Poisson Distribution Example_Likelihood Ratio testing_1

Example) Mathematical Statistics and Data Analysis 3ED, Chpater 9, Q7. 

Let $X_{1}, \cdots ,X_{n}$ be a sample from a Poisson distribution. Find the likelihood ratio for testing $H_{0}: \lambda = \lambda_{0}$ versus $H_{A}: \lambda = \lambda_{1}$, where $\lambda_{1}> \lambda_{0}$. Use the fact that the sum of independent Poisson random variables follows a Poisson distribution to explain how to determine a rejection region for a test at level $\alpha$   


$\triangleright$ Think First
the LRT reject $H_{o}\Leftrightarrow \frac {L(data|H_{o})}{L(data|H_{1})}$ < C 

If $X_{1}$ ~Poisson($\lambda_{1}$ ) and $X_{2}$ ~ Poisson($\lambda_{2}$), and $X_{1}$ & $X_{2}$ independent. 
then ($X_{1}$+$X_{2}$) ~ Poisson ($\lambda_{1}$+$\lambda_{2}$)
Thus, $\sum x_{i}$ ~ (under $H_{0}$) Poisson ($n \cdot \lambda_{0}$)

$\triangleright$ Solution
First, the likelihood ratio is...
$\rightarrow \frac {L(data|H_{o})}{L(data|H_{1})}=\frac {\prod e^{-\lambda_{0}}\lambda_{0}^{x_{i}}}{x_{i}!} / \frac {\prod e^{-\lambda_{1}}\lambda_{1}^{x_{i}}}{x_{i}!} = e^{n(\lambda_{1}-\lambda_{0})} \cdot (\frac {\lambda_{0}}{\lambda_{1}})^{\sum x_{i}}$ 

Second, LRT is...
Reject $H_{o}\Leftrightarrow$ $e^{n(\lambda_{1}-\lambda_{0})} \cdot (\frac {\lambda_{0}}{\lambda_{1}})^{\sum x_{i}}$ < C
                  $(\sum x_{i}) \ln \frac {\lambda_{0}}{\lambda_{1}} < \ln C-n(\lambda_{1}-\lambda_{0})$ 
                   $\sum x_{i} > \frac {\ln C-n(\lambda_{1}-\lambda_{0})}{\ln \lambda_{0}-\ln \lambda_{1}} = C$ Some constant C. (why? $\ln \frac {\lambda_{0}}{\lambda_{1}}< 0$ )

Thus, reject $H_{o}\Leftrightarrow \sum x_{i}> C$ 

Finally significant level $\alpha$ is...
$\alpha$ = P(reject $H_{0}$ | $H_{0}$ ) = $P(\sum x_{i} > C | \lambda = \lambda_{0})$ 

So, we have the following eqtn on C 
$\alpha = P(Y> C)$ where Y~ Poisson ($n \cdot \lambda_{0}$)
$\alpha = 1-F_{(n \cdot \lambda_{0})}(c)$, where $F_{(n \cdot \lambda_{0})}$ is the CDF of Y. 

$\therefore C= F^{-1}_{n\cdot \lambda_{0}}(1-\alpha)$  

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