Exercise


May 13'23

Answer

Key: B

The likelihood is:

[[math]]L=\prod_{j=1}^{n} \frac{r(r+1) \cdots\left(r+x_{j}-1\right) \beta^{x_{j}}}{x_{j} !(1+\beta)^{r+x_{j}}} \propto \prod_{j=1}^{n} \beta^{x_{j}}(1+\beta)^{-r-x_{j}} .[[/math]]

The loglikelihood is:

[[math]]\begin{aligned} & l=\sum_{j=1}^{n}\left[x_{j} \ln \beta-\left(r+x_{j}\right) \ln (1+\beta)\right] \\ & l^{\prime}=\sum_{j=1}^{n}\left[\frac{x_{j}}{\beta}-\frac{r+x_{j}}{1+\beta}\right]=0 \\ & 0=\sum_{j=1}^{n}\left[x_{j}(1+\beta)-\left(r+x_{j}\right) \beta\right]=\sum_{j=1}^{n} x_{j}-r n \beta=n \bar{x}-r n \beta \\ & \hat{\beta}=\bar{x} / r . \end{aligned}[[/math]]

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

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