MLE for a Poisson mean

Question

You observe X1, \dots, Xn i.i.d. from a Poisson distribution with mean \lambda, so P(X = x) = \frac{e^{-\lambda}\lambda^x}{x!}. Derive the MLE of \lambda. Is it unbiased, and what is its variance?

Accepted Answer

Log-likelihood. \ell(\lambda) = \sum{i=1}^n \big(-\lambda + Xi \log\lambda - \log Xi!\big) = -n\lambda + \Big(\sumi Xi\Big)\log\lambda - 	ext{const}. Maximize. Differentiate with respect to \lambda: \frac{d\ell}{d\lambda} = -n + \frac{\sumi Xi}{\lambda} = 0 \implies \hat\lambda = \frac{1}{n}\sum{i=1}^n Xi = \bar X. The second derivative -\sumi Xi / \lambda^2 < 0 confirms a maximum. The MLE of the Poisson mean is just the sample mean. Bias. Since each \mathbb{E}[Xi] = \lambda, \mathbb{E}[\hat\lambda] = \mathbb{E}[\bar X] = \lambda, so \hat\lambda is unbiased, no correction needed. Variance. F…

MLE for a Poisson mean

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