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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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