# Definition of Corrected Contingency Coefficient

I quote RACHEV, HÖCHSTÖTTER, FABOZZI, FOCARDI (2010).

Starting from the bivariate variable $$(x,y)$$, with the component variable $$x$$ taking values in $$(v_i, i=1,\dots,r)$$ and the component variable $$y$$ taking values in $$(w_j, j=1,\dots,s)$$, one can define the chi-square test statistic as: $$$$\chi^2=n\sum_{i=1}^r\sum_{j=1}^s\frac{\left(f_{x,y}\left(v_i, w_j\right)-f_x\left(v_i\right)f_y\left(w_j\right)\right)^2}{f_x\left(v_i\right)f_y\left(w_j\right)}\tag{1}$$$$ with $$f_{x,y}$$ denoting the relative frequency of the bivariate variable $$(x,y)$$, $$f_x$$ denoting the relative frequency of the variable $$x$$ and $$f_y$$ denoting the relative frequency of the variable $$y$$.

One problem arises with $$\chi^2$$ as to its dependence on the data size $$n$$. For increasing $$n$$, the statistic can grow beyond any bound such that there is no theoretical maximum. The solution to this problem is given by the so-called Pearson contingency coefficient (or simply contingency coefficient) defined by: $$$$C=\sqrt{\frac{\chi^2}{\chi^2+n}}\tag{2}$$$$ Clearly, $$C$$ is such that $$0\leq C< 1$$. Consequently, it assumes values which are strictly less than one but may become arbitrarily close to one. $$\color{red}{\textrm{This is still not satisfactory enough for our purpose to design a measure that can}}$$ $$\color{red}{\textrm{uniquely determine}}$$ $$\color{red}{\textrm{the respective degrees of dependence of different data sets.}}$$

There is another coefficient that can be based on the following. In the extreme case of total dependence of $$x$$ and $$y$$, each variable will assume a certain value if and only if the other variable assumes a particular corresponding value. Hence, we have $$k=\min\left\{r,s\right\}$$ unique pairs that occur with positive frequency. $$\color{red}{\textrm{Then one can show that:}}$$ $$$$\color{red}{C=\sqrt{\frac{k-1}{k}}\tag{3}}$$$$ such that, generally, $$0\leq C\leq \sqrt{\frac{k-1}{k}}<1$$. $$\color{red}{\textrm{Now, the standardized coefficient can be given by:}}$$ $$$$\color{red}{C_{\text{corr}}=\sqrt{\frac{k}{k-1}}C\tag{4}}$$$$ which is called the corrected contingency coefficient with $$0\leq C\leq 1$$.

Could you please help me understand the $$\color{red}{\textrm{three parts in red above?}}$$. In general, which is the "aim" pursued in passing from $$(2)$$ to $$\color{red}{(4)}$$?

The Pearson contingency coefficient is intended to provide us some measure regarding the dependence between data columns (which sometimes, when discussing regression, we call collinearity). $$C$$ takes values in $$[0,1)$$ when values near 0 indicate column independence, values far from 0 indicate dependence. But how far is far? Can we achieve a tighter upper bound? That's the general idea.

Red part (a): notice the term "degrees of dependence". If there are zero of them, columns are independent.

Red part (c): as the upper bound for $$C$$ was found, dividing by this bound (/multiplying by its reciprocal) $$C_{corr}=\sqrt{\frac{k}{k-1}}C$$ yields $$0\le C_{corr} \le 1$$.

Red part (b): Let $$k=\min\{r,s\}$$ and assume the extreme condition described (e.g. $$x,y$$ is a pair of letter and digit. Although there are 10 digits and 26 letters, The only pairs which appear are $$(a,0), (s,1), (h,2), (i,3), (r,4), (g,5), (u,6), (n,7), (y,8), (t,9)$$). Each of the $$k=10$$ possible pairs has relative frequency $$f_{xy}$$, as well as each of the components (i.e each letter and digit have frequencies $$f_x,f_y$$).

For letters outside the $$k$$ used above, $$f_x=0$$ and obviously $$f_{xy}=0$$ so they're not summed. Our sum is now: $$\chi^2=n\sum_{i=1}^k\sum_{j=1}^k\frac{(f_{x,y}(v_i, w_j)-f_x(v_i)f_y(w_j))^2}{f_x\left(v_i\right)f_y(w_j)}\\=n\sum_{i=1}^k\sum_{j=1}^k\left[\frac{f^2_{xy}(v_i,w_j)}{f_x(v_i)f_y(w_j)}\right]-2n\sum_{i=1}^k\sum_{j=1}^k\left[f_{xy}(v_i,w_j)\right]+n\sum_{i=1}^k\sum_{j=1}^k\left[f_x(v_i)f_y(w_j)\right]$$

The rightmost sum equals 1 as well as the middle sum (second probability axiom), so we only need to solve

$$\chi^2=n\sum_{i=1}^k\sum_{j=1}^k\left[\frac{f^2_{xy}(v_i,w_j)}{f_x(v_i)f_y(w_j)}\right]-2n+n=n\sum_{i=1}^k\sum_{j=1}^k\left[\frac{f^2_{xy}(v_i,w_j)}{f_x(v_i)f_y(w_j)}\right]-n$$

If the pair $$(v_i,w_j)$$ does not appear on our list, $$f_{xy}(v_i,w_j)=0$$ is not summed. The pair matching is one-to-one (a digit always appears with the same letter and vice versa), so now assume that a pair $$(v_i,w_j)$$ does exist. In fact, it appears $$a$$ times (out of $$n$$), so its relative pair frequency is $$f_{xy}(v_i,w_j)=\frac{a}{n}$$. However, these are also the only $$a$$ occurrences of $$v_i$$ and $$w_j$$ so $$f_x(v_i)=f_y(w_j)=f_{xy}(v_i,w_j)=\frac{a}{n}$$. This means that when a pair does exist, the fraction equals $$1$$. We have $$k$$ such cases, so overall $$\chi^2=nk-n=n(k-1)$$

Now getting the boundary for $$C$$ is easy:

$$C=\sqrt{\frac{\chi^2}{\chi^2+n}}=\sqrt{\frac{n(k-1)}{n(k-1)+n}}=\sqrt{\frac{k-1}{k}} \qquad\blacksquare.$$

• First of all, thank you a lot. Just three questions: 1) Above, you meant "The rightmost sum equals 1" (instead of "The leftmost sum equals 1"), didn't you? 2) I cannot see why $\sum_{i=1}^k\sum_{j=1}^k\left[f_x(v_i)f_y(w_j)\right]$ can be proven to be equal to $1$ thanks to the second probability axiom. Could you please give me an insight? 3) Why does $C_{\text{corr}}=\sqrt{\frac{k}{k-1}}C$ yield that $0\leq C_{\text{corr}}\leq1$? Oct 20, 2021 at 8:29
• 1) Thanks, corrected it. 2) The sum of $f_y(w_j)$ over all $j$ values is 1. That's the axiom. The sum $\sum_{j=1}^{k}{f_x(v_i)f_y(w_j)}$ is thus $f_x(v_i)$. The sum of $f_x(v_i)$ over all $i$ values is 1. That's the axiom again. We therefore get that $\sum_{i=1}^k\sum_{j=1}^k\left[f_x(v_i)f_y(w_j)\right]=1$. That's (also) the 2D integral for the product of 2 marginal PDFs (which is the axiom once again). 3) If $\max(C)=\sqrt{\frac{k-1}{k}}$ then $\max(C_{corr})=\sqrt{\frac{k}{k-1}}\cdot\max(C)=\sqrt{\frac{k}{k-1}}\cdot\sqrt{\frac{k-1}{k}}=1$. Oct 20, 2021 at 9:18