## Priliminaries

- A Simple Linear Regression
- Least Squares Estimation
- linear algebra

## Square Loss Function for Regression^{1}

For any input \(\mathbf{x}\), our goal in a regression task is to give a prediction \(\hat{y}=f(\mathbf{x})\) to approximate target \(t\) where the function \(f(\cdot)\) is the chosen hypothesis or model as mentioned in the post https://anthony-tan.com/A-Simple-Linear-Regression/.

The difference between \(t\) and \(\hat{y}\) can be called ‘error’ or more precisely ‘loss’. Because in an approximation task, ‘error’ occurs by chance and always exists, and ‘loss’ is a good word to represent the difference. The loss can be written generally as function \(\ell(f(\mathbf{x}),t)\). Intuitively, the smaller the loss, the better the approximation.

So the expectation of loss:

\[ \mathbb E[\ell]=\int\int \ell(f(\mathbf{x}),t)p(\mathbf{x},t)d \mathbf{x}dt\tag{1} \]

should be as small as possible.

In probability viewpoint, the input vector \(\mathbf{x}\), target \(t\) and parameters in function(model) \(f(\cdot)\) are all random variables. Then the expectation of loss function may exist.

Considering the square error loss function \(e=(f(\mathbf{x})-t)^2\), it is a usual measure of the difference between the prediction and the target. And substitute the loss function into equation (1), we have:

\[ \mathbb E[\ell]=\int\int (f(\mathbf{x})-t)^2p(\mathbf{x},t)d \mathbf{x}dt\tag{2} \]

To minimize this function, we could use Euler-Lagrange equation, Fundamental theorem of calculus and Fubini’s theorem:

Fubini’s theorem told us that we can change the order of integration: \[ \begin{aligned} \mathbb E[\ell]&=\int\int (f(\mathbf{x})-t)^2p(\mathbf{x},t)d \mathbf{x}dt\\ &=\int\int (f(\mathbf{x})-t)^2p(\mathbf{x},t)dtd \mathbf{x} \end{aligned}\tag{3} \]

According to the Euler-Lagrange equation, we first create a new function \(G(x,f,f')\): \[ G(x,f,f')= \int (f(\mathbf{x})-t)^2p(\mathbf{x},t)dt\tag{4} \]

The Euler-Lagrange equation is used to minimize the equation (2): \[ \frac{\partial G}{\partial f}-\frac{d}{dx}\frac{\partial G}{\partial f'}=0\tag{5} \]

Because there is no \(y'\) component in function \(G()\). Then the equation: \[ \frac{\partial G}{\partial f}=0\tag{6} \] becomes the necessary condition to minimize the equation (2):

\[ 2\int (f(\mathbf{x})-t)p(\mathbf{x},t)dt=0 \tag{7} \]

Rearrange the equation (7), and we get a good predictor that can minimize the square loss function :

\[ \begin{aligned} \int (f(\mathbf{x})-t)p(\mathbf{x},t)dt&=0\\ \int f(\mathbf{x})p(\mathbf{x},t)dt-\int tp(\mathbf{x},t)dt&=0\\ f(\mathbf{x})\int p(\mathbf{x},t)dt&=\int tp(\mathbf{x},t)dt\\ f(\mathbf{x})&=\frac{\int tp(\mathbf{x},t)dt}{\int p(\mathbf{x},t)dt}\\ f(\mathbf{x})&=\frac{\int tp(\mathbf{x},t)dt}{p(\mathbf{x})}\\ f(\mathbf{x})&=\int tp(t|\mathbf{x})dt\\ f(\mathbf{x})&= \mathbb{E}_t[t|\mathbf{x}] \end{aligned}\tag{8} \]

We finally find the expectation of \(t\) given \(\mathbf{x}\) is the optimum solution. The expectation of \(t\) given \(\mathbf{x}\) is also called the **regression function**.

A small summary: \(\mathbb{E}[t| \mathbf{x}]\) is a good estimate of \(f(\mathbf{x})\)

## Maximum Likelihood Estimation

Generally, we assume that there is a generator behind the data:

\[ t=g(\mathbf{x},\mathbf{w})+\epsilon\tag{9} \]

where the function \(g(\mathbf{x},\mathbf{w})\) is a deterministic function, \(t\) is the target variable and \(\epsilon\) is zero mean Gaussian random variable with percision \(\beta\) which is the inverse variance. Because of the property of Gaussian distribution, \(t\) has a Gaussian distribution, with mean(expectation) \(g(\mathbf{x},\mathbf{w})\) and percesion \(\beta\). And recalling the standard form of Gaussian distribution:

\[ \begin{aligned} \Pr(t|\mathbf{x},\mathbf{w},\beta)&=\mathcal{N}(t|g(\mathbf{x},\mathbf{w}),\beta^{-1})\\ &=\frac{\beta}{\sqrt{2\pi}}\mathrm{e}^{-\frac{1}{2}(\beta(x-\mu)^2)} \end{aligned}\tag{10} \]

Our task here is to approximate the generator in equation (9) with a linear function. Somehow, when we use the square loss function, the optimum solution for this task is \(\mathbb{E}[t|\mathbf{x}]\) to equation (8).

the solution to equation (10) is:

\[ \mathbb{E}[t|\mathbf{x}]=g(\mathbf{x},\mathbf{w})\tag{11} \]

We set the linear model as: \[ f(x)=\mathbf{w}^T\mathbf{x}+b\tag{12} \]

and this can be converted to:

\[ f(x)= \begin{bmatrix} b&\mathbf{w}^T \end{bmatrix} \begin{bmatrix} 1\\ \mathbf{x} \end{bmatrix}=\mathbf{w}_a^T\mathbf{x}_a \tag{13} \]

for short, we just write the \(\mathbf{w}_a\) and \(\mathbf{x}_a\) as \(\mathbf{w}\) and \(\mathbf{x}\). Then the linear model becomes:

\[ f(x)=\mathbf{w}^T\mathbf{x}\tag{14} \]

As we mentioned above we consider all the parameter as a random variable, then the conditioned distribution of \(\mathbf{w}\) is \(\Pr(\mathbf{w}|\mathbf{t},\beta)\). \(X\) or \(\mathbf{x}\) was omitted in the condition because it does not affect the result at all. And the Bayesian theorem told us:

\[ \Pr(\mathbf{w}|\mathbf{t},\beta)=\frac{\Pr( \mathbf{t}|\mathbf{w},\beta) \Pr(\mathbf{w})} {\Pr(\mathbf{t})}=\frac{\text{Likelihood}\times \text{Prior}}{\text{Evidence}}\tag{15} \]

We want to find the \(\mathbf{w}^{\star}\) that maximise the posterior probability \(\Pr(\mathbf{w}|\mathbf{t},\beta)\). Because \(\Pr(\mathbf{t})\) and \(\Pr(\mathbf{w})\) are constant. Then the maximum of likelihood \(\Pr(\mathbf{t}|\mathbf{w},\beta)\) maximise the posterior probability.

\[ \begin{aligned} \Pr(\mathbf{t}|\mathbf{w},\beta)&=\Pi_{i=0}^{N}\mathcal{N}(t_i|\mathbf{w}^T\mathbf{x}_i,\beta^{-1})\\ \ln \Pr(\mathbf{t}|\mathbf{w},\beta)&=\sum_{i=0}^{N}\ln \mathcal{N}(t_i|\mathbf{w}^T\mathbf{x}_i,\beta^{-1})\\ &=\sum_{i=0}^{N}\ln \frac{\beta}{\sqrt{2\pi}}\mathrm{e}^{-\frac{1}{2}(\beta(t_i-\mathbf{w}^T\mathbf{x}_i)^2)}\\ &=\sum_{i=0}^{N} \ln \beta - \sum_{i=0}^{N} \ln \sqrt{2\pi} - \frac{1}{2}\beta\sum_{i=0}^{N}(t_i-\mathbf{w}^T\mathbf{x}_i)^2 \end{aligned}\tag{16} \]

This gives us a wonderful result.

We can only control the component \(\frac{1}{2}\beta\sum_{i=0}^{N}(t_i-\mathbf{w}^T\mathbf{x}_i)^2\) of the last line of equation(16), because \(\sum_{i=0}^{N} \ln \beta\) and \(- \sum_{i=0}^{N} \ln \sqrt{2\pi}\) were decided by the assumptions. In other words, to maximise the likelihood, we just need to minimise:

\[ \sum_{i=0}^{N}(t_i-\mathbf{w}^T\mathbf{x}_i)^2\tag{17} \]

This was just to minimize the sum of squares. Then this optimization problem went back to the least square problem.

## Least Square Estimation and Maximum Likelihood Estimation

When we assume there is a generator:

\[ t=g(\mathbf{x},\mathbf{w})+\epsilon\tag{18} \]

behind the data, and \(\epsilon\) has a zero-mean Gaussian distribution with any precision \(\beta\), the maximum likelihood estimation finally converts to the least square estimation. This is not only worked for linear regression because we did not assume what \(g(\mathbf{x},\mathbf{w})\) is.

However, when the \(\epsilon\) has a different distribution but not Gaussian distribution, the least square estimation will not be the optimum solution for maximum likelihood estimation.

## References

Bishop, Christopher M. Pattern recognition and machine learning. springer, 2006.↩︎