# Prove that the Deviance and the Generalised Pearson Statistic are asymptotically equivalent

I am reading the paper Exponential Dispersion Models from Jørgesen and at page $$137$$ I have encountered a claim that I don't know how to prove.

The author claims that the Generalised Pearson Statistic, defined as $$(y-\hat{\mu})^T V(\hat{\mu})^{-1}(y-\hat{\mu})$$, and the Deviance $$D(y,\hat{\beta})=2\big(\sup_{\theta\in \Theta}\{ y \theta - K(\theta)\}- \{y \,\theta(\hat{\beta})-K(\theta(\hat{\beta})\}\big)$$ are asymptotically equivalent.

I have spent the last couple of hours looking for some references but was unsuccessful. Does anyone know how to do it or where I could look?

In $$\rm [I] ~p.68,$$ Jørgensen notes

The saddlepoint approximation may be viewed as refinement of the normal approximation. The deviance $$D(y, \mu)$$ is approximately a quadratic form in $$y$$ for $$\sigma^2$$ small. Thus a quadratic expansion of $$D(y, \mu)$$ as a function of $$y$$ around $$\mu$$ yields $$D(y,\mu)\simeq (y-\mu)^2/V(\mu),~~\sigma^2\to 0.$$

In section $$3.5.5.,$$ he shows using this (theorem $$3.3.3$$), the estimate based on the generalized Pearson statistic $$\bar{\sigma}^2=\frac{1}{n-k_1}\sum_{i=1}^n(y_i-\hat \mu_i)^2w_i/V(\hat \mu_i)\simeq \frac{1}{n-k_1}D(\mathbf y, \mu(\hat\beta))$$ as $$\mathbf w\to\infty.$$

Refer to $$\rm [I]$$ for the sketch of the argument.

## Reference:

$$\rm [I]$$ The Theory of Exponential Dispersion Models and Analysis of Deviance, Bent Jørgensen, $$1992.$$

• Thank you very much the book is great. For what concerns my specific question, it looks like it doesn't even use Theorem 3.33 but only the easier (3.5). Nov 23, 2023 at 1:59
• Yes @No-one. It ultimately all boils down to the saddlepoint approximation and its use. He has another book on analysis of deviance although I have not had time to check that. Nov 23, 2023 at 2:08