Actuarial Notes

Coverage Modifications

Study guides: Exam FAM, Exam ASTAM, Exam MAS-I

Coverage modifications are the policy provisions — a deductible, a policy limit, coinsurance — that turn a ground-up loss X into the amount the insurer pays. Together with inflation they decide how loss severity becomes insurer cost, and they change both the payment severity and the number of payments.

Y^L = \alpha\big[(X \wedge u) - (X \wedge d)\big]

E[Y^L] = \alpha\big(E[X \wedge u] - E[X \wedge d]\big)

E[Y^P] = \frac{E[Y^L]}{1 - F_X(d)}

  • d is an ordinary deductible, u the maximum covered loss (the loss above which the payment stops growing) and \alpha the coinsurance share, so the largest payment is \alpha(u - d). Some pages, such as Benefit Limit, write u for the maximum payment instead — check which one a question means. E[X \wedge u] is the limited expected value.
  • Per loss vs per payment. Y^L, the per-loss variable, is 0 whenever X \le d and averages over every loss. Y^P = Y^L \mid Y^L > 0, the per-payment variable, averages only over losses that produce a payment — hence the division by S(d) = 1 - F_X(d). See Payment Random Variable.
  • Franchise deductible. Once X > d it pays the whole loss, so with no limit or coinsurance E[Y^L] = E[(X-d)_+] + d\,S(d). The share of expected loss an ordinary deductible removes is the Loss Elimination Ratio, E[X \wedge d]/E[X].
  • Inflation. If losses grow to (1+r)X while d and u stay fixed, deflate the limits instead of inflating the losses (for E[Y^P], divide by 1 - F_X\big(\frac{d}{1+r}\big)):
\begin{aligned}
E[Y^L] = \alpha(1+r)\Big(&E\big[X \wedge \tfrac{u}{1+r}\big] \\
-\ &E\big[X \wedge \tfrac{d}{1+r}\big]\Big)
\end{aligned}
  • Frequency. Only losses above d become payments, so each loss becomes a payment with probability v = S(d). A Poisson(\lambda) loss count gives a Poisson(\lambda v) payment count — see Poisson Thinning. Negative binomial \beta becomes \beta v, and binomial q becomes qv. Expected aggregate payments are the same either way, E[N^L]\,E[Y^L] = E[N^P]\,E[Y^P] — see Aggregate Loss Model.

Per-Loss and Per-Payment Cost of a Layer Ground-up losses are Pareto with \alpha = 3, \theta = 2{,}000 (mean 1,000), so E[X \wedge x] = 1{,}000\left[1 - \left(\frac{2000}{2000+x}\right)^2\right]. A policy has an ordinary deductible of 500, a maximum covered loss of 5,000 and 80% coinsurance. Losses per policy are Poisson with mean 0.3. Find E[Y^L], E[Y^P] and the expected number of payments.

Answer

\begin{align*}
E[X \wedge 500] &= 1{,}000(1 - 0.64) \\
&= 360 \\
E[X \wedge 5000] &= 1{,}000(1 - 0.081633) \\
&= 918.367 \\
E[Y^L] &= 0.8\,(918.367 - 360) \\
&= 446.694
\end{align*}

With S(500) = (2000/2500)^3 = 0.512:

\begin{align*}
E[Y^P] &= \frac{446.694}{0.512} \\
&= 872.45 \\
E[N^P] &= 0.3 \times 0.512 \\
&= 0.1536
\end{align*}

Expected payments per policy come to 134.01 whichever way they are counted: 0.3 \times 446.694 per loss, or 0.1536 \times 872.45 per payment.

Leveraged Effect of 10% Inflation In the example above, losses inflate by 10% while the deductible and limit stay fixed. Recompute E[Y^L] and E[Y^P].

Answer A Pareto scales by its \theta, so inflated losses are Pareto with \alpha = 3, \theta = 2{,}200, and E[X' \wedge x] = 1{,}100\left[1 - \left(\frac{2200}{2200+x}\right)^2\right].

\begin{align*}
E[X' \wedge 500] &= 369.684 \\
E[X' \wedge 5000] &= 997.299 \\
E[Y^L] &= 0.8\,(997.299 - 369.684) \\
&= 502.09
\end{align*}

With S'(500) = (2200/2700)^3 = 0.54097:

\begin{align*}
E[Y^P] &= \frac{502.09}{0.54097} \\
&= 928.13
\end{align*}

Ground-up severity rose 10%, but the cost per loss rose 12.4%: the fixed deductible leverages inflation more than the limit dampens it. The cost per payment rose only 6.4%, because 5.7% more losses now pierce the deductible. Those new small payments pull the per-payment average down.