Common Errors In Aggregate Probability And Stop-Loss Problems

Why Aggregate Calculations Need an Error Check

A compound aggregate calculation combines a claim count with individual claim amounts. Several errors can produce plausible-looking numbers: using severity variance where the severity second moment is required, leaving out the probability of no claims, shifting a recursion index, or treating a lattice threshold as though it were a claim count. Tail truncation adds another risk: a sum over a few claim counts may be close to the full stop-loss value without being the full value.

The purpose of an error check is not merely to redo arithmetic. It is to verify that the model, indices, support, and threshold all match the quantity being calculated. The examples use hypothetical loss models and state those assumptions explicitly.

Definition: Let \(N\) be a nonnegative integer-valued claim count, and let \(X_1,X_2,\ldots\) be independent and identically distributed nonnegative claim severities, independent of \(N\). The aggregate loss is \(S_N=\sum_{i=1}^{N}X_i\), with \(S_0=0\). The aggregate stop-loss expectation at retention \(d\geq 0\) is \(E[(S_N-d)_+]\), where \((x)_+=\max(x,0)\).

Separate Frequency and Severity Moments

For a random claim count \(N\) and independent severities with finite second moment, conditioning on \(N\) gives \(E[S_N]=E[N]E[X]\) and \(\operatorname{Var}(S_N)=E[N]\operatorname{Var}(X)+\operatorname{Var}(N)(E[X])^2\). For Poisson frequency with mean \(\lambda\), the frequency variance also equals \(\lambda\), so the aggregate variance simplifies to \(\lambda E[X^2]\).

\operatorname{Var}(S_N) =E[N]\operatorname{Var}(X)+\operatorname{Var}(N)(E[X])^2

The distinction between \(E[X^2]\) and \(\operatorname{Var}(X)\) is essential. For Poisson frequency, \(\lambda E[X^2]=\lambda\{\operatorname{Var}(X)+(E[X])^2\}\), not \(\lambda\operatorname{Var}(X)\). The extra term reflects variability in the number of claims.

Justification: Conditional on \(N=n\), the sum has mean \(nE[X]\) and variance \(n\operatorname{Var}(X)\). The law of total variance therefore gives \(\operatorname{Var}(S_N)=E[N\operatorname{Var}(X)]+\operatorname{Var}(NE[X])\), which is the stated formula. For Poisson \(N\), \(E[N]=\operatorname{Var}(N)=\lambda\), yielding \(\lambda\{\operatorname{Var}(X)+(E[X])^2\}=\lambda E[X^2]\).

Worked Example: Checking the Compound Poisson Variance

A hypothetical portfolio has \(N\sim\operatorname{Poisson}(2)\). Each claim severity is independently \(1\) or \(3\) loss units, each with probability \(1/2\). Find the aggregate mean, variance, and probability of zero aggregate loss.

Method. Calculate the severity moments first, then use the compound Poisson formulas. This avoids substituting the severity variance for the second moment.

The severity mean and second moment are \(E[X]=(1+3)/2=2\) and \(E[X^2]=(1^2+3^2)/2=(1+9)/2=5\). Thus \(\operatorname{Var}(X)=E[X^2]-(E[X])^2=5-4=1\).

The aggregate mean is \(E[S_N]=2(2)=4\) loss units. The aggregate variance is \(\operatorname{Var}(S_N)=2E[X^2]=2(5)=10\) square loss units. As a check, the general formula gives \(2(1)+2(2^2)=2+8=10\).

Because every claim severity is strictly positive, aggregate loss is zero exactly when \(N=0\). Therefore \(P(S_N=0)=P(N=0)=e^{-2}=0.1353\), to four decimal places.

Common mistake and correction. Using \(\lambda\operatorname{Var}(X)=2(1)=2\) as the aggregate variance omits the frequency contribution. The correct value is \(10\), and the equality of the two variance calculations above checks it.

Zero Atoms and Panjer Recursion

The aggregate distribution has a probability at zero even when every individual claim is positive: it is the probability of no claims. For a Poisson count and positive severities, that probability is \(e^{-\lambda}\). More generally, if the severity distribution has probability \(f_0=P(X=0)\) at zero, the Poisson compound model has \(P(S_N=0)=\exp\{\lambda(f_0-1)\}\). A calculation of \(P(S_N\leq t)\) must include \(P(S_N=0)\) whenever zero is in the event.

For a nonnegative integer-valued severity, write \(f_j=P(X=j)\) and \(p_k=P(S_N=k)\). If \(N\) is Poisson with mean \(\lambda\), Panjer recursion is \(p_0=\exp\{\lambda(f_0-1)\}\), followed, for \(k\geq1\), by the recursion below.

p_k=\frac{\lambda}{k}\sum_{j=1}^{k}j f_j p_{k-j}

The upper limit is \(k\) because a severity larger than \(k\) cannot contribute to aggregate value \(k\). The factor \(j\) multiplies the severity probability \(f_j\), and the corresponding aggregate probability is \(p_{k-j}\). For positive severities, \(f_0=0\), so the initialization is \(p_0=e^{-\lambda}\).

Derivation: Let \(F(z)=\sum_{j\geq0}f_jz^j\) be the severity probability generating function. The compound Poisson probability generating function is \(P(z)=\exp\{\lambda(F(z)-1)\}\). Differentiation gives \(zP'(z)=\lambda zF'(z)P(z)\). Equating coefficients of \(z^k\), the left side is \(kp_k\), while the right side is \(\lambda\sum_{j=1}^{k}j f_jp_{k-j}\). Dividing by \(k\) gives the recursion. Its initial value is \(P(0)=\exp\{\lambda(f_0-1)\}\).

Worked Example: Including Zero and Auditing Panjer Indices

Suppose \(N\sim\operatorname{Poisson}(1)\), and each hypothetical claim is \(1\) or \(2\) loss units with probability \(1/2\) each. Use Panjer recursion to find \(P(S_N\leq2)\). Also check the calculation directly.

Method. Initialize at \(p_0=e^{-1}\), then compute the required probabilities in order. This method avoids both omitting the zero atom and shifting the recursion indices.

Here \(f_1=f_2=1/2\), and \(f_j=0\) for \(j>2\). The initial probability is \(p_0=e^{-1}\). For \(k=1\), \(p_1=(1/1)(1\cdot(1/2)p_0)=0.5e^{-1}\).

For \(k=2\), both severity values contribute: \(p_2=(1/2)\{1\cdot(1/2)p_1+2\cdot(1/2)p_0\}\) \(=(1/2)\{0.25p_1+p_0\}=(1/2)\{0.125p_0+p_0\}=0.5625p_0\).

Therefore \(P(S_N\leq2)=p_0+p_1+p_2=(1+0.5+0.5625)e^{-1}=2.0625e^{-1}=0.7588\), to four decimal places.

For a direct check, aggregate loss 1 occurs only with one claim of size 1, with probability \(e^{-1}(1)(1/2)=0.5e^{-1}\). Aggregate loss 2 occurs either with one claim of size 2, probability \(0.5e^{-1}\), or two claims both of size 1, probability \(e^{-1}(1^2/2!)(1/2)^2=0.125e^{-1}\). Hence \(p_2=0.625e^{-1}\), not \(0.5625e^{-1}\).

This direct check reveals an arithmetic error in the displayed recursion evaluation: its \(k=2\) expression must use \(2f_2p_0=1p_0\), while the \(j=1\) term is \(0.5p_1=0.25p_0\). Their sum is \(1.25p_0\), divided by 2, so \(p_2=0.625p_0\). Thus the corrected cumulative probability is \((1+0.5+0.625)e^{-1}=2.125e^{-1}=0.7812\).

Common mistake and correction. A miscopied coefficient or an incomplete arithmetic check can survive a formally correct recursion. The direct claim-count decomposition confirms the coefficient: the two-claim contribution is \(0.125e^{-1}\), and the one-claim contribution is \(0.5e^{-1}\). Their sum is \(0.625e^{-1}\). The final answer is \(0.7812\), and the explicit sum includes the zero atom \(e^{-1}\).

Lattice Thresholds and Stop-Loss Tails

For a lattice aggregate with probabilities \(p_k=P(S_N=k)\), the stop-loss expectation is \(E[(S_N-d)_+]=\sum_{k\geq0}(k-d)_+p_k\). The positive-part operation matters: values at or below \(d\) contribute zero, and a value strictly above \(d\) contributes its excess over \(d\). If the threshold is not an integer, the first contributing lattice point is the smallest integer strictly greater than \(d\). Replacing \(d\) by a rounded integer changes the payment and is generally wrong.

Worked Example: A Noninteger Retention and a Truncated Tail

Let \(N\sim\operatorname{Poisson}(1)\), with each hypothetical claim equal to exactly one loss unit. Thus \(S_N=N\). Find \(E[(S_N-2.5)_+]\), and measure the error from retaining only claim counts \(3,4,5\).

Method. Sum the lattice stop-loss contributions and then evaluate the omitted tail independently. The check is important because a finite partial sum is not the full expectation.

Since \(N\) is integer-valued, \((N-2.5)_+=0\) for \(N=0,1,2\), while the contributions at \(N=3,4,5\) are \(0.5,1.5,2.5\). Using \(P(N=n)=e^{-1}/n!\), the partial sum is

e^{-1}\left(\frac{0.5}{3!}+\frac{1.5}{4!}+\frac{2.5}{5!}\right) =e^{-1}\left(\frac{1}{12}+\frac{1}{16}+\frac{1}{48}\right) =\frac{e^{-1}}{6}=0.0613132.

To find the full value, use \(E[N\mathbf{1}_{\{N\geq3\}}]=P(N\geq2)=1-2e^{-1}\). This follows by shifting the Poisson sum: \(nP(N=n)=P(N=n-1)\) when the mean is 1. Also, \(P(N\geq3)=1-2.5e^{-1}\). Consequently,

E[(N-2.5)_+] =E[N\mathbf{1}_{\{N\geq3\}}]-2.5P(N\geq3) =-1.5+\frac{4.25}{e} =0.0634876.

The omitted contribution from \(N\geq6\) is \(0.0634876-0.0613132=0.0021744\). The three-term calculation understates the full stop-loss value by about \(0.0022\) loss units. The partial sum is close, but it is not the full answer.

Common mistake and correction. Rounding the retention 2.5 to 3 would incorrectly make the payment at \(N=3\) zero. The correct contribution there is \(3-2.5=0.5\). A second trap is arithmetic: the three terms inside the parentheses sum to \(1/6\), not \(0.15\); the corrected partial sum is \(e^{-1}/6=0.0613132\).

A Practical Audit Before Finalizing

An aggregate answer is more reliable when its inputs and support are checked in a fixed order. The following process catches structural errors before they become polished but incorrect numerical results.

State the model.
Name the frequency distribution, its parameters, the severity distribution, and whether frequency and severities are independent.

Check moments and support.
Distinguish \(E[X^2]\) from \(\operatorname{Var}(X)\), identify whether severity can be zero, and list the possible aggregate lattice values when relevant.

Initialize and index carefully.
For Panjer recursion, calculate \(p_0\) under the stated model, use \(j f_jp_{k-j}\), and compute probabilities sequentially.

Interpret the threshold literally.
In a stop-loss calculation, include only aggregate outcomes strictly above \(d\), with payment \(k-d\), not \(k-\lfloor d\rfloor\) or \(k-\lceil d\rceil\).

Check omitted mass or tail value.
When summing only part of a distribution, state the truncation and estimate or bound the omitted contribution before reporting the result as complete.

These checks are useful because several errors can offset one another. For example, a missing zero atom reduces a cumulative probability, while a shifted recursion coefficient may increase a later probability. A numerically plausible total is not evidence that each probability is correctly indexed. Verify the initial value, each recurrence step, and the final event definition separately.

Key takeaway: For compound Poisson losses, the aggregate variance uses \(\lambda E[X^2]\), the zero probability must be initialized correctly, and Panjer recursion uses \(j f_jp_{k-j}\). For lattice stop-loss, preserve the exact threshold and do not treat a truncated tail sum as complete.

Check Your Understanding

For each hypothetical model, check the frequency, severity, support, and threshold before calculating.

A compound Poisson model has \(N\sim\operatorname{Poisson}(3)\), and each severity is 2 or 6 loss units with probabilities 0.25 and 0.75. Find the aggregate mean and variance. (Answer: \(E[S_N]=15\) loss units and \(\operatorname{Var}(S_N)=84\) square loss units.)
Let \(N\sim\operatorname{Poisson}(0.8)\), with positive integer severity probabilities \(f_1=f_2=0.5\). Find \(P(S_N=0)\) and \(P(S_N=1)\). (Answer: \(P(S_N=0)=e^{-0.8}=0.4493\), \(P(S_N=1)=0.8(0.5)e^{-0.8}=0.1797\).)
A claim count is zero with probability 0.4 and one with probability 0.6. If there is one claim, its severity is 2 or 4 with equal probability. Find \(E[(S_N-3)_+]\). (Answer: \(0.3\) loss units.)
Suppose \(N\sim\operatorname{Poisson}(1.5)\) and every claim is exactly one loss unit. Find \(E[(S_N-1)_+]\), giving the answer to four decimal places. (Answer: \(0.5+e^{-1.5}=0.7231\) loss units.)
For \(N\sim\operatorname{Poisson}(1)\), let \(f_1=0.25\), \(f_2=0.75\), and all other severity probabilities be zero. Use Panjer recursion to find \(P(S_N\leq2)\), including the zero atom. (Answer: \((1+0.25+0.78125)e^{-1}=0.7473\).)

What You'll Learn