You are given:

1. The number of claims follows a Poisson distribution with mean [math]\lambda [/math] .
2. Observations other than 0 and 1 have been deleted from the data.
3. The data contain an equal number of observations of 0 and 1.

Calculate the maximum likelihood estimate of [math]\lambda [/math] .

• 0.50
• 0.75
• 1.00
• 1.25
• 1.50

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Key: C

There are n/2 observations of N = 0 (given N = 0 or 1) and n/2 observations of N = 1 (given N = 0 or 1). The likelihood function is

[[math]] L = \left( \frac{e^{-\lambda}}{e^{-\lambda} + \lambda e^{-\lambda}} \right)^{n/2}\left( \frac{\lambda e^{-\lambda}} {e^{-\lambda} + \lambda e^{-\lambda}} \right)^{n/2} = \frac{\lambda^{n/2}e^{-n\lambda}}{e^{-\lambda} + \lambda e^{-\lambda}}= \frac{\lambda^{n/2}}{(1+\lambda)^n} [[/math]]

Taking logs, differentiating and solving provides the answer.

[[math]] \begin{aligned} &l = \ln L = (n / 2) \ln \lambda − n \ln(1 + \lambda ) \\ &l^{'} = \frac{n}{2\lambda} - \frac{n}{1+\lambda} = 0 \\ &n(1+\lambda) -2 n\lambda = 0 \\ &1-\lambda = 0, \lambda = 1. \end{aligned} [[/math]]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.