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

Question

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?

Answer 1

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.

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Reference:

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