May 13'23

Exercise

You are given:

  1. The random variable X has probability density function
    [[math]] f(x) = \alpha (1500)^{\alpha} (1500 + x)^{-(\alpha+1)}, \, \alpha \gt 0, \, x \gt 0[[/math]]
  2. Five sample observations are:
    50 250 450 650 850

Calculate the maximum likelihood estimate of [math]\alpha [/math]

  • 0.16
  • 0.79
  • 1.85
  • 2.91
  • 3.97

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

May 13'23

Key: E

The likelihood function is

[[math]] L(\alpha ) = \alpha^51500^{5\alpha} \prod (1500 + x_i)^{-(\alpha + 1)} [[/math]]

The loglikelihood function is

[[math]] l (\alpha ) = 5 \ln \alpha + 5\alpha \ln1500 − (\alpha + 1) \alpha \ln (1500 + x_i ) [[/math]]

[[math]] l^{'}(\alpha ) = \frac{5}{\alpha} + 5\ln 1500 - \sum\ln(1500 + x_i) [[/math]]

Setting [math] l^{'}(\alpha ) = 0 [/math], we find:

[[math]] \frac{5}{\alpha} = -5\ln(1500) + \sum \ln(1500 + x_i) \Rightarrow \hat{\alpha} = 3.974. [[/math]]

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

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