Limited Expected Value

Study guides: Exam MAS-I, Exam 5

The Limited Expected Value (LEV) of a random variable $X$ at limit $u$ is the expected value of the payment under a policy that pays $X$ up to a maximum of $u$ . It equals the expected value of $\min(X, u)$ .

$E[X \wedge u] = \int_0^u [1 - F(x)]\,dx \quad \text{(for non-negative } X\text{)}$

$E[X \wedge u] = E[X] - E[\max(X - u, 0)] = E[X] - e(u)\cdot[1-F(u)]$

Also written $E[\min(X, u)]$ ; the limited loss variable is $Y = \min(X, u)$
$e(u)$ is the mean excess loss at $u$ : $e(u) = \dfrac{E[X] - E[X \wedge u]}{1 - F(u)}$
As $u \to \infty$ , $E[X \wedge u] \to E[X]$
Used to price policy limits and excess-of-loss reinsurance: the insurer pays $E[X \wedge u]$ and the reinsurer pays $E[X] - E[X \wedge u]$
Exam 5 (Werner & Modlin) writes the limit as $H$ and calls $E[X \wedge H]$ the limited average severity $\text{LAS}(H)$ . Under its simplifying assumptions, the increased limit factor for limit $H$ over basic limit $B$ is $\text{ILF}(H) = \text{LAS}(H)/\text{LAS}(B)$ (see Increased Limits)

Common formulas:

Distribution	$E[X \wedge u]$
Exponential $(\theta)$	$\theta(1 - e^{-u/\theta})$
Pareto $(\alpha, \theta)$ , $\alpha \neq 1$	$\dfrac{\theta}{\alpha-1}\!\left[1 - \left(\dfrac{\theta}{\theta+u}\right)^{\alpha-1}\right]$
Pareto $(\alpha = 1, \theta)$	$-\theta \ln\!\left(\dfrac{\theta}{\theta+u}\right)$

Insurer Payment with a Policy Limit Losses $X \sim \text{Exponential}(\theta = 1{,}000)$ . A policy pays losses up to a limit of $u = 2{,}000$ . Find the expected payment per loss.

Answer $E[X \wedge 2{,}000] = 1{,}000\left(1 - e^{-2{,}000/1{,}000}\right) = 1{,}000(1 - e^{-2}) \approx 1{,}000(0.8647) = 864.7$ The insurer expects to pay $ $864.70$ per loss.