linear regression
premise
- $X$ is a
matrixwhich hasmrows andncolumns, that means it is a $m \times n$ matrix, represents for training set. - $\theta$ is a $1 \times n$
vector, stands for hypothesis parameter. - $y$ is a $m \times 1$
vector, stands for real value of training set. - $\alpha$ named
learning ratefor defining learning or descending speed. - $S(X_j)$ means to get standard deviation of the j feature from training set.
1. Hypothesis
\[h_{\theta}(X) = X \times \theta^T\]Draw hypothesis of a pattern.
2. Cost
\[Cost(X^{(i)},y^{(i)})=[h_{\theta}(X^{(i)}) - y^{(i)}]^2\]Calculate the Cost for single training point.
3. Cost function
\[\begin{aligned} J(\theta) &=\left(\frac{1}{2m}\right)\sum_{i=1}^m Cost(X^{(i)},y^{(i)}) \\ &=\left(\frac{1}{2m}\right)\sum_{i=1}^m[h_{\theta}(X^{(i)}) - y^{(i)}]^2 \end{aligned}\]Draw cost function for iterating whole training set.
4. Get optimized parameter
Learn from training set to get optimized parameter for proposed algorithm.
Gradient Descend###
\[\begin{aligned} grad(j) &= \frac{\partial}{\partial \theta_j} J(\theta) \\ &= \frac{1}{m}\sum_{i=1}^m[(h_{\theta}(X^{(i)}) - y^{(i)}) X_{j}^{(i)}] \end{aligned}\] \[\theta_j := \theta_j - \alpha \times grad(j) \quad \text{Repeat many times}\]Complicate to implement.
suitable for any senario.
Normal equation###
\[\theta = (X^{T}X)^{\prime}X^{T}y\]Convenient, but performance bad while
mgrow large than 100000.
Unable to conquer non-invertable matrix.
Feature scaling
\[X_j=\frac{X_j - \mu}{a}\] \[\begin{aligned} a &= max(X_j)-min(X_j) \\ & or \\ &= S(X_j) \end{aligned}\]Use feature scaling to optimize training set.
Make gradient descend converge much faster.