Does $\chi^{2}$ statistic always works for multinomial hypothesis testing? The $\chi^2$ statistic has been frequently employed in the hypothesis testing of the multinomial distributions. But after looking at the derivation of such a process (where they use Taylor expansion and discard all the remaining terms after the power of 3), I become suspicious and wonder if there's any counterexample to the statement (in which a contradiction arises after directly applying $\chi^2$ statistic to the multinomial distribution). Is there any way to strictly prove the validity of such an application of $\chi^{2}$ statistic (with mathematical rigor), or is there any counterexample to the argument?
 A: In statistical genetics, "empirical p-value testing" is often used to get around distributional assumptions.  For example, consider a multiway table with $r$ rows and $c$ columns.  For each row, the multinomial probability of the row total $\{ n_{i.} \}$ is
\begin{equation}
P(\{n_{i.}\})= \frac{n_{i.}!}{\prod_j n_{ij}!} \prod_j \pi_{j|i}^{n_{ij}},
\end{equation}
where $\pi_{j|i}$ is the conditional probability in column $j$ ($j=1,2,...,c$) given row $i$.
The probability of the table $\{n_{ij}\}$ is the product of all of the multinomial row probabilities
\begin{equation}
P(\{n_{ij}\})= \prod_i \left( \frac{n_{i.}!}{\prod_j n_{ij}!} \prod_j \pi_{j|i}^{n_{ij}}\right).
\end{equation}
Under independence, we assume that in each column the conditional cell probabilities are
equal (including the row total), shown as $\pi_{j|1}=\pi_{j|2}= \cdots = \pi_{j|r}=\pi_{.j}$.
Thus, let's substitute $\pi_{.j}$ for $\pi_{j|i}$ and $n_{.j}$ for $n_{ij}$ in the above equation and obtain
\begin{equation}
\begin{split}
P(\{n_{ij}\})&= \prod_i \left( \frac{n_{i.}!}{\prod_j n_{ij}!} \prod_j \pi_{.j}^{n_{.j}}
\right)\\
&= \frac{\prod_i n_{i.}! \prod_j \pi_{.j}^{n_{.j}} } {\prod_i \prod_j n_{ij}!}. \\
\end{split}
\end{equation}
The distribution of the cell counts in each row, however, also depend also on $\pi_{.j}$.
Fisher assumed these to be sufficient nuisance parameters, and conditioned on them so
that the resulting conditional probability does not depend on them.  The contribution of
$\pi_{.j}$ to the product multinomial distribution depends on the data through $n_{.j}$,
which are the sufficient statistics.  Each column total $n_{.j}$ has the multinomial distribution
\begin{equation}
P(\{n_{.j}\})= \frac{n!}{\prod_j n_{.j}!} \prod_j \pi_{.j}^{n_{.j}}.
\end{equation}
The probability of $\{n_{ij}\}$ conditional on $\{n_{.j}\}$ equals the probability of
$\{n_{ij}\}$ divided by the probability of $\{n_{.j}\}$, so we invert the above ratio and
multiply as
\begin{equation}
\begin{split}
P(\{n_{ij}\}|\{n_{.j}\}) &= \left( \frac{\prod_i n_{i.}! \prod_j \pi_{.j}^{n_{.j}} } {\prod_i \prod_j n_{ij}!} \right)
\left( \frac{\prod_j n_{.j}!}{ \prod_j \pi_{.j}^{n_{.j}}{n!}}   \right)\\
&= \frac{\prod_i n_{i.}! \prod_j n_{.j}} {n!\prod_i \prod_j n_{ij}!}. \\
\end{split}
\end{equation}
$P$ above is first calculated from the observed data. Permutations are then used by re-shuffling categories for one of the two categorical variables being considered, each time recalculating $P^{(b)}$ $(b=1,2,...,B)$ repeatedly in order to simulate the null distribution.  After $B$ iterations, the exact p-value is
\begin{equation}
P=\frac{\# \{b: P^{(b)}< P\}}{B} .
\end{equation}
Randomization Test for $\chi^2$. You will notice that even for small tables, e.g. 3x4, 4x4, 4x5, 5x5, the number of iterations required for fully implementing the exact form of the test can exceed than $10^4$ configurations.  Therefore, one can use the following randomization test:


*

*Calculate the raw chi-squared value for the table, call this $\chi^2$.

*Randomly shuffle (permute) one of the grouping variable's values.

*Calculate chi-squared using the grouping variables (with one of them shuffled), call this $\chi^2_{(b)}$.

*Repeat steps 2-3 $B=10,000$ times (default).
The p-value for this randomization test is
\begin{equation}
P=\frac{\#\{b: \chi^2_{(b)} \geq \chi^2\}}{B},
\end{equation}
that is, the number of times the chi-squared value for the permuted configurations exceeds or is equal to the chi-squared value for the raw data configuration (table), divided by the number of iterations.
