find asympto. distribution of MLE

**pengchao1024** · November 23rd 2009, 09:37 AM

give me a path to do the last part

**matheagle** · November 24th 2009, 11:37 PM

I get for the log likelihood function

$\ln L=-n\ln(1-e^{-\lambda})-n\lambda+\sum X_i\ln\lambda-\ln(\prod (X_i!))$

The derivative wrt lambda is

${-ne^{-\lambda} \over 1-e^{-\lambda}}-n+{\sum X_i\over \lambda}$

set this equal to zero and divide by n ...

${\bar X\over \lambda}={1 \over 1-e^{-\lambda}}$

I don't think we can find the MLE of $\lambda$ say $\hat \lambda$

But I don't think they want you to either.

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