计算器使用

问题如下：

An insurance company specializing in inexperienced drivers is building a decision-tree model to classify drivers that it has previously insured as to whether they made a claim or not in their first year as policyholders. They have the following data on whether a claim was made (“Claim_made”) and two features (for the label and the features, in each case, “yes” = 1 and “no” = 0): whether the policyholder is a car owner and on whether they have a college degree:

a. Calculate the “base entropy” of the Claim_made series.

b. Build a decision tree for this problem.

选项：

解释：

a. The base entropy is the entropy of the output series before any splitting. There are four policyholders who made claims and six who did not. The base entropy is therefore:

b. Both of the features are binary, so there are no issues with having to determine a threshold as there would be for a continuous series. The first stage is to calculate the entropy if the split was made for each of the two features.

Examining the Car_owner feature first, among owners (feature = 1), two made a claim while four did not, leading to entropy for this sub-set of:

Among non-car owners (feature = 0), two made a claim and two did not, leading to an entropy of 1. The weighted entropy for splitting by car ownership is therefore given by

and the information gain is information gain = 0.971 - 0.951 = 0.020

We repeat this process by calculating the entropy that would occur if the split was made via the College_degree feature. If we did so, we would observe that the weighted entropy would be 0.551, with an information gain of 0.420. Therefore, because the entropy is maximized when the sample is first split by College_degree, this becomes the root node of the decision tree.

For policyholders with a college degree (i.e., the feature=1), there is already a pure split as four of them have not made claims while none have made claims (in other words, nobody with college degrees made claims). This means that no further splits are required along this branch. The other branch can be split using the Car_ownership feature, which is the only one remaining.

The tree structure is given below: