Fisher.AggExcess

Reading: Fisher, G. et al, "Individual Risk Study Note," CAS Study Note, Version 3, October 2019. Chapter 3. Section 1

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Based on past exams, the main things you need to know (in rough order of importance) are:

• fact A...
• fact B...

reference	part (a)	part (b)	part (c)	part (d)
E (2018.Spring #1)
E (2018.Spring #1)
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In Plain English!

When modelling an insurance contract you need to know who pays and how much they pay out. Risk sharing at an aggregate level makes it challenging to determine overall coverage responsibilities. A familiar example (to US students at least) of aggregate loss cost coverage is health insurance. An insured has a deductible, a co-payment (another type of deductible), and a maximum annual out of pocket amount (the aggregate limit for your deductible and co-pay).

Let's look at the following example from the text. Suppose we have a Commercial General Liability large deductible policy with a per-occurrence deductible of $100,000 and an aggregate limit on the deductible of $500,000. The claim payment pattern for this policy may look like:

Date	(1) Dollars of loss on claims that are each less than $100k	(2) Number of claims over $100k	(3) Dollars of loss on claims over $100k	(4) Deductible	(5) Insurance payment	(6) Insured's cost so far this year
Q1	$132,500	0	0	$132,500	0	$132,500
Q2	$93,000	2	$350,000	$293,000	$150,000	$425,500
Q3	$105,000	0	0	$74,500	$30,500	$500,000
Q4	$122,500	1	$150,000	0	$272,500	$500,000

The main thing to do is to keep track of the cumulative deductible amount (6) in relation to the aggregate limit. Claims below the per-occurrence limit are covered by the insured's deductible unless the aggregate deductible is reached. Claims above the per-occurrence limit have the insurer cover the difference between the claim and the per-occurrence deductible. If a claim would cause the aggregate deductible to be exceeded, then the insured is only responsible for the amount up to the aggregate deductible.

Put differently, (4) = min( (1) + (2)*[Per-occurrence deductible] , [Aggregate deductible] - [Prior row (6)]). Then (5) = (1) + (3) - (4).

Here's a practice problem for you to try. Insert Fisher.AggDed1 PDF.

Question: What are three key questions which help price an insurance contract?

Solution:

1. How much does the insured pay before the insurer is responsible?
2. How much does the insurer pay before hitting its policy limits?
3. How much is the insured responsible for above the policy limits?

Key Definitions and Notation

Let A be a random variable for the actual loss experience and [math]E=E[A][/math] be the expected losses. The entry ratio is the ratio of actual to expected losses, and is written as [math]r=\frac{A}{E}[/math]. It may equivalently be defined as the ratio of actual loss ratio to expected loss ratio.

Pop Quiz    :-o

• A policy is expected to have $300,000 in losses. At the end of the policy, actual losses were $198,000. What is the entry ratio for this policy?

We may view the entry ratio as a random variable. Similarly sized policies having the same entry ratio and coverage should behave similarly. It's common practice to estimate aggregate excess losses based on the entry ratio.

The entry ratio is often used to look up various risk characteristics that have been derived from a group of similar risks. There are several tables that may be used and the correct one depends on the underlying conditions of the policy you're asked to price.

One particular table is Table M. This consists of entry ratios with corresponding Table M charges and Table M savings and are usually grouped by policy size and limit. Table M is used for aggregate excess policies without a per-occurrence deductible.

A Table M Charge corresponds to an entry ratio, r, and is denoted by [math]\phi(r)[/math]. It is the ratio of excess losses over rE to all expected losses. It may also referred to as the Aggregate Excess Loss Factor, Aggregate Excess Ratio, or Excess Pure Premium Ratio.

Let's use a small example to understand the Table M Charge.

Question: An insurer has five similar policies, each with an expected loss of $150,000. In normal year, the actual losses on those policies are: $132,000; $141,000; $150,000; $159,000; and $168,000. Calculate [math]\phi(1), \phi(0.6), \mbox{ and } \phi(1.12)[/math].

Solution:

Since the expected loss is $150,000 for each policy, [math]\phi(1)[/math] is the ratio of excess losses over $150,000 to the total expected losses. That is [math]\phi(1) = \frac{0+0+0+9,000+18,000}{5\cdot 150,000}=0.036[/math].

Similarly, [math]\phi(0.6)[/math] requires us to aggregate the loss dollars in excess of [math]0.6E=0.6\cdot\$150,000=\$90,000[/math]. This yields [math]\begin{align} \phi(0.6)&=\frac{132,000+141,000+150,000+159,000+168,000 - 5\cdot 90,000}{5\cdot 150,000}\\ &=0.4\end{align}[/math].

Lastly, [math]\phi(1.12)[/math] requires us to consider the loss dollars in excess of [math]1.12E=1.12\cdot\$150,000=\$168,000[/math]. Since the largest actual loss is $168,000, there are no losses over this threshold, so [math]\phi(1.12)=0[/math]. In fact, [math]\phi(r)=0 \mbox{ for } r\ge 1.12[/math].

Note that a risk contributed 0 to the Table M Charge, [math]\phi(r)[/math] if its actual loss was lower than rE.

The insurance charge is [math]\phi(r)\cdot E[/math]. That is, the expected loss multiplied [math]\phi(r)[/math]. For a retrospectively rated policy the insurance charge is the part of the premium which is fixed and pays for losses.

A Table M Savings is very similar to a Table M Charge but in the opposite direction. The Table M Savings measures the gap between the actual loss and rE and then divides this by the total expected loss. It is denoted by [math]\psi(r)[/math].

A risk which has its actual loss in excess of rE contributes 0 to the Table M Savings, [math]\psi(r)[/math].

Let's return to our example.

Question: Using the same five risks as the previous example, compute [math]\psi(1), \psi(0.6) \mbox{ and } \psi(1.12)[/math].

Solution>:

To calculate [math]\psi(1)[/math] we need to measure the gap between the actual loss and [math]1\cdot \$150,000 =\$150,000[/math]. We get [math]\psi(1)=\frac{(150,000-132,000)+(150,000-141,000)+0+0+0}{5\cdot150,000}=0.036[/math].

Similarly, [math]\psi(0.6)[/math] uses the gap between the actual loss and [math]0.6E=90,000[/math]. Since none of the actual losses are lower than $132,000 we deduce [math]\psi(0.6)=0[/math].

Lastly, [math]\psi(1.12)[/math] uses the gap between the actual loss and [math]1.12E=168,000[/math]. We get [math]\begin{align}\psi(1.12)&=\frac{5\cdot 168,000-(132,000+141,000+150,000+159,000+168,000)}{5\cdot 150,000}\\ &=0.12\end{align}[/math].

A neat trick to associate the symbols with the terminology is the Table M Savings uses the Greek letter Psi.

Pop Quiz Answers

The entry ratio is [math]r=\frac{A}{E}=\frac{198,000}{300,000}=0.66[/math].