# Modification of Pearson's Chi-square test

At first glance, Pearson's chi-squared test seems flawed in a major way. Can you help me identify the error in my logic?

I have a multinomial distribution with $$k$$ outcomes, and $$p_i$$ denotes the hypothesized probability of outcome $$i$$. I obtain $$n$$ samples from this distribution, and let $$n_i$$ denote the number of times I observe outcome $$i$$ (so $$\sum n_i=n$$).

We have $$E[n_i]=np_i$$, $$Var[n_i]=np_i(1-p_i)$$ and $$Cov[n_i,n_j]=-np_ip_j$$ for $$i\ne j$$. Setting $$x_i:=\frac{n_i-np_i}{\sqrt{np_i(1-p_i)}}$$, we have $$E[x_i]=0$$, $$Var[x_i]=1$$ and $$Cov[x_i,x_j]=-\sqrt{\frac{p_ip_j}{(1-p_i)(1-p_j)}}$$.

Let $$\vec{x}\in\mathbb{R}^k$$ be the vector of $$x_i$$s, so $$E[\vec{x}]=\vec{0}$$ and $$Var[\vec{x}]_{i,j}=Cov[x_i,x_j]$$, which was calculated above. This covariance matrix has corank $$1$$ (this is easier to show before rescaling, and can include a proof in future edits if requested). If the $$n_i$$ are all sufficiently large, the distribution of each $$x_i$$ is roughly $$\mathcal{N}(0,1)$$. But $$Var[\vec{x}]$$ does not depend on $$n$$, and is really not that close to the identity matrix (especially if $$k$$ is small), so the $$x_i$$ are not uncorrelated and do not become any less correlated if $$n$$ is taken to be larger and larger.

And yet we estimate $$\sum\limits_{i=1}^{k-1} x^2_i\sim \chi^2_{k-1}$$ to test this hypothesis. The choice of not including $$x^2_k$$ is itself quite strange, because while it is true that $$x_k$$ completely depends on the other $$x_i$$, we've ignored their correlations in every other context! And isn't there an easy fix here?

Let $$A$$ be a $$(k-1)\times k$$ matrix with $$ACov[\vec{x}]A^T=I_{k-1}$$, the $$(k-1)\times (k-1)$$ identity matrix (which can be computed via e.g. SVD). Then $$Var[A\vec{x}]=I_{k-1}$$. As $$n$$ gets very large the distribution of the components of $$A\vec{x}$$ individually approach $$\mathcal{N}(0,1)$$, and their correlation matrix is the identity. This doesn't imply that their joint distribution approaches the joint gaussian distribution (as uncorrelated guassians are not necessarily independent), but I would imagine this is provably true.

And so why don't we test $$\|A\vec{x}\|^2\sim \chi_{k-1}^2$$ instead? This $$A$$ is not hard to compute as its complexity is in terms of $$k$$.

As I know the Pearson test uses:

$$X_i = \frac{n_i - np_i}{\sqrt{np_i}}$$ and not the $$X_i$$ that you mentioned. To prove the result of the Pearson test, let $$\widehat{\boldsymbol{\theta}}_n = \frac1n \begin{pmatrix}n_1\\\vdots\\n_k\end{pmatrix}$$ and $$\boldsymbol{\theta} = \begin{pmatrix}p_1\\\vdots\\p_k\end{pmatrix}$$ you can easily prove that

$$\sqrt{n} \left(\widehat{\boldsymbol{\theta}}_n - \boldsymbol{\theta}\right) \overset{d}\to \mathcal N\left(\boldsymbol{0}, \Sigma\right)$$

where $$\Sigma = \begin{bmatrix} p_1\left(1-p_1\right) & -p_1p_2 & \cdots & -p_1p_k\\ -p_1p_2 & p_2\left(1-p_2\right) & \cdots & -p_2p_k\\ \vdots & \vdots & \ddots & \vdots\\ -p_1p_n & -p_2p_k & \cdots & p_k\left(1-p_k\right) \end{bmatrix}$$

Let $$\boldsymbol{A} = \begin{bmatrix} \frac1{\sqrt{p_1}} & \\ & \ddots \\ &&\frac1{\sqrt{p_k}}\end{bmatrix}$$ then, $$\sqrt{n}\boldsymbol{A} \left(\widehat{\boldsymbol{\theta}}_n - \boldsymbol{\theta}\right) \overset{d}\to \mathcal N\left(\boldsymbol{0}, \boldsymbol{A}\Sigma\boldsymbol{A}^\intercal\right)$$

You can observe that $$\boldsymbol{X} = \sqrt{n}\boldsymbol{A} \left(\widehat{\boldsymbol{\theta}}_n - \boldsymbol{\theta}\right)$$

and if $$\boldsymbol{u} = \begin{pmatrix}\sqrt{p_1}\\\vdots\\\sqrt{p_k}\end{pmatrix}$$ then $$\boldsymbol{A}\Sigma\boldsymbol{A}^\intercal = \mathbf I - \boldsymbol{u}\boldsymbol{u}^\intercal$$

Finally

$$\boldsymbol{X} \overset{d}\to \mathcal N\left(\boldsymbol{0}, \mathbf I - \boldsymbol{u}\boldsymbol{u}^\intercal\right)$$ Since $$\boldsymbol{u}^{\intercal} \boldsymbol{X}= 0$$ then if $$Z\sim \mathcal N(0, 1)$$ then $$\boldsymbol{X} + Z\boldsymbol{u} \overset{d}\to \mathcal N(0, \mathbf I).$$

By Cochran's theorem, $$\left\|X\right\|^2 = \left\|\pi_{\boldsymbol{u}^\perp}\left(\boldsymbol{X} + Z\boldsymbol{u}\right)\right\|^2 \sim \chi^2_{k-1}$$

• Beautiful; thank you. I suspect this implies that there is a very nice diagonal" $A$ satisfying the conditions of my question, to make it equivalent to the Pearson chi-square test Commented Feb 5 at 23:10
• Probably. So your question has another test different from the one of Pearson? Is that right? Commented Feb 5 at 23:11
• Based on your answer, I'm sure that the two are equivalent. $A$ will probably correct the $\frac{1}{\sqrt{1-p_i}}$ difference. Commented Feb 5 at 23:15