Index: The Book of Statistical ProofsStatistical Models ▷ Poisson data ▷ Poisson distribution with exposure values ▷ Maximum likelihood estimation

Theorem: Consider data $y = \left\lbrace y_1, \ldots, y_n \right\rbrace$ following a Poisson distribution with exposure values:

$\label{eq:Poiss-exp} y_i \sim \mathrm{Poiss}(\lambda x_i), \quad i = 1, \ldots, n \; .$

Then, the maximum likelihood estimate for the rate parameter $\lambda$ is given by

$\label{eq:Poiss-exp-MLE} \hat{\lambda} = \frac{\bar{y}}{\bar{x}}$

where $\bar{y}$ and $\bar{x}$ are the sample means

$\label{eq:xy-mean} \begin{split} \bar{y} &= \frac{1}{n} \sum_{i=1}^n y_i \\ \bar{x} &= \frac{1}{n} \sum_{i=1}^n x_i \; . \end{split}$

Proof: With the probability mass function of the Poisson distribution, the likelihood function for each observation implied by \eqref{eq:Poiss-exp} is given by

$\label{eq:Poiss-exp-LF-s1} p(y_i|\lambda) = \mathrm{Poiss}(y_i; \lambda x_i) = \frac{(\lambda x_i)^{y_i} \cdot \exp[-\lambda x_i]}{y_i !}$

and because observations are independent, the likelihood function for all observations is the product of the individual ones:

$\label{eq:Poiss-exp-LF-s2} p(y|\lambda) = \prod_{i=1}^n p(y_i|\lambda) = \prod_{i=1}^n \frac{(\lambda x_i)^{y_i} \cdot \exp[-\lambda x_i]}{y_i !} \; .$

Thus, the log-likelihood function is

$\label{eq:Poiss-LL} \mathrm{LL}(\lambda) = \log p(y|\lambda) = \log \left[ \prod_{i=1}^n \frac{(\lambda x_i)^{y_i} \cdot \exp[-\lambda x_i]}{y_i !} \right]$

which can be developed into

$\label{eq:Poiss-LL-der} \begin{split} \mathrm{LL}(\lambda) &= \sum_{i=1}^n \log \left[ \frac{(\lambda x_i)^{y_i} \cdot \exp[-\lambda x_i]}{y_i !} \right] \\ &= \sum_{i=1}^n \left[ y_i \cdot \log(\lambda x_i) - \lambda x_i - \log(y_i !) \right] \\ &= - \sum_{i=1}^n \lambda x_i + \sum_{i=1}^n y_i \cdot \left[ \log(\lambda) + \log(x_i) \right] - \sum_{i=1}^n \log(y_i !) \\ &= - \lambda \sum_{i=1}^n x_i + \log(\lambda) \sum_{i=1}^n y_i + \sum_{i=1}^n y_i \log(x_i) - \sum_{i=1}^n \log(y_i !) \\ &= - n \bar{x} \lambda + n \bar{y} \log(\lambda) + \sum_{i=1}^n y_i \log(x_i) - \sum_{i=1}^n \log(y_i !) \\ \end{split}$

where $\bar{x}$ and $\bar{y}$ are the sample means from equation \eqref{eq:xy-mean}.

The derivatives of the log-likelihood with respect to $\lambda$ are

$\label{eq:Poiss-dLLdl-d2LLdl2} \begin{split} \frac{\mathrm{d}\mathrm{LL}(\lambda)}{\mathrm{d}\lambda} &= - n \bar{x} + \frac{n \bar{y}}{\lambda} \\ \frac{\mathrm{d}^2\mathrm{LL}(\lambda)}{\mathrm{d}\lambda^2} &= -\frac{n \bar{y}}{\lambda^2} \; . \\ \end{split}$

Setting the first derivative to zero, we obtain:

$\label{eq:Poiss-dLLdl} \begin{split} \frac{\mathrm{d}\mathrm{LL}(\hat{\lambda})}{\mathrm{d}\lambda} &= 0 \\ 0 &= - n \bar{x} + \frac{n \bar{y}}{\hat{\lambda}} \\ \hat{\lambda} &= \frac{n \bar{y}}{n \bar{x}} = \frac{\bar{y}}{\bar{x}} \; . \end{split}$

Plugging this value into the second derivative, we confirm:

$\label{eq:Poiss-d2LLdl2} \begin{split} \frac{\mathrm{d}^2\mathrm{LL}(\hat{\lambda})}{\mathrm{d}\lambda^2} &= -\frac{n \bar{y}}{\hat{\lambda}^2} \\ &= -\frac{n \cdot \bar{y}}{(\bar{y}/\bar{x})^2} \\ &= -\frac{n \cdot \bar{x}^2}{\bar{y}} < 0 \; . \end{split}$

This demonstrates that the estimate $\hat{\lambda} = \bar{y}/\bar{x}$ maximizes the likelihood $p(y \vert \lambda)$.

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Metadata: ID: P224 | shortcut: poissexp-mle | author: JoramSoch | date: 2021-04-16, 11:42.