# Thread: find asympto. distribution of MLE

1. ## find asympto. distribution of MLE

give me a path to do the last part

2. 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.