In Logistic Regression, MLE is used to develop a mathematical function to estimate the model parameters, optimization techniques like Gradient Descent are used to solve this function.

Can somebody please provide a resource that explains this process with an example?


2 Answers 2


In short, Maximum Likelihood estimation is used to find parameters given target values y and x. The Maximum likelhood estimation finds the parameters maximises the probability of y given x. It has been proved that MLE estimation problem caan be solved by finding the parametrs which gives least cross entropy in case of binary classification.

Gradients descent is an optimisation algorithm get helps you update the parameters iteratively to find the parameters which gives highest probability of y.

For more details:https://www.google.com/amp/s/glassboxmedicine.com/2019/12/07/connections-log-likelihood-cross-entropy-kl-divergence-logistic-regression-and-neural-networks/amp/


Maximum Likelihood

Maximum likelihood estimation involves defining a likelihood function for calculating the conditional probability of observing the data sample given probability distribution and distribution parameters. This approach can be used to search a space of possible distributions and parameters.

The logistic model uses the sigmoid function (denoted by sigma) to estimate the probability that a given sample y belongs to class $1$ given inputs $X$ and weights $W$,

\begin{align} \ P(y=1 \mid x) = \sigma(W^TX) \end{align}

where the sigmoid of our activation function for a given $n$ is:

\begin{align} \large y_n = \sigma(a_n) = \frac{1}{1+e^{-a_n}} \end{align}

The accuracy of our model predictions can be captured by the objective function $L$, which we are trying to maximize.

\begin{align} \large L = \displaystyle\prod_{n=1}^N y_n^{t_n}(1-y_n)^{1-t_n} \end{align}

If we take the log of the above function, we obtain the maximum log-likelihood function, whose form will enable easier calculations of partial derivatives. Specifically, taking the log and maximizing it is acceptable because the log-likelihood is monotonically increasing, and therefore it will yield the same answer as our objective function.

\begin{align} \ L = \displaystyle \sum_{n=1}^N t_nlogy_n+(1-t_n)log(1-y_n) \end{align}

In our example, we will actually convert the objective function (which we would try to maximize) into a cost function (which we are trying to minimize) by converting it into the negative log-likelihood function:

\begin{align} \ J = -\displaystyle \sum_{n=1}^N t_nlogy_n+(1-t_n)log(1-y_n) \end{align}

Gradient Descent

Gradient descent is an iterative optimization algorithm, which finds the minimum of a differentiable function. In this process, we try different values and update them to reach the optimal ones, minimizing the output.

Once we have an objective function, we can generally take its derivative with respect to the parameters (weights), set it equal to zero, and solve for the parameters to obtain the ideal solution. However, in the case of logistic regression (and many other complexes or otherwise non-linear systems), this analytical method doesn’t work. Instead, we resort to a method known as gradient descent, whereby we randomly initialize and then incrementally update our weights by calculating the slope of our objective function. When applying the cost function, we want to continue updating our weights until the slope of the gradient gets as close to zero as possible. We can show this mathematically:

\begin{align} \ w:=w+\triangle w \end{align}

where the second term on the right is defined as the learning rate times the derivative of the cost function with respect to the weights (which is our gradient):

\begin{align} \ \triangle w = \eta\triangle J(w) \end{align}

Thus, we want to take the derivative of the cost function with respect to the weight, which, using the chain rule, gives us:

\begin{align} \frac{J}{\partial w_i} = \displaystyle \sum_{n=1}^N \frac{\partial J}{\partial y_n}\frac{\partial y_n}{\partial a_n}\frac{\partial a_n}{\partial w_i} \end{align}

Thus, we are looking to obtain three different derivatives. Let us start by solving for the derivative of the cost function with respect to $y$:

\begin{align} \frac{\partial J}{\partial y_n} = t_n \frac{1}{y_n} + (1-t_n) \frac{1}{1-y_n}(-1) = \frac{t_n}{y_n} - \frac{1-t_n}{1-y_n} \end{align}

Next, let us solve for the derivative of $y$ with respect to our activation function:

\begin{align} \large y_n = \sigma(a_n) = \frac{1}{1+e^{-a_n}} \end{align}

\begin{align} \frac{\partial y_n}{\partial a_n} = \frac{-1}{(1+e^{-a_n})^2}(e^{-a_n})(-1) = \frac{e^{-a_n}}{(1+e^-a_n)^2} = \frac{1}{1+e^{-a_n}} \frac{e^{-a_n}}{1+e^{-a_n}} \end{align}

\begin{align} \frac{\partial y_n}{\partial a_n} = y_n(1-y_n) \end{align}

And lastly, we solve for the derivative of the activation function with respect to the weights:

\begin{align} \ a_n = W^TX_n \end{align}

\begin{align} \ a_n = w_0x_{n0} + w_1x_{n1} + w_2x_{n2} + \cdots + w_Nx_{NN} \end{align}

\begin{align} \frac{\partial a_n}{\partial w_i} = x_{ni} \end{align}

Now we can put it all together and simply.

\begin{align} \frac{\partial J}{\partial w_i} = - \displaystyle\sum_{n=1}^N\frac{t_n}{y_n}y_n(1-y_n)x_{ni}-\frac{1-t_n}{1-y_n}y_n(1-y_n)x_{ni} \end{align}

\begin{align} = - \displaystyle\sum_{n=1}^Nt_n(1-y_n)x_{ni}-(1-t_n)y_nx_{ni} \end{align}

\begin{align} = - \displaystyle\sum_{n=1}^N[t_n-t_ny_n-y_n+t_ny_n]x_{ni} \end{align}

\begin{align} \frac{\partial J}{\partial w_i} = \displaystyle\sum_{n=1}^N(y_n-t_n)x_{ni} = \frac{\partial J}{\partial w} = \displaystyle\sum_{n=1}^{N}(y_n-t_n)x_n \end{align}

We can get rid of the summation above by applying the principle that a dot product between two vectors is a summover sum index. That is:

\begin{align} \ a^Tb = \displaystyle\sum_{n=1}^Na_nb_n \end{align}

Therefore, the gradient with respect to w is:

\begin{align} \frac{\partial J}{\partial w} = X^T(Y-T) \end{align}

If you are asking yourself where the bias term of our equation (w0) went, we calculate it the same way, except our x becomes 1.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.