# Explanation of min_child_weight in xgboost algorithm

The definition of the min_child_weight parameter in xgboost is given as the:

minimum sum of instance weight (hessian) needed in a child. If the tree partition step results in a leaf node with the sum of instance weight less than min_child_weight, then the building process will give up further partitioning. In linear regression mode, this simply corresponds to minimum number of instances needed to be in each node. The larger, the more conservative the algorithm will be.

I have read quite a few things on xgboost including the original paper (see formula 8 and the one just after equation 9), this question and most things to do with xgboost that appear on the first few pages of a google search. ;)

Basically I'm still not happy as to why we are imposing a constraint on the sum of the hessian? My only thought at the minute from the original paper is that it relates to the weighted quantile sketch section (and the reformulation as of equation 3 weighted squared loss) which has $$h_i$$ as the 'weight' of each instance.

A further question relates to why it is simply the number of instances in linear regression mode? I guess this is related to the second derivative of the sum of squares equation?

For a regression, the loss of each point in a node is

$\frac{1}{2}(y_i - \hat{y_i})^2$

The second derivative of this expression with respect to $\hat{y_i}$ is $1$. So when you sum the second derivative over all points in the node, you get the number of points in the node. Here, min_child_weight means something like "stop trying to split once your sample size in a node goes below a given threshold".

For a binary logistic regression, the hessian for each point in a node is going to contain terms like

$\sigma(\hat{y_i})(1 - \sigma(\hat{y_i}))$

where $\sigma$ is the sigmoid function. Say you're at a pure node (e.g., all of the training examples in the node are 1's). Then all of the $\hat{y_i}$'s will probably be large positive numbers, so all of the $\sigma(\hat{y_i})$'s will be near 1, so all of the hessian terms will be near 0. Similar logic holds if all of the training examples in the node are 0. Here, min_child_weight means something like "stop trying to split once you reach a certain degree of purity in a node and your model can fit it".

The Hessian's a sane thing to use for regularization and limiting tree depth. For regression, it's easy to see how you might overfit if you're always splitting down to nodes with, say, just 1 observation. Similarly, for classification, it's easy to see how you might overfit if you insist on splitting until each node is pure.

• Thank you for the reply, I can't upvote you due to low reputation. Jan 17, 2018 at 17:21
• Hi @maw501: No problems, I can. Good answer Hahdawg ! Oct 11, 2018 at 10:05
• So in a case of highly imbalanced data, what is your proposed range for the min_child_weight? Feb 16, 2019 at 6:30
• When in unbalanced dataset, should min_child_weight include weights as well? Thank you! @hahdawg Mar 22, 2019 at 18:45
• @MahdiBaghbanzadeh, it is difficult to give recommendations on such starting parameters, as differences in the datasets being used are vital to this determination. I would recommend performing a cross-validated grid search over either 1) randomly generated sets of hyperparameters or 2) a hand-picked set of probably useful hyperparameters to determine what helps from a predictive PoV. Sep 25, 2020 at 12:00

When there is little information, gradients of the loss function will tend to change slower, hence a smaller hessian. In the MLE framework, the negative of the hessian is known as the observed Fisher information. Ignoring the sign, a larger hessian will mean that more information is available. You don't want splits to happen when there is too little information. This shortage of information manifests in different ways for different loss functions, some of which were already described in another answer: smaller sample size for ordinary least squares regression and similar for logistic regression but now also weighted by the impurity $$p(1-p)$$ expected by the current model (so smaller and purer samples will be the less informative ones). Also notice that since the score of a leaf is related to $$\frac{\sum grad}{\sum hess}$$, a very small $$\sum hess$$ will make the ratio unstable, which is another way this lack of information manifests.