Decision Tree Learning Deviation - Russell and Norvig I am working through the Russell and Norvig AI book and came across the following on the top of page 706.

The section concerns Decision Tree pruning and testing a given attribute against the null hypothesis. The example at hand deals with a binary decision tree and $p_k$ denotes the number of positive ("yes") responses in the training data and $n_k$ the number of negative ("no") responses in the training data with respect to the training examples that are in the subset of values that took on the value $v_k$ associated with the $k$th attribute of a vector $\vec{x}$ of length $n \times 1$. Furthermore $p$ and $n$ simply denote the number of positive and negative outputs of the training data.

The authors state the following:


  We can measure the deviation by comparing the actual numbers of positive and negative examples in each subset $p_k$ and $n_k$ with the expected numbers 
  $\hat{p}_k$ and $\hat{n}_k$, assuming true irrelevance:
  
  \begin{align*}
\hat{p}_k&amp;=p\times \frac{p_k+n_k}{p+n}\\
\hat{n}_k&amp;=n\times \frac{p_k+n_k}{p+n}\,.
\end{align*}
  
  A convenient measure of the total deviation is given by
  
  $$\Delta = \sum_{k=1}^d \frac{(p_k-\hat{p}_k)^2}{\hat{p}_k} + \frac{(n_k-\hat{n}_k)^2}{\hat{n}_k}\,.$$


My questions:

I think I understand the use of the estimators. We are basically assuming a uniform distribution on $p$ and $n$ and so within each subset of training examples that have the attribute value $A_k$ equal to $v_k$ we simply scale the overall $p$ and $n$ by the total proportion of samples in that set. In essence, we are assuming that, for instance, if $p=10$ and $n=5$, then those proportions hold in our subsets, and so the expected value of $\hat{p}_k$ is simply scaled by the size of the subset.

This leads me to also understanding the numerator of the measure of deviation $\Delta$. We use the squared deviation of the expectation from the observed value. My question then is why do the estimators appear in the denominator? This seems very close to an equation for variance, and I assume that the estimators are somehow scaling $\Delta$, but how can we justify that?
TL;DR: They are basically doing a chi-squared test to compare the distributions.

In more detail: Let's break this down into two parts, $\Delta = \Delta_1 + \Delta_2$ where

$$\begin{align*}
\Delta_1 &amp;= \sum_{k=1}^d \frac{(p_k-\hat{p}_k)^2}{\hat{p}_k}\\
\Delta_2 &amp;= \sum_{k=1}^d \frac{(n_k-\hat{n}_k)^2}{\hat{n}_k}.
\end{align*}$$

Then $\Delta_1$ is the test statistic for a chi-squared test of the goodness of fit between the two distributions $p_k$ and $\hat{p}^k$.  Similarly, $\Delta_2$ is the test statistic for the goodness of fit between $n_k$ and $\hat{n}_k$.

If the distribution $p_k$ is similar to $\hat{p}_k$, and $n_k$ is similar to $\hat{n}_k$, then $\Delta$ will be small (the former means that $\Delta_1$ will be small, and the latter means that $\Delta_2$ will be small).  Conversely, if $p_k$ is not similar to $\hat{p}_k$, or $n_k$ is not similar to $\hat{n}_k$, then $\Delta$ will be large (in the former case, $\Delta_1$ will be large; in the latter case,  $\Delta_2$ will be large).

So this is a convenient measure of the total deviation between these distributions: a convenient way to get a single number to help us measure the extent to which both $p_k$ is similar to $\hat{p}_k$  and $n_k$ is similar to $\hat{n}_k$.

The chi-squared test is a standard test for goodness of fit, so its test statistic is a reasonable choice here.  As for why the chi-squared test  is a good test for measuring goodness of fit, you can read the standard literature on the chi-squared test.