Expected payout after a deductible

Question

To my understanding E[X-1] = E[x] - 1. So why doesn't the following work?

A policy has a deductible of 1 and a capped payment of 5 and losses are distributed exponentially with mean 2. The expected loss in the range of 1-6 (where variable payment amounts are made) would be $\int_{1}^{6}.5xe^{-.5x}dx$ which is about 1.42. I would have thought the expected payout was then .42 = E[x] - 1, but it's not. If you replace .5x by .5(x-1) in the integral you get roughly .86.

I see how the math works, but why doesn't using expectation as a linear operator work here?

Answer 1

We first have to understand what $$\int_{x=1}^6 x f_X(x) \, dx$$ represents. It is the conditional expectation of the ground-up loss between $1$ and $6$, multiplied by the probability of the loss being in that range. In other words, it is $$\int_{x=1}^6 x f_X(x) \, dx = \operatorname{E}[X \mid 1 \le X \le 6]\Pr[1 \le X \le 6]. \tag{1}$$ So the expression $$\begin{align} \int_{x=1}^6 (x-1) f_X(x) \, dx &= \operatorname{E}[X - 1 \mid 1 \le X \le 6]\Pr[1 \le X \le 6] \\ &= (\operatorname{E}[X \mid 1 \le X \le 6] - 1)\Pr[1 \le X \le 6], \tag{2} \end{align}$$ and unless $\Pr[1 \le X \le 6] = 1$, we see why subtracting $1$ from $(1)$ does not yield $(2)$.

Answer 2

$P(x) = \begin{cases} 0, & x < 1 \\ x-1, & 1 \le x < 6 \\ 5,& 6 \le x \end{cases}$

Let $f$ be the pdf, then $E[P] = \int_0^\infty P(x) f(x) dx =\int_1^6 (x-1){ 1\over 2} e^{- {x \over 2}} dx + \int_6^\infty (5){ 1\over 2} e^{- {x \over 2}} dx$.