Calculating the probability of total payout for an insurance company

Question

Automobile losses reported to an insurance company are independent and uniformly distributed between $0$ and $20,000$. The company covers each loss subject to a deductible of $5,000$.

Calculate the probability that the total payout on 200 reported losses is between $1,000,000$ and $1,200,000$.

My attempt:

Let $S$ denote the random variable for all losses reported to the insurance company. Then, using CLT, $S \sim N\left(200 \times \dfrac{20,000}{2}, 200 \times \dfrac{20,000}{12}\right) = (2\times 10^6, 577.35^2)$. Now let $Y$ be the random variable for the payout on all claims. Then, $Y=200(S-5000)$ if $10^6\le s<\infty$ and $Y=0$ otherwise.

The desired probability is given by $$P[10^6<Y<1.2\times 10^6] = P[10^6<200(S-5000)<1.2\times 10^6] = P[10^4<S<1.1\times 10^4]$$

Now, I computed $$P\left[Z\le\frac{(1.1\times 10^4) - 2\times 10^6}{577.35}\right] - P\left[Z\le\frac{10^4 - (2\times 10^6)}{577.35}\right]$$

but this didn't lead me to the correct answer. Can someone please suggest what I did wrong in this question? Thanks.

Answer 1

As pointed out by BruceET, you deduce $5000$ from every claim, including those who where valued less than $5000$. This fact changes the total payout.

Here is an example for why it didn't work. Let's say that we have $100$ claim of $20\ 000$ and $100$ claim of $0$. The total loss is $2\cdot10^6$. According to your reasonning, we deduce $5000$ for every claim, since it is a total loss of over $10^6$. After the deduction, we have a total payout of $10^6$. $100$ people recieved $15\ 000$ for the insurance company while the other $100$ people will give back $5000$ to the insurance company! In reality, the insurance company will pay $15\ 000$ to a $100$ people, and $0$ to the other, for a total of $1.5\cdot10^6$.

Let $X_i$ be an automobile loss. Then $X_i\sim \text{Unif}(0, 20\ 000)$. Let $P_i$ be the payout from the insurance company, then $$P_i = \begin{cases}0 &\text{if }X_i \leq 5000\\X_i-5000 &\text{if } 5000<X_i\leq 20\ 000\end{cases}$$

Since it is not an uniform distribution anymore, we have to compute the mean and variance. $$E[P_i] = \int_0^{15\ 000} P_i\times \frac{1}{20\ 000}\ \mathrm{d}P_i = 5625$$ $$E[P_i^2]\int_0^{15\ 000} P_i^2\times \frac{1}{20\ 000}\ \mathrm{d}P_i = 5.625\times10^7$$ $$Var[X_i] = 24\ 609\ 375$$

Now, let $Y$ be the total payout by the insurance company, for $200$ claims, then $$Y\sim N(200\times 5625, 200\times 24\ 609\ 375)$$ $$Y\sim N(1\ 125\ 000, 70\ 156.076^2)$$ All is left if to evaluate the probability as you did.

Answer 2

Not sure, because you give few reasons along with your computations, but it seems you're deducting \$5000 from all claims, including those for less than \$5000. [About $1/4$ of claims result in no payout because of the deductible; then payout is \$0, not -\$5000.]

Here is a simulation (of 100,000 lots of 200 claims) that may be helpful: Simulation uses thousands of dollars.

If $T$ is the total payout for 200 claims, then the simulation illustrates that $E(T)\approx \$1,125,000$ and $P(\$1\,000\,000 < T < \$1\,200\,000) \approx 0.82.$ With 100,000 iterations it is reasonable to expect about 2-place accuracy.

set.seed(2020)
m = 10^5;  tot.pay = numeric(m)
for(i in 1:m){
 loss = runif(200, 0, 20)
 tot,pay[i] = sum(pmax(0, loss-5))
}
mean(tot,pay)
[1] 1125.617
mean(tot.pay > 1000 & pay < 1200)
[1] 0.81652  $ aprx P(Total pay btw $1m & 1.2m)

summary(pay)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  825.1  1077.9  1125.5  1125.6  1173.0  1428.6 

hdr = "Payouts for 200 Claims"
hist(pay, prob=T, col="skyblue", xlab="$1000s", main=hdr)
 abline(v = c(1000, 1200), col="red")

Answer 3

Your calculation is incorrect because you are applying an aggregate deductible when in fact the deductible is applied on an individual basis. This means that if a ground-up loss is below $5000$, no claim payment is made. This is not adequately expressed in your assumption about your aggregate claim random variable $Y$.

To understand why your approach is flawed, consider the toy model where you have $10$ random ground-up losses, each uniformly distributed on $[0,1]$. The deductible on each loss is $0.5$. Then according to your logic, the aggregate claim amount on the $10$ losses is $Y = 10(S - 0.5)$ if $S > 5$, and $0$ otherwise. But this does not work because the losses could look like this: $$\{0.2, 0.7, 0.8, 0.9, 0.7, 0.8, 0.9, 0.7, 0.8, 0.9\}$$ and only the first loss is less than the deductible. The sum of these losses is greater than $5$ but the aggregate claim amount is not $$\{-0.3, 0.2, 0.3, 0.4, 0.2, 0.3, 0.4, 0.2, 0.3, 0.4\}$$ as your model would suggest: the first claim amount is $0$.

Rather, the model behaves like this. Let all numbers be expressed in units of thousands. $X_i \sim \operatorname{Uniform}(0,20)$ is the ground-up loss for the $i^{\rm th}$ reported loss. Then $$Y_i = \max(X_i - 5, 0)$$ is the claim size for the $i^{\rm th}$ reported loss. The aggregate claim on $n = 200$ reported losses is $$S = \sum_{i=1}^n Y_i.$$ It is on $S$ that the central limit theorem may be used; to this end, we would need to compute $\operatorname{E}[Y_i]$ and $\operatorname{Var}[Y_i]$. This is not difficult when $X_i$ is uniform. We have $$\begin{align} \operatorname{E}[Y_i] &= \operatorname{E}[X_i - 5 \mid X_i > 5]\Pr[X_i > 5] + \operatorname{E}[0 \mid X_i \le 5]\Pr[X_i \le 5] \\ &= \left(\frac{20+5}{2} - 5 \right)\frac{3}{4} + (0)\frac{1}{4} \\ &= \frac{45}{8}. \end{align}$$ The variance calculation I leave to you as an exercise (hint: calculate $\operatorname{E}[(X_i - 5)^2 \mid X_i > 5]$); the result is $$\operatorname{Var}[Y_i] = \frac{1575}{64}.$$ Now we can use a normal asymptotic approximation: $S$ is approximately normal with mean $$\mu = n \operatorname{E}[Y_i] = 1125,$$ and variance $$\sigma^2 = n \operatorname{Var}[Y_i] = \frac{39375}{8}.$$ Then the desired probability is $$\begin{align} \Pr[1000 \le S \le 1200] &= \Pr\left[\frac{1000 - 1125}{70.1561} \le \frac{S - \mu}{\sigma} \le \frac{1200 - 1125}{70.1561}\right] \\ &\approx \Pr\left[-1.78174 \le Z \le 1.06904\right] \\ &\approx 0.82008. \end{align}$$