Home page for accesible maths 15 Deviance and the LRT Exam Question 15.3 Summary

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Solution:

a
1. i
  
  For the model, the likelihood function is
  
  $\displaystyle L(\theta|{\bf X})$ $\displaystyle=\prod_{i=1}^{n}\theta^{X_{i}}(1-\theta)$
  
  $\displaystyle=(1-\theta)^{n}\theta^{\sum X_{i}}.$
  
  The log-likelihood is then
  
  $l(\theta|{\bf X})=n\log(1-\theta)+\sum X_{i}\log(\theta),$
  
  with derivative
  
  $l^{\prime}(\theta|{\bf X})=-\frac{n}{1-\theta}+\frac{\sum X_{i}}{\theta}.$
  
  A candidate MLE solves $l^{\prime}(\hat{\theta})=0$ , giving
  
  $\hat{\theta}=\frac{\sum X_{i}}{n+\sum X_{i}}.$
  
  Moreover,
  
  $l^{\prime\prime}(\theta|{\bf X})=-\frac{n}{(1-\theta)^{2}}-\frac{\sum X_{i}}{% \theta^{2}}<0,$
  
  so this is indeed the MLE.
  
  For the Fisher Information,
  
  $\displaystyle I_{E}(\theta)$ $\displaystyle=\mathbb{E}\left[-l^{\prime\prime}(\theta|{\bf X})|_{\theta=% \theta}\right]$
  
  $\displaystyle=\mathbb{E}\left[\frac{n}{(1-\theta)^{2}}-\frac{\sum X_{i}}{% \theta^{2}}\right]$
  
  $\displaystyle=\frac{n}{(1-\theta)^{2}}+\frac{n}{\theta^{2}}\mathbb{E}[X_{1}]$
  
  $\displaystyle=\frac{n}{\theta_{\text{true}}(1-\theta)^{2}},$
  
  after simplification, since $\mathbb{E}[X_{1}]=\theta/(1-\theta)$ .
2. ii
  
  Using the Fisher information, the asymptotic distribution of the MLE is
  
  $\hat{\theta}({\bf X})\sim N(\theta_{\text{true}},I_{E}^{-1}(\theta_{\text{true% }}))\approx N(\theta_{\text{true}},I_{0}^{-1}(\hat{\theta})).$
b
1. i
  
  Using the data, the MLE is $\hat{\theta}=\frac{10}{10+10}=\frac{1}{2}$ . The observed information is
  
  $I_{O}(\hat{\theta})=\frac{10}{(1-1/2)^{2}}+\frac{10}{(1/2)^{2}}=80.$
  
  Therefore a $95\%$ confidence interval is
  
  $(1/2-\frac{1.96}{\sqrt{80}},1/2+\frac{1.96}{\sqrt{80}})=(0.281,0.719).$
2. ii
  
  The deviance is given by
  
  $\displaystyle D(\theta)$ $\displaystyle=2\{l(\hat{\theta})-l(\theta)\}$
  
  $\displaystyle=2\{10\log(1/2)+10\log(1/2)-10\log(1-\theta)-10\log(\theta)\}$
  
  $\displaystyle=20\left(-2\log 2-\log(\theta(1-\theta))\right).$
  
  To plot the deviance calculate $D(0.1)$ and $D(0.9)$ , and note that $D(0.5)=D(\hat{\theta})=0$ . A $95\%$ confidence interval is obtained by drawing a horizontal line at 3.84; the interval is all $\theta$ with $D(\theta)\leq 3.84$ .
3. iii
  
  To construct a confidence interval for the mean, we would use the mean function on the deviance-based confidence interval just calculated, as this is invariant to re-parametrization.

	$\displaystyle L(\theta\|{\bf X})$	$\displaystyle=\prod_{i=1}^{n}\theta^{X_{i}}(1-\theta)$
		$\displaystyle=(1-\theta)^{n}\theta^{\sum X_{i}}.$

	$\displaystyle I_{E}(\theta)$	$\displaystyle=\mathbb{E}\left[-l^{\prime\prime}(\theta\|{\bf X})\|_{\theta=% \theta}\right]$
		$\displaystyle=\mathbb{E}\left[\frac{n}{(1-\theta)^{2}}-\frac{\sum X_{i}}{% \theta^{2}}\right]$
		$\displaystyle=\frac{n}{(1-\theta)^{2}}+\frac{n}{\theta^{2}}\mathbb{E}[X_{1}]$
		$\displaystyle=\frac{n}{\theta_{\text{true}}(1-\theta)^{2}},$

	$\displaystyle D(\theta)$	$\displaystyle=2\{l(\hat{\theta})-l(\theta)\}$
		$\displaystyle=2\{10\log(1/2)+10\log(1/2)-10\log(1-\theta)-10\log(\theta)\}$
		$\displaystyle=20\left(-2\log 2-\log(\theta(1-\theta))\right).$