Python/machine_learning/local_weighted_learning/local_weighted_learning.md

Locally Weighted Linear Regression

It is a non-parametric ML algorithm that does not learn on a fixed set of parameters such as linear regression.
So, here comes a question of what is linear regression?
Linear regression is a supervised learning algorithm used for computing linear relationships between input (X) and output (Y). \

Terminology Involved

number_of_features(i) = Number of features involved.
number_of_training_examples(m) = Number of training examples.
output_sequence(y) = Output Sequence.
\theta ^T x = predicted point.
J($\theta$) = COst function of point.

The steps involved in ordinary linear regression are:

Training phase: Compute \theta to minimize the cost.
J($\theta$) = \sum_{i=1}^m (($\theta$)^T x^i - $y^i$)^2

Predict output: for given query point x,
return: ($\theta$)^T x

This training phase is possible when data points are linear, but there again comes a question can we predict non-linear relationship between x and y ? as shown below

So, here comes the role of non-parametric algorithm which doesn't compute predictions based on fixed set of params. Rather parameters $\theta$ are computed individually for each query point/data point x.

While Computing $\theta$ , a higher preference is given to points in the vicinity of x than points farther from x.

Cost Function J($\theta$) = \sum_{i=1}^m w^i (($\theta$)^T x^i - $y^i$)^2

w^i is non-negative weight associated to training point x^i.
w^i is large fr $x^i$'s lying closer to query point x_i.
w^i is small for $x^i$'s lying farther to query point x_i.

A Typical weight can be computed using \

w^i = $\exp$(-$\frac{(x^i-x)(x^i-x)^T}{2\tau^2}$)

Where \tau is the bandwidth parameter that controls w^i distance from x.

Let's look at a example :

Suppose, we had a query point x=5.0 and training points $x^1$=4.9 and $x^2$=5.0 than we can calculate weights as :

w^i = $\exp$(-$\frac{(x^i-x)(x^i-x)^T}{2\tau^2}$) with $\tau$=0.5

w^1 = $\exp$(-$\frac{(4.9-5)^2}{2(0.5)^2}$) = 0.9802

w^2 = $\exp$(-$\frac{(3-5)^2}{2(0.5)^2}$) = 0.000335

So, J($\theta$) = 0.9802*(\theta ^T x^1 - $y^1$) + 0.000335*(\theta ^T x^2 - $y^2$)

So, here by we can conclude that the weight fall exponentially as the distance between x & x^i increases and So, does the contribution of error in prediction for x^i to the cost.

Steps involved in LWL are :
Compute \theta to minimize the cost. J($\theta$) = \sum_{i=1}^m w^i (($\theta$)^T x^i - $y^i$)^2
Predict Output: for given query point x,
return : \theta ^T x