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:
\begin{equation}
\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}
\end{equation}
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:
\begin{equation}
C=\sqrt{\frac{\chi^2}{\chi^2+n}}\tag{2}
\end{equation}
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. $\color{red}{\textrm{Then one can show that:}}$
\begin{equation}
\color{red}{C=\sqrt{\frac{k-1}{k}}\tag{3}}
\end{equation}
such that, generally, $0\leq C\leq \sqrt{\frac{k-1}{k}}<1$. $\color{red}{\textrm{Now, the standardized coefficient can be given by:}}$
\begin{equation}
\color{red}{C_{\text{corr}}=\sqrt{\frac{k}{k-1}}C\tag{4}}
\end{equation}
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 $\color{red}{(4)}$?
 A: 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:
$$C=\sqrt{\frac{\chi^2}{\chi^2+n}}=\sqrt{\frac{n(k-1)}{n(k-1)+n}}=\sqrt{\frac{k-1}{k}} \qquad\blacksquare.$$
