Practice Problem Set 2 – finding median losses

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This post has two practice problems to find the median of a distribution that models insurance losses.

Practice Problem 2a
Losses have a distribution with the following density function:

    \displaystyle f(x)=\frac{1}{6} e^{-\frac{x}{12}}-\frac{1}{6} e^{-\frac{x}{6}} \ \ \ x>0.

Calculate the median loss amount.

Practice Problem 2b
Losses are modeled by a distribution that is a mixture of two exponential distributions, one with mean 6 and the other with mean 12. The weight of each distribution is 50%. Calculate the median loss amount.

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Solutions

Problem 2a

    Median \displaystyle =-12 \times \log \biggl(\frac{2-\sqrt{2}}{2} \biggr)=14.7354

    log is logarithm to base e = 2.718281828…

Problem 2b

    Median \displaystyle =-12 \times \log \biggl(\frac{-1+\sqrt{5}}{2} \biggr)=5.7745

    log is logarithm to base e = 2.718281828…

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\copyright 2017 – Dan Ma

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