Use the fact that (x log x=(x-1)+(x-1)^{2} / 2+oleft((x-1)^{2} ight)) to show that the deviance test statistic

Use the fact that \(x \log x=(x-1)+(x-1)^{2} / 2+o\left((x-1)^{2}\right)\) to show that the deviance test statistic \(D^{2}\) in (4.43) has the same asymptotic distribution as the general Pearson chi-square statistic in (4.44).

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XGP = K j=1 (n; - Ej) Ej (4.43)