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## Monday, October 21, 2019

A kind of boosting using functional gradient descent. It uses weak learner to fit gradient residuals and then add to original learner, thus making an ensemble.
 Source: Wikipedia

### XGBOOST

$$\text{obj}(\theta) = L(\theta) + \Omega(\theta)$$

where $L$ is the training loss function, and $Ω$ is the regularization term. The training loss measures how predictive our model is with respect to the training data. A common choice of L is the mean squared error, which is given by
$$L(\theta) = \sum_i (y_i-\hat{y}_i)^2$$

$$\text{obj} = \sum_{i=1}^n l(y_i, \hat{y}_i^{(t)}) + \sum_{i=1}^t\Omega(f_i)$$

$$\begin{split}\hat{y}_i^{(0)} &= 0\\ \hat{y}_i^{(1)} &= f_1(x_i) = \hat{y}_i^{(0)} + f_1(x_i)\\ \hat{y}_i^{(2)} &= f_1(x_i) + f_2(x_i)= \hat{y}_i^{(1)} + f_2(x_i)\\ &\dots\\ \hat{y}_i^{(t)} &= \sum_{k=1}^t f_k(x_i)= \hat{y}_i^{(t-1)} + f_t(x_i)\end{split}$$

For step t:
$$\begin{split}\text{obj}^{(t)} & = \sum_{i=1}^n l(y_i, \hat{y}_i^{(t)}) + \sum_{i=1}^t\Omega(f_i) \\ & = \sum_{i=1}^n l(y_i, \hat{y}_i^{(t-1)} + f_t(x_i)) + \Omega(f_t) + \mathrm{constant}\end{split}$$

For L2 loss:
$$\begin{split}\text{obj}^{(t)} & = \sum_{i=1}^n (y_i - (\hat{y}_i^{(t-1)} + f_t(x_i)))^2 + \sum_{i=1}^t\Omega(f_i) \\ & = \sum_{i=1}^n [2(\hat{y}_i^{(t-1)} - y_i)f_t(x_i) + f_t(x_i)^2] + \Omega(f_t) + \mathrm{constant}\end{split}$$

$$\text{obj}^{(t)} = \sum_{i=1}^n [l(y_i, \hat{y}_i^{(t-1)}) + g_i f_t(x_i) + \frac{1}{2} h_i f_t^2(x_i)] + \Omega(f_t) + \mathrm{constant}$$

where
$$\begin{split}g_i &= \partial_{\hat{y}_i^{(t-1)}} l(y_i, \hat{y}_i^{(t-1)})\\ h_i &= \partial_{\hat{y}_i^{(t-1)}}^2 l(y_i, \hat{y}_i^{(t-1)})\end{split}$$

So:
$$\textrm{obt}^{(t)}=\sum_{i=1}^n [g_i f_t(x_i) + \frac{1}{2} h_i f_t^2(x_i)] + \Omega(f_t)$$

Regularization

Redefine each tree as:
$$f_t(x) = w_{q(x)}, w \in R^T, q:R^d\rightarrow \{1,2,\cdots,T\}$$

Here w is the vector of scores on leaves, q is a function assigning each data point to the corresponding leaf, and T is the number of leaves. In XGBoost, we define the complexity as

$$\Omega(f) = \gamma T + \frac{1}{2}\lambda \sum_{j=1}^T w_j^2$$

Reformatting:

$$\begin{split}\text{obj}^{(t)} &\approx \sum_{i=1}^n [g_i w_{q(x_i)} + \frac{1}{2} h_i w_{q(x_i)}^2] + \gamma T + \frac{1}{2}\lambda \sum_{j=1}^T w_j^2\\ &= \sum^T_{j=1} [(\sum_{i\in I_j} g_i) w_j + \frac{1}{2} (\sum_{i\in I_j} h_i + \lambda) w_j^2 ] + \gamma T\end{split}$$

where Ij={i|q(xi)=j} is the set of indices of data points assigned to the j-th leaf. Notice that in the second line we have changed the index of the summation because all the data points on the same leaf get the same score. We could further compress the expression by defining:

$G_j = \sum_{i\in I_j} g_i$

$H_j = \sum_{i\in I_j} h_i$

$$\text{obj}^{(t)} = \sum^T_{j=1} [G_jw_j + \frac{1}{2} (H_j+\lambda) w_j^2] +\gamma T$$

The theoretical best solution is:
$$\begin{split}w_j^\ast &= -\frac{G_j}{H_j+\lambda}\\ \text{obj}^\ast &= -\frac{1}{2} \sum_{j=1}^T \frac{G_j^2}{H_j+\lambda} + \gamma T\end{split}$$

Splitting criterion:
$$Gain = \frac{1}{2} \left[\frac{G_L^2}{H_L+\lambda}+\frac{G_R^2}{H_R+\lambda}-\frac{(G_L+G_R)^2}{H_L+H_R+\lambda}\right] - \gamma$$