penalty values until we find the right values, but I don't see how we came to choose them in the cost function.

  • 1
    $\begingroup$ The cost function is, up to a sign, the log- likelihood function to be maximized in the MLE procedure. Does it sound reasonable to you? Logarithms are used because they convert products into sums and do not alter the maximization search, as they are monotone increasing functions. You can find an introduction here: en.wikipedia.org/wiki/Likelihood_function $\endgroup$
    – Avitus
    Aug 3, 2014 at 20:27
  • $\begingroup$ @Avitus. Thanks. What are the benefits of converting products to sums? Is it easier computationally(computer chip instructions) or faster to reach the solution(in terms of search space)? $\endgroup$
    – RAbraham
    Apr 13, 2020 at 13:41
  • $\begingroup$ Perhaps the answer is: ""Since concavity plays a key role in the maximization, and as the most common probability distributions—in particular the exponential family—are only logarithmically concave,[33][34] it is usually more convenient to work with the log-likelihood function. Also, the log-likelihood is particularly convenient for maximum likelihood estimation. Because logarithms are strictly increasing functions, maximizing the likelihood is equivalent to maximizing the log-likelihood. ": en.wikipedia.org/wiki/Likelihood_function#Log-likelihood $\endgroup$
    – RAbraham
    Apr 13, 2020 at 13:58

4 Answers 4


Lets try to derive why the logarithm comes in the cost function of logistic regression from first principles.

So we have a dataset X consisting of m datapoints and n features. And there is a class variable y a vector of length m which can have two values 1 or 0.

Now logistic regression says that the probability that class variable value $y_i =1$ , $i=1,2,...m$ can be modelled as follows

$$ P( y_i =1 | \mathbf{x}_i ; \theta) = h_{\theta}(\mathbf{x}_i) = \dfrac{1}{1+e^{(- \theta^T \mathbf{x}_i)}} $$

so $y_i = 1$ with probability $h_{\theta}(\mathbf{x}_i)$ and $y_i=0$ with probability $1-h_{\theta}(\mathbf{x}_i)$.

This can be combined into a single equation as follows, ( actually $y_i$ follows a Bernoulli distribution)

$$ P(y_i ) = h_{\theta}(\mathbf{x}_i)^{y_i} (1 - h_{\theta}(\mathbf{x}_i))^{1-y_i}$$

$P(y_i)$ is known as the likelihood of single data point $\mathbf{x}_i$, i.e. given the value of $y_i$ what is the probability of $\mathbf{x}_i$ occurring. it is the conditional probability $P(\mathbf{x}_i | y_i)$.

The likelihood of the entire dataset $\mathbf{X}$ is the product of the individual data point likelihoods. Thus

$$ P(\mathbf{X}|\mathbf{y}) = \prod_{i=1}^{m} P(\mathbf{x}_i | y_i) = \prod_{i=1}^{m} h_{\theta}(\mathbf{x}_i)^{y_i} (1 - h_{\theta}(\mathbf{x}_i))^{1-y_i}$$

Now the principle of maximum likelihood says that we find the parameters that maximise likelihood $P(\mathbf{X}|\mathbf{y})$.

As mentioned in the comment, logarithms are used because they convert products into sums and do not alter the maximization search, as they are monotone increasing functions. Here too we have a product form in the likelihood.So we take the natural logarithm as maximising the likelihood is same as maximising the log likelihood, so log likelihood $L(\theta)$ is now:

$$ L(\theta) = \log(P(\mathbf{X}|\mathbf{y}) = \sum_{i=1}^{m} y_i \log(h_{\theta}(\mathbf{x}_i)) + (1-y_i) \log(1 - h_{\theta}(\mathbf{x}_i)) $$.

Since in linear regression we found the $\theta$ that minimizes our cost function , here too for the sake of consistency, we would like to have a minimization problem. And we want the average cost over all the data points. Currently, we have a maximimzation of $L(\theta)$. Maximization of $L( \theta)$ is equivalent to minimization of $ -L(\theta)$. And using the average cost over all data points, our cost function for logistic regresion comes out to be,

$$ J(\theta) = - \dfrac{1}{m} L(\theta)$$

$$ = - \dfrac{1}{m} \left( \sum_{i=1}^{m} y_i \log (h_{\theta}(\mathbf{x}_i)) + (1-y_i) \log (1 - h_{\theta}(\mathbf{x}_i)) \right )$$

Now we can also understand why the cost for single data point comes as follows:

the cost for a single data point is $ = -\log( P(\mathbf{x}_i | y_i))$, which can be written as $ - \left ( y_i \log (h_{\theta}(\mathbf{x}_i)) + (1 - y_i) \log (1 - h_{\theta}(\mathbf{x}_i) \right )$.

We can now split the above into two depending upon the value of $y_i$. Thus we get

$J(h_{\theta}(\mathbf{x}_i), y_i) = - \log (h_{\theta}(\mathbf{x}_i)) , \text{ if } y_i=1$, and

$J(h_{\theta}(\mathbf{x}_i), y_i) = - \log (1 - (h_{\theta}(\mathbf{x}_i) ) , \text{ if } y_i=0 $.


For calculate the gradient is used the chain rule:

$$ \dfrac{dJ(\theta)}{d\theta} = \dfrac{dJ(\theta)}{dh_\theta (x^{(i)})}\dfrac{h_\theta (x^{(i)})}{dZ}\dfrac{dZ}{d\theta} $$ In your case, your are using the Cross-Entropy as Cost Function: $$ J(\theta) = - \frac{1}{m} \sum_{i=1}^m [y^{(i)}\ln (h_\theta (x^{(i)})) + (1 - y^{(i)})\log (1 - h_\theta(x^{(i)}))] $$

Below is the explanation that how to derivative the Cross-Entropy.

$$ \dfrac{dJ(\theta)}{dy} = \dfrac{1}{m}. \Big[\dfrac{y}{h_\theta (x^{(i)})}-\dfrac{(1-y)}{(1-h_\theta (x^{(i)}))}\Big] $$

$$ = \dfrac{1}{m}. \Big[\dfrac{y.(1-h_\theta (x^{(i)})) - (1-y).h_\theta (x^{(i)})}{h_\theta (x^{(i)}).(1-h_\theta (x^{(i)}))}\Big] $$

$$ = \dfrac{1}{m}. \Big[\dfrac{y-yh_\theta (x^{(i)}) - h_\theta (x^{(i)})+y h_\theta (x^{(i)})}{h_\theta (x^{(i)}).(1-h_\theta (x^{(i)}))}\Big] $$ $$ = \dfrac{1}{m}. \Big[\dfrac{y - h_\theta (x^{(i)})}{h_\theta (x^{(i)}).(1-h_\theta (x^{(i)}))} \Big] $$ Considering that the derivative of the Sigmoid Function is: $$ \dfrac{h_\theta (x^{(i)})}{dZ} = h_\theta (x^{(i)}).(1-h_\theta (x^{(i)})) $$ The result of the gradient is: $$ \dfrac{dJ(\theta)}{dh_\theta (x^{(i)})}\dfrac{h_\theta (x^{(i)})}{dZ}\dfrac{dZ}{d\theta} = \dfrac{1}{m}. \Big[\dfrac{y - h_\theta (x^{(i)})}{h_\theta (x^{(i)}).(1-h_\theta (x^{(i)}))} \Big] . h_\theta (x^{(i)}).(1-h_\theta (x^{(i)})) .x^{(i)} $$ $$ =(y - h_\theta (x^{(i)})). x^{(i)} $$


@user76170 should be correct. But I don't understand a thing and as I am new to here I am not allowed to comment. I hope it is all right to write here.

I think, what @user76170 means by :

$\ p(y_i)=h_\theta(x_i)^{y_i}*(1−h_\theta(x_i))^{1−y_i}$

is:$$ p(y_i|x_i;\theta)=h_\theta(x_i)^{y_i}*(1−h_\theta(x_i))^{1−y_i} $$ And I don't understand why he/she changes the above to conditonal probability $\ P(x_i|y_i)$. As I have read from here http://www.stat.cmu.edu/~cshalizi/mreg/15/lectures/06/lecture-06.pdf

The expression of the log likelihood function of parameter value can be: $$ L(\theta) =log (\prod_{i=1}^{m}p(y_i|x_i;\theta))\\ L(\theta) =\sum_{i=1}^{m}log(p(y_i|x_i;\theta))\\ L(\theta) = \sum_{i=1}^{m}y_i\log(h_\theta(x_i))+(1-y_i)*log(1-h_\theta(x_i)) $$ After that, in my schoolwork, we will use first and second derivative to find the value of $\ \theta$ when the above log-likelihood reaches its maximum value and we call that the MLE method.

As here, we want to use the Cost Function method so we need a cost function which while minimising it, the log-likelihood will be maximised. Hence, we just need to add a minus to it as suggested by @user76170. For reason I don't know, the Cost Function given below and in the lecture is the average of cost over all data points. $$ J(\theta)=-\frac{1}{m}*L(\theta) $$ Why do we have have to divide it by m. Will that make the gradient descent method goes quicker? Probably, I should raise a question.

  • $\begingroup$ I have the same question as this. $\endgroup$
    – YQ.Wang
    Dec 5, 2019 at 13:27

Maybe, we can try to imagine how the values returned by the sigmoid function would look. Even if there are a million inputs, all of them will be converted to values between 0 and 1 by the sigmoid function.

On top of that, if we use the square mean error function to find the cost, we would get extremely small fractions. You would basically be squaring a fraction and then dividing it by a relatively large number. Larger the size of the input sample, smaller will be the mean square error. Beyond a certain training set size, the cost will be so small that it would become impossible to be handled by our computers, and would essentially be converted to zero. Just imagine, your algorithm fails to find the cost, and returns zero all the time.

In order to avoid this situation, we take the logs of the actual sigmoid, because log of even the tiniest fraction will be a manageable and sufficiently sized number.


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