Bayesian Linear Regression

#Machine Learning #Regression #Bayesian

Linear Regression and Likelihood

The linear estimator $y$ is

$$ \begin{equation} y^n = \beta^m X_m^{\phantom{m}n}. \label{eq-linear-model} \end{equation} $$

As usual, we have redefined our data to get rid of the intercept $\beta^0$.

In ordinary linear models, we find the error being the difference between the target $\hat y$ and the estimator $y$

$$ \epsilon = \hat y - y, $$

which is required to have a minimum absolute value.

In linear regressions, we use least squares to solve the problem. In Bayesian linear regression, instead of using a deterministic estimator $\beta^m X_m^{\phantom{m}n}$, we assume a Gaussian random estimator

$$ \begin{equation} \mathcal{N}(\mu, \sigma^2) = \mathcal{N}(\beta^m X_m^{\phantom{m}n}, \sigma^2), \end{equation} $$

where we have used the knowledge of linear regression, that the mean of the estimator should be a linear model $\beta^m X_m^{\phantom{m}n}$. The likelihood becomes

$$ \begin{equation} P(\hat y^n \mid [X_m^{\phantom{m}n}, \beta^m] ) = \frac{1}{\sqrt{2 \sigma^2 \pi}} \exp \left( -\frac{(\hat{y^n} - \beta^m X_m^{\phantom{m}n})^2}{2 \sigma^2} \right) \end{equation} $$

It is not surprising that requiring the maximum likelihood will lead to the same result as the least-squares since the log operation takes out the exponential operation.

Bayesian Linear Model

Applying Bayes’ theorem to this problem,

$$ P( [X_m^{\phantom{m}n}, \beta^m] \mid \hat y^n ) {\color{red}P(\hat y^n)} = P(\hat y^n \mid [X_m^{\phantom{m}n}, \beta^m] ) P([X_m^{\phantom{m}n}, \beta^m]). $$

Since ${\color{red}P(\hat y^n)}$ doesn’t depend on the parameters and is a constant, we will ignore it for the sake of optimization.

$$ P( [X_m^{\phantom{m}n}, \beta^m] \mid \hat y^n ) \propto P(\hat y^n \mid [X_m^{\phantom{m}n}, \beta^m] ) P([X_m^{\phantom{m}n}, \beta^m]). $$

Fall back to Maximum Likelihood

If $$P([X_m^{\phantom{m}n}, \beta^m]) = 1.$$

We will assume a least information model for $P([X_m^{\phantom{m}n}, \beta^m])$, that is

$$ P([X_m^{\phantom{m}n}, \beta^m]) = \mathcal{N} (0, \sigma_\beta^2) = \frac{1}{\sqrt{2 \sigma_\beta^2 \pi}} \exp \left( -\frac{(\beta^m )^2}{2 \sigma_\beta^2} \right). $$

Our posterior becomes

$$ \log P( [X_m^{\phantom{m}n}, \beta^m] \mid \hat y^n ) = -\frac{(\hat{y^n} - \beta^m X_m^{\phantom{m}n})^2}{2 \sigma^2} -\frac{(\beta^m )^2}{2 \sigma_\beta^2} + \mathrm{Const.} $$

This is nothing but Ridge loss with coefficient $\lambda$, where

$$ \frac{\sigma^2}{\sigma_\beta^2} = \lambda. $$

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