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by Sameer NegiJuly 6th, 2018

Linear regression is a simple solution to our classification problems but what happens when it fails. As we will see in below problem

Suppose we want to classify **Y= {0,1}** and **X** are **data samples**. It is binary classification. Let’s try it with linear Regression

Wow, Linear Regression has done the job.It is really a good fit but what happens if i add a new data to given data set.

It’s really a bad solution. Linear Regression didn’t worked. What should we do next? If there is problem, there is solution and the solution is **Logistic Regression****.** Let’s start with formal definition and have an idea what is linear and logistic Regression are:

In ** linear regression**, the outcome (dependent variable) is continuous. It can have any one of an infinite number of possible values. In

Intuitively, it also doesn’t make sense for h(x) to take values larger than 1 or smaller than 0 when we know that y ∈ {0, 1}.To fix this, lets change the form for our hypotheses h(x). We will choose hypothesis as follows :

g(z) is called ** Sigmoid function** or L**

Let’s assume that

if we combine above equations together it can be re written as follows:

As in above equation when **y =0**, h(x) will become 1, p(y|x;theta) = (1-h(x)) and when **y =1**,(1- h(x)) will become 1, p(y|x;theta) = h(x)

Now, **likelihood of the parameters** are given by

and when we *Maximize log likelihood,**it* will be given by :

Now we can use **Gradient Descent** , we already know that h(x) is given by sigmoid function.

For ease, the partial derivative of g(z) with respect to z is given by

Now if apply gradient descent by taking the *partial derivative of log likelihood with respect to theta*

If we compare the above rule to the least mean square, it looks identical but it’s not. This is a different learning algorithm because h(x) is now defined as non-linear function of theta transpose * x[i].

If you find any inconsistency in my post, please feel free to point out in the comments. Thanks for reading.

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