Exercise 3.6 - MLE for the Poisson distribution

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The Poisson distribution plays a central role in the stochastic process, e.g., the queueing theory. If data are assumed to be generated from a similar process then the Bayesian analysis of the Poisson distribution derived in this exercise and the next can be applied directly. The likelihood of data for a Poisson distribution is (assuming i.i.d.):

p (𝒟 | λ) = \prod_{n = 1}^{N} Poi (x_{n} | λ) = \exp (- 𝜆𝑁) \cdot λ^{\sum_{n = 1}^{N} x_{n}} \cdot \frac{1}{\prod_{n = 1}^{N} x_{n}!} .

Setting the derivative of the likelihood w.r.t. $λ$ to zero:

\frac{\partial}{∂𝜆} p (𝒟 | λ) = \frac{\exp (- 𝜆𝑁) \cdot λ^{(\sum_{n = 1}^{N} x_{n}) - 1}}{\prod_{n = 1}^{N} x_{n}!} {- 𝑁𝜆 + \sum_{n = 1}^{N} x_{n}} .

Thus:

λ_{MLE} = \frac{\sum_{n = 1}^{N} x_{n}}{N} .

The formulation could be made easier by taking logarithm (since the Poisson distribution can be considered an element of the exponential family as well):

\log p (𝒟 | λ) = - λ \cdot N + (\sum_{n = 1}^{N} x_{n}) \cdot \log λ,

where we have omitted the term independent of $λ$ .

solour_lfq

2021-03-24 13:42

Exercise 3.6 - MLE for the Poisson distribution

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