Asymptotic distribution of the Pearson chi-square statistic

Imagen tomada de ResearchGate.

Statistical Odds & Ends

I recently learned of a fairly succinct proof for the asymptotic distribution of the Pearson chi-square statistic (from Chapter 9 of Reference 1), which I share below.

First, the set-up: Assume that we have $latex n$ independent trials, and each trial ends in one of $latex J$ possible outcomes, which we label (without loss of generality) as $latex 1, 2, dots, J$. Assume that for each trial, the probability of the outcome being $latex j$ is $latex p_j > 0$. Let $latex n_j$ denote that number of trials that result in outcome $latex j$, so that $latex sum_{j=1}^J n_j = n$. Pearson’s $latex chi^2$-statistic is defined as

$latex begin{aligned} chi^2 = sum_{text{cells}} dfrac{(text{obs} – text{exp})^2}{text{exp}} = sum_{j=1}^J dfrac{(n_j – np_j)^2}{np_j}. end{aligned}$

Theorem. As $latex n rightarrow infty$, $latex chi^2 stackrel{d}{rightarrow} chi_{J-1}^2$, where $latex stackrel{d}{rightarrow}$ denotes convergence in distribution.

Before proving the theorem, we prove a lemma that we will…

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