Consider a random sample of size $n$ from a Poisson distribution with mean $\mu$. Let $\theta=P(X=0)$. Find the MLE of $\theta$ and show that it is a consistent estimator.
–We have $\theta=P(X=0)=e^{-\mu}$. To find the MLE, I took the log likelihood, $\ell(\mu,\mathbf{x})=-n\mu$, which has a derivative $-n$ with respect to $\mu$. Therefore the MLE would be $0$. Is this calculation correct? It seems too simple…
Best Answer
You have to take the joint likelihood of the $n$ samples. If $X_1,X_2,\ldots,X_n$ are the samples you write $$\log P(X_1,X_2,\ldots,X_n)=\log\prod P(X=X_i)\\=\sum\log P(X=X_i)\\=\sum (X_i\log\lambda -\lambda-\log(X_i!)).$$ To find the MLE of $\theta$ you write the above expression in terms of $\theta$ and the maximizer with respect to $\theta$ would be the desired MLE.