Question
You are given the following:
- the amount of an individual loss in the year $2022$ follows an exponential distribution with mean $15000$
- Between $2022$ and $2025$, losses will be multiplied by an inflation factor $C$. For example, if $C = 1.2$, then losses in $2025$ will be $20\%$ more than that in $2022$.
- You are uncertain of what $C$ will be, but estimate that it will be a random draw from a inverse gamma distribution with parameters $\alpha = 2.4$ and $\delta = 3.2$. The inverse gamma distribution has pdf $f(x; \alpha, \delta) = \frac {\delta^{\alpha}} {\Gamma(\alpha)} x^{-\alpha + 1} e^{-\frac {\delta} x}$.
Estimate the probability that a loss in $2025$ will exceed $100000$.
My thoughts
I know that the mixture of an exponential and an inverse gamma will result in a Pareto distribution. However, I am not sure how to account for $C$ in this case. I mean, if the exponential had a mean of $C$ and $C$ followed an inverse gamma, I would know how to compute the integral. Does the computation of the resulting distribution require me to take into account $C$, or does it not matter? How should I go about evaluating this?
Any intuitive explanations will be greatly appreciated :)
Let $L \sim \operatorname{Exponential}(\mu = 15000)$ be the individual loss. Then the inflated loss will be $T = CL$, and we want $$\Pr[T > 10^5] = \Pr[L > 10^5/C].$$ Since $C$ is inverse gamma, $1/C$ will be gamma with shape $\alpha$ and rate $\delta$, with PDF $$\frac{\delta^\alpha x^{\alpha - 1} e^{-\delta x}}{\Gamma(\alpha)}.$$ Hence $$\begin{align} \Pr[T > 10^5] &= \int_{x = 0}^\infty \Pr[L > 10^5/C \mid 1/C = x]f_{1/C}(x) \, dx \\ &= \int_{x = 0}^\infty e^{-10^5 x/\mu} \frac{\delta^\alpha x^{\alpha-1} e^{-\delta x}}{\Gamma(\alpha)} \, dx \\ &= \int_{x = 0}^\infty \frac{\delta^\alpha x^{\alpha-1} e^{-(\delta + 10^5/\mu)x}}{\Gamma(\alpha)} \, dx \\ &= \left(\frac{\delta}{\delta + 20/3}\right)^{\!\alpha} \int_{x=0}^\infty \frac{(\delta + 20/3)^\alpha x^{\alpha-1} e^{-(\delta + 20/3)x}}{\Gamma(\alpha)} \, dx \\ &= \left(\frac{3\delta}{3\delta + 20}\right)^{\!\alpha}. \end{align}$$