linear regression

premise

1. $X$ is a matrix which has m rows and n columns, that means it is a $m \times n$ matrix, represents for training set.
2. $\theta$ is a $1 \times n$ vector, stands for hypothesis parameter.
3. $y$ is a $m \times 1$ vector, stands for real value of training set.
4. $\alpha$ named learning rate for defining learning or descending speed.
5. $S(X_j)$ means to get standard deviation of the j feature from training set.

---

1. Hypothesis

Draw hypothesis of a pattern.

$$
h_{\theta}(X) = X \times \theta^T
$$

2. Cost

Calculate the Cost for single training point.

$$Cost(X^{(i)},y^{(i)})=[h_{\theta}(X^{(i)}) - y^{(i)}]^2$$

3. Cost function

Draw cost function for iterating whole training set.

$$
\begin{aligned}
J(\theta) &=\left(\frac{1}{2m}\right)\sum_{i=1}^m Cost(X^{(i)},y^{(i)}) \
&=\left(\frac{1}{2m}\right)\sum_{i=1}^m[h_{\theta}(X^{(i)}) - y^{(i)}]^2
\end{aligned}
$$

4. Get optimized parameter

Learn from training set to get optimized parameter for proposed algorithm.

Gradient Descend###

Complicate to implement.
suitable for any senario.

$$
\begin{aligned}
grad(j) &= \frac{\partial}{\partial \theta_j} J(\theta) \
&= \frac{1}{m}\sum_{i=1}^m[(h_{\theta}(X^{(i)}) - y^{(i)}) X_{j}^{(i)}]
\end{aligned}
$$

$$\theta_j := \theta_j - \alpha \times grad(j) \quad \text{Repeat many times} $$

Normal equation###

Convenient, but performance bad while m grow large than 100000.
Unable to conquer non-invertable matrix.

$$
\theta = (X^{T}X)^{\prime}X^{T}y
$$

Feature scaling

Use feature scaling to optimize training set.
Make gradient descend converge much faster.

$$X_j=\frac{X_j - \mu}{a}$$

$$
\begin{aligned}
a &= max(X_j)-min(X_j) \
& or \
&= S(X_j)
\end{aligned}
$$