Coinsurance and Treatment Thereof in Combination with Ordinary Deductibles, Policy Limits, and Inflation

In an arrangement of coinsurance, an insurance company pays a proportion of the loss, denoted α, and the policyholder pays (1-α) of the loss, i.e., the remainder. For a policy with coinsurance and loss random variable X, the random variable for the coinsurance payment becomes Y = αX.

This random variable is treated as any other multiple of a random variable, and this treatment is discussed in depth in Section 13.

A useful set of formulas applies to a case where a policy has all of the following: coinsurance with proportion α, an ordinary deductible of amount d, and a limit of amount u. There is also inflation at a uniform rate r. Note that for the equations below to apply, the coinsurance must apply last, after the deductible and limit have already been considered. In that case, the per-loss payment random variable is denoted Y^L and defined as follows:
Y^L = 0 when X < d/(1+r).

Y^L = α((1+r)X - d) when d/(1+r) ≤ X < u/(1+r).

Y^L = α(u - d) when u/(1+r) ≤ X.

E(Y^L) = α(1+r)(E(X Λ u/(1+r)) - E(X Λ d/(1+r))).

E((Y^L)²) = α²(1+r)²(E((X Λ u/(1+r))²)- E((X Λ d/(1+r))²) - 2(d/(1+r))E(X Λ u/(1+r)) + 2(d/(1+r))E(X Λ d/(1+r)))

For the per-payment payment random variable, Y^P, the following is true.

E(Y^P) = E(Y^L)/(1-F_X(d/(1+r))).

E((Y^P)²) = E((Y^L)²)/(1-F_X(d/(1+r))).

Source:

Loss Models: From Data to Decisions, (Third Edition), 2008, by Klugman, S.A., Panjer, H.H. and Willmot, G.E., Chapter 8, pp. 189-190.

Original Problems and Solutions from The Actuary's Free Study Guide

Problem S4C29-1. Losses from falling cows follow an exponential distribution with mean θ = 340. Sophisticated Cow Insurance Company has a policy in place with the following features:
- A deductible of 100;

- A policy limit of 400;

- Coinsurance of 60%, applied after the deductible and limit have been considered.

Now the Fed begins to print vast amounts of money, and the annual rate of inflation is 30%.

After one year, an insured under this policy suffers a falling cow loss of 230. What is the value of Y^L, the per-loss payment random variable associated with this loss?

Solution S4C29-1. Here, u = 400, d = 100, r = 0.3, and α = 0.6. We find u/(1+r) = 400/1.3 = 307.6923077 and d/(1+r) = 100/1.3 = 76.92307692.

The loss X = 230, which is between d/(1+r) and u/(1+r). Thus, the following formula applies:

Y^L = α((1+r)X - d) = 0.6((1.3)230-100) = Y^L = 119.4.

Problem S4C29-2. Losses from falling cows follow an exponential distribution with mean θ = 340. Sophisticated Cow Insurance Company has a policy in place with the following features:
- A deductible of 100;

- A policy limit of 400;

- Coinsurance of 60%, applied after the deductible and limit have been considered.

Now the Fed begins to print vast amounts of money, and the annual rate of inflation is 30%.

Find E(Y^L) for this policy after one year.

Relevant properties for exponential distributions: S_X(x) = e^-x/θ;

E(X) = θ; E(X Λ K) = θ(1 - e^-K/θ).

Solution S4C29-2. Here, u = 400, d = 100, r = 0.3, and α = 0.6.

We use the formula E(Y^L) = α(1+r)(E(X Λ u/(1+r)) - E(X Λ d/(1+r))).

From Solution S4C29-1, we know that d/(1+r) = 76.92307692 and u/(1+r) = 307.6923077.

E(X Λ u/(1+r)) = E(X Λ 307.6923077) = 340(1 - e^{-307.6923077/340}) = 202.4526472.

E(X Λ d/(1+r)) = E(X Λ 76.92307692) = 340(1 - e^{-76.92307692/340}) = 68.8421093.

Thus, E(Y^L) = 0.6*1.3(202.4526472-68.8421093) = E(Y^L) = 104.2162196.

Problem S4C29-3. Losses from falling cows follow an exponential distribution with mean θ = 340. Sophisticated Cow Insurance Company has a policy in place with the following features:
- A deductible of 100;

- A policy limit of 400;

- Coinsurance of 60%, applied after the deductible and limit have been considered.

Now the Fed begins to print vast amounts of money, and the annual rate of inflation is 30%.

Find E(Y^P) for this policy after one year.

Relevant properties for exponential distributions: S_X(x) = e^-x/θ;

E(X) = θ; E(X Λ K) = θ(1 - e^-K/θ).

Solution S4C29-3. We use the formula E(Y^P) = E(Y^L)/(1-F_X(d/(1+r))).

We know from Solution S4C29-2 that E(Y^L) = 104.2162196.

We also know from Solution S4C29-1 that d/(1+r) = 76.92307692.

1-F_X(d/(1+r)) = S_X(d/(1+r)) = e^{-76.92307692/340} = 0.7975232079.

Thus, E(Y^P) = 104.2162196/0.7975232079 = E(Y^P) = 130.6748425.

Problem S4C29-4. Losses from falling cows follow an exponential distribution with mean θ = 340. Sophisticated Cow Insurance Company has a policy in place with the following features:
- A deductible of 100;

- A policy limit of 400;

- Coinsurance of 60%, applied after the deductible and limit have been considered.

Now the Fed begins to print vast amounts of money, and the annual rate of inflation is 30%.

Find Var(Y^L) for this policy after one year. You may use the following Excel input format for any incomplete Gamma function computations required in this problem:

"=GAMMADIST(x, α, 1, TRUE)"

Relevant properties for exponential distributions: S_X(x) = e^-x/θ;

E(X) = θ; E(X Λ K) = θ(1 - e^-K/θ); E((X Λ K)²) = 2θ²*Г(3; K/θ) + K²e^-K/θ.

Solution S4C29-4. Var(Y^L) = E((Y^L)²)- E((Y^L))².

We know from Solution S4C29-2 that E(Y^L) = 104.2162196.

Thus, E((Y^L))² = 10861.02042.

To find E((Y^L)²), we use the formula E((Y^L)²) = α²(1+r)²(E((X Λ u/(1+r))²)- E((X Λ d/(1+r))²) - 2(d/(1+r))E(X Λ u/(1+r)) + 2(d/(1+r))E(X Λ d/(1+r)))

From Solution S4C29-2, we know that E(X Λ u/(1+r)) = 202.4526472 and E(X Λ d/(1+r)) = 68.8421093.

From Solution S4C29-1, we know that d/(1+r) = 76.92307692 and u/(1+r) = 307.6923077.

We need to find E((X Λ u/(1+r))²) = (E(X Λ 307.6923077)²) =

2*340²*Г(3; 307.6923077/340) + 307.6923077²e^{-307.6923077/340}.

To find Г(3; 307.6923077/340), we use the Excel input

"=GAMMADIST(307.6923077/340, 3, 1, TRUE)", which gives the result of 0.063679002. Thus,

(E(X Λ 307.6923077)²) = 2*340²*0.063679002 + 38300.68996 = E((X Λ u/(1+r))²) = 53023.27523.

We need to find E((X Λ d/(1+r))²) = (E(X Λ 76.92307692)²) =

2*340²*Г(3; 76.92307692/340) + 76.92307692²e^{-76.92307692/340}.

To find Г(3; 76.92307692/340), we use the Excel input

"=GAMMADIST(76.92307692/340, 3, 1, TRUE)", which gives the result of 0.001630465.

Thus, (E(X Λ 76.92307692)²) = 2*340²*0.001630465 + 4719.072236 = E((X Λ d/(1+r))²) = 5096.035744.

Therefore, E((Y^L)²) = 0.6²(1.3)²(53023.27523 - 5096.035744 - 2*76.92307692*202.4526472 + 2*76.92307692*68.8421093) = E((Y^L)²) = 16652.98616.

Thus, Var(Y^L) = 16652.98616 - 10861.02042 = Var(Y^L) = 5719.965736.

Problem S4C29-5. Losses from falling cows follow an exponential distribution with mean θ = 340. Sophisticated Cow Insurance Company has a policy in place with the following features:
- A deductible of 100;

- A policy limit of 400;

- Coinsurance of 60%, applied after the deductible and limit have been considered.

Now the Fed begins to print vast amounts of money, and the annual rate of inflation is 30%.

Find Var(Y^P) for this policy after one year.

Solution S4C29-5. Var(Y^P) = E((Y^P)²)- E((Y^P))².

We know from Solution S4C29-3 that E(Y^P) = 130.6748425.

Thus, E((Y^P))² = 17075.91446.

We now find E((Y^P)²) = E((Y^L)²)/(1-F_X(d/(1+r))).

We know from Solution S4C29-4 that E((Y^L)²) = 16652.98616.

We know from Solution S4C29-3 that 1-F_X(d/(1+r)) = 0.7975232079

Thus, E((Y^P)²) = 16652.98616/0.7975232079 = 20880.8797.

Hence, Var(Y^P) = 20880.8797 - 17075.91446 = Var(Y^P) = 3804.965237.

See other sections of The Actuary's Free Study Guide for Exam 4 / Exam C.