Why Peak Exposure Allocation Matters
A reinsurance program may have a fixed amount of available limit even when the insurer has several locations, lines, or event scenarios that could use it. The allocation question is not simply which exposure has the largest possible loss. It is where the next unit of limit is expected to reduce retained loss most, subject to contractual and portfolio constraints. A small exposure with a high probability of reaching its limit may offer more expected protection per unit than a larger but less exposed one.
This analysis is especially useful when losses are concentrated: a single region or peril may account for a large share of potential retained loss. A reinsurer may impose a limit on capacity for that region, or the insurer may set its own concentration cap. The resulting allocation must balance marginal protection value with those limits. It is a decision about the placement of capacity, not a substitute for evaluating the entire reinsurance program.
For an allocation based on expected recovery, each exposure has a sequence of marginal values. If the first $1 million of limit on an exposure has expected recovery of $0.50 million and the next $1 million has expected recovery of $0.30 million, those are different values for different blocks. The first block may protect losses that occur frequently; later blocks may respond only to more severe losses. Comparing successive blocks, rather than assigning one average value to an entire exposure, is central to a sound allocation.
How to Measure a Block’s Protection Value
Suppose a hypothetical exposure has annual loss \(X\), and a layer pays above an attachment point \(a\), up to a further limit \(h\). Its payment is \(\min((X-a)^+,h)\), where \((z)^+=\max(z,0)\). For a layer of width $1 million from \(a\) to \(a+1\) million, expected recovery can be calculated from the loss scenarios or from the loss exceedance curve. The value is the expected payment, not the maximum possible payment.
Here \(S_X(t)=P(X>t)\) is the loss survival function, and loss and limit are expressed in the same units. To see why the formula holds, write the payment as the length of the part of the interval from \(a\) to \(a+h\) lying below the realized loss. Equivalently, for each \(t\) in that interval, the payment includes one unit of width exactly when \(X>t\). Taking expectations of these indicators gives \(P(X>t)=S_X(t)\); integrating over the interval gives the formula. For a one-unit layer, a loss distribution with substantial probability above the layer’s attachment generally produces a larger expected recovery than one that rarely reaches it.
When losses are given as discrete scenarios, calculate the payment in each scenario and multiply by its probability. Add those probability-weighted payments. Expected recovery values for different exposures can then be added even if the losses are dependent, because expectation is additive. However, dependence can matter for other program objectives and constraints; a ranking based only on expected recovery does not capture every portfolio consideration.
Worked Example: Allocating Capacity When Later Blocks Have Equal Value
A hypothetical insurer has $9 million of limit to assign among three regions. Maximum available limits are $2 million for Region A, $5 million for Region B, and $6 million for Region C. Each region’s first $1 million of limit has expected recovery of $0.48 million, $0.42 million, and $0.35 million, respectively. Every subsequent $1 million block in any region has expected recovery of $0.18 million. Allocate the capacity to maximize expected recovery and calculate the total.
Method. Treat each $1 million block as a separate allocation unit. Rank the first blocks by their marginal values: A at $0.48 million, B at $0.42 million, and C at $0.35 million. Allocate one block to each first, for a total of $1.25 million expected recovery and $3 million of capacity used.
There are \(9-3=6\) million dollars of capacity left. All subsequent blocks have the same value, $0.18 million each. The remaining region capacities are $1 million in A, $4 million in B, and $5 million in C, so one feasible allocation of the six remaining blocks is $1 million to A, $4 million to B, and $1 million to C.
The resulting total limits are $2 million in A, $5 million in B, and $2 million in C. They use all $9 million of capacity and respect each region cap. Since every subsequent block is worth the same amount, other feasible distributions of the six remaining blocks have the same expected recovery. Common mistake avoided: the stated values apply only to each region’s first block, not to every block assigned there.
Applying Concentration Limits
A concentration limit restricts how much capacity can be assigned to a specified exposure or group of exposures. The limit may be expressed as a maximum dollar amount or as a share of total capacity. In either case, first identify the actual constraint and translate it into an allocation bound. Do not assume that the exposure with the highest initial value can receive unlimited capacity.
Worked Example: Reallocating Capacity Under a Regional Cap
A hypothetical insurer has six $1 million blocks to allocate. Each region offers three successive blocks with the following expected recoveries, in millions of dollars per block: Region A, $0.60, $0.50, and $0.42; Region B, $0.55, $0.45, and $0.30; Region C, $0.52, $0.40, and $0.35. A concentration rule limits Region A to $2 million. Determine the allocation that maximizes expected recovery.
Method. Rank marginal values, respecting the fact that a later block in a region cannot be selected without allocating its earlier blocks. Without the concentration rule, the six largest values would be $0.60, $0.55, $0.52, $0.50, $0.45, and $0.42 million. That selection gives Region A three blocks and violates its $2 million cap.
Under the cap, Region A’s $0.42 million third block is unavailable. The six highest feasible blocks are A’s $0.60 and $0.50 blocks, B’s $0.55 and $0.45 blocks, and C’s $0.52 and $0.40 blocks. These selections are prefixes of each region’s sequence, so each selected block is available after its preceding block.
The allocation uses six blocks, gives Region A exactly its $2 million maximum, and produces expected recovery of $3.02 million. Without the concentration cap, the best six blocks would be worth $3.04 million, but that allocation is infeasible. The relevant answer is the best feasible allocation, not the unconstrained ranking.
Why the Marginal-Value Rule Works
When each exposure has diminishing marginal values, a simple priority rule is justified under a specific set of assumptions. Suppose limit is divisible into equal blocks, the total capacity is fixed, expected recovery is additive across exposures, each exposure has its own upper bound, and each successive block in an exposure is worth no more than the preceding block. There are no additional interactions, minimum allocations, or group constraints beyond those represented in the block availability and caps.
For every exposure, record the expected recovery from each successive block of limit, in order, and identify the maximum number of blocks available.
Remove blocks that exceed exposure caps or concentration limits. Preserve the order within each exposure: a later block is not available unless earlier blocks are allocated.
Allocate capacity to the highest-value available blocks until the capacity is exhausted or no further block is available.
Check the total capacity, every cap, the selected block sequence for each exposure, and the sum of marginal expected recoveries.
The justification is an exchange argument. Consider a feasible allocation that selects a lower-value available block while leaving an available higher-value block unselected. If the higher-value block can replace the lower-value block without violating a cap or a block-order requirement, the replacement preserves the amount of capacity and increases or leaves unchanged expected recovery. Repeating such exchanges leads to an allocation consisting of the highest feasible marginal values. Diminishing values ensure that selected blocks within an exposure form a prefix: a later block cannot outrank an unselected earlier block.
This argument does not establish that a simple ranking solves every program-design problem. For example, a constraint limiting the combined capacity of several regions may couple their allocations. Non-diminishing values, indivisible blocks, minimum participation requirements, or interactions between covers can also invalidate the simple exchange. In those cases, formulate the allocation as a constrained optimization problem and check feasibility across the full set of restrictions.
Worked Example: Deriving Marginal Values from Loss Scenarios
A hypothetical insurer can allocate at most $2 million to each of Regions K and L, with total capacity of $3 million. In one year, Region K’s loss is $4 million with probability 0.10, $2 million with probability 0.20, and zero otherwise. Region L’s loss is $3 million with probability 0.15, $1 million with probability 0.25, and zero otherwise. Each allocated block is a $1 million layer beginning at zero. Find the allocation maximizing expected recovery.
Method. For each layer, calculate its payment in each loss scenario. In K’s first layer, both positive-loss scenarios pay $1 million, so expected recovery is \(0.10(1)+0.20(1)=\$0.30\) million. K’s second layer also pays in both positive-loss scenarios, giving $0.30 million. K’s third layer pays only when the loss is $4 million, giving \(0.10(1)=\$0.10\) million; this block is beyond K’s $2 million cap and is unavailable.
In L, the first layer pays in both positive-loss scenarios, so its expected recovery is \(0.15(1)+0.25(1)=\$0.40\) million. The second layer pays only when the loss is $3 million, giving \(0.15(1)=\$0.15\) million. This is the full set of available blocks under L’s $2 million cap.
The three largest available values are L’s first block at $0.40 million and both of K’s blocks at $0.30 million each. Thus allocate $2 million to K and $1 million to L. Expected recovery is \(0.30+0.30+0.40=\$1.00\) million. The allocation uses all $3 million of capacity and meets both region caps. The unused second block in L is worth $0.15 million, less than either selected K block.
Common Exam Trap and Practical Checks
A frequent error is to apply the first-block value to all capacity assigned to an exposure. In the first example, a tempting but incorrect calculation for the $2 million, $5 million, and $2 million allocation is \(2(0.48)+5(0.42)+2(0.35)=\$3.76\) million. That calculation treats each region’s first-block value as though it applied to every block. It contradicts the stated marginal schedule: only the first block in each region has its listed value, and the other six blocks are each worth $0.18 million. The correct calculation is \(0.48+0.42+0.35+6(0.18)=\$2.33\) million.
Before finalizing an allocation, check the following:
- Units: values per $1 million of limit and total expected recoveries must be combined consistently.
- Block order: a later layer cannot be counted without the preceding limit that makes it available.
- Capacity: the sum of allocated limits must not exceed the total capacity.
- Concentration: verify every exposure cap and any group-level limit.
- Scope of value: expected recovery is one measure of protection value; it is not automatically a complete assessment of all program objectives.
The allocation method is aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus: define the objective and constraints, calculate marginal protection values on a consistent basis, and demonstrate that the proposed allocation is both feasible and optimal under the stated assumptions.
Check Your Understanding
For each problem, treat each listed value as the expected recovery from that particular $1 million block. Unless stated otherwise, blocks must be selected in order within each exposure.
- Capacity is $4 million. Region A has marginal values of $0.50, $0.30, and $0.10 million; Region B has values of $0.40, $0.35, and $0.20 million. Each has a $3 million cap. What allocation maximizes expected recovery, and what is the total expected recovery? (Answer: A $2 million, B $2 million; $1.55 million expected recovery.)
- Capacity is $5 million. Regions A, B, and C have successive marginal values of, respectively, $0.70/$0.50/$0.25 million, $0.60/$0.45/$0.30 million, and $0.55/$0.40/$0.20 million. A concentration cap limits A to $1 million. Find the optimal allocation and expected recovery. (Answer: A $1 million, B $2 million, C $2 million; $2.70 million expected recovery.)
- Region M has annual loss of $3 million with probability 0.20, $1 million with probability 0.30, and zero otherwise. Region N has annual loss of $2 million with probability 0.35, $1 million with probability 0.10, and zero otherwise. Each can receive at most $2 million of zero-attachment limit, and total capacity is $3 million. Find the allocation maximizing expected recovery. (Answer: M $1 million, N $2 million; $1.30 million expected recovery.)
- Capacity is $4 million. Region P’s first two $1 million blocks are worth $0.50 million and $0.25 million; Region Q’s are worth $0.45 million and $0.35 million; Region R’s are worth $0.30 million and $0.20 million. A group concentration rule limits the combined allocation to P and Q to $3 million. Find an optimal allocation and its expected recovery. (Answer: P $1 million, Q $2 million, R $1 million; $1.60 million expected recovery.)
- A one-million-dollar block pays $1 million when the annual loss exceeds its attachment and zero otherwise. For a loss distribution with \(P(X>1)=0.30\) and \(P(X>2)=0.12\), calculate the expected recoveries of the layers from $1 million to $2 million and from $2 million to $3 million. (Answer: $0.30 million and $0.12 million, respectively.)