Expected Value of Insurance Pmt with Policy Limit

actuarial-scienceexpected valuestatistics

I'm trying to answer the following SOA Practice Question:

Michael is a professional stuntman who performs dangerous motorcycle
jumps at extreme sports events around the world. The annual cost of
repairs to his motorcycle is modeled by a two parameter Pareto
distribution with $θ=5000$ and $α=2$. An insurance reimburses
Michael's motorcycle repair costs subject to the following provisions:

(i) Michael pays an annual ordinary deductible of 1000 each year.

(ii) Michael pays 20% of repair costs between 1000 and 6000 each year.

(iii) Michael pays 100% of the annual repair costs above 6000 until Michael has paid 10,000 in out-of-pocket repair costs each year.

(iv)Michael pays 10% of the remaining repair costs each year.

Calculate the expected annual insurance reimbursement.

Attempt:

$X = \text{Repair Cost}$
$$
\text{Reimbursement} = Y \\ = \begin{cases}
0,\ X<1000\\
.8(X-1000),\ 1000\le X<6000 \\
.8(6000-1000),\ 6000\le X<14000 \\
.8(6000-1000) + .9(X-14000),\ X\ge 14000 \end{cases} \\ $$

$$ = \begin{cases} 0,\ X<1000 \\
.8X-800,\ 1000\le X<6000 \\
4000,\ 6000\le X<14000 \\
.9X-8600,\ X\ge 14000 \end{cases}
$$

I'll spare you the rest of it because the problem I had was integrating the last interval. Note that $f(x) = \left(\frac{5000}{x+5000}\right)^2$

$$ \mathsf E[.9X-8600|X\ge14000] = \int_{14000}^\infty (.9x-8600)\left(\frac{5000}{x+5000}\right)^2 \mathrm dx \div \mathrm{Pr}(X\ge14000)$$

I got that this integral does not converge and Wolfram Alpha agrees. What did I mess up?

Best Answer

Since $\alpha > 1$, we know that the expectation of the ground-up loss $X$ where $$X \sim \operatorname{Pareto}(\theta = 5000, \alpha = 2), \\ f_X(x) = \frac{\alpha \theta^\alpha}{(x+\theta)^{\alpha + 1}} = \frac{\alpha}{\theta}\left(1 + \frac{x}{\theta}\right)^{-(\alpha + 1)}$$ is $$\operatorname{E}[X] = \frac{\theta}{\alpha - 1} < \infty.$$ Since $0 < Y < X$, it follows that $\operatorname{E}[Y] < \operatorname{E}[X]$.

The error you have made is that the expression you wrote for the density is actually the survival function, $S_X(x) = \Pr[X > x]$. You can't use that. Instead, the correct calculation for the conditional expectation is $$\begin{align} \operatorname{E}[Y \mid X \ge 14000] &= \frac{1}{S_X(14000)}\int_{x=14000}^\infty (\tfrac{9}{10}(x - 14000) + 4000) f_X(x) \, dx \\ &= \frac{1}{S_X(14000)}\left(\frac{9}{10} \int_{x=0}^\infty x f_X(x+14000) \, dx + 4000 \int_{x=14000}^\infty f_X(x) \, dx \right) \\ &= \frac{9}{10S_X(14000)} \int_{x=0}^\infty x f_X(x+14000) \, dx + 4000. \end{align}$$ Now observe $$f_X(x + L) = \frac{\alpha \theta^\alpha}{(x+L+\theta)^{\alpha + 1}} = \frac{\theta^\alpha}{(\theta + L)^\alpha} \cdot \frac{\alpha (\theta + L)^\alpha}{(x + (\theta + L))^{\alpha+1}} = S_X(L) f_{X^*}(x),$$ where $$X^* \sim \operatorname{Pareto}(\theta^* = \theta + L, \alpha^* = \alpha),$$ so $$\int_{x=0}^\infty x f_X(x+14000) \, dx = S_X(14000) \operatorname{E}[X^*] = S_X(14000) \frac{5000 + 14000}{2 - 1} = 19000 S_X(14000).$$ This simplifies our computation substantially:

$$\operatorname{E}[Y \mid X \ge 14000] = \frac{9}{10} 19000 + 4000 = 21100.$$

Of course this is not the full answer; it is simply the conditional expectation of the insurer's payment for ground-up losses at or above $14000$.

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