Question

a. What is the maximum likelihood estimator for the parameter 2 of the poisson distribution for a sample of n poisson random variables?

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Answer #1

l (\lambda ; x) = \prod e-\lambda\lambdaX / X1

= e-n\lambda\lambda\sumXi / Xi !

Taking log on both sides

ln (l(\lambda ; x) ) = - n\lambda + \sum Xi ln \lambda + ln (xi ! )

Take deriavative and equate to 0

d/dx ( ln (l(\lambda ; x) )) = - n + \sum Xi / \lambda + 0

- n + \sum Xi / \lambda = 0

\sumXi / \lambda = n

Solve for \lambda

\lambda = \sum Xi / n

Thereofore,

Maximum likelihood estimator of \lambda is

\hat{\lambda } = \sum Xi / n ( OR \hat{\lambda } =  \bar{x} )

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