From Claim Counts to Aggregate Loss
In a compound Poisson model, the total loss over a fixed time interval combines two sources of randomness: the number of claims and the severity of each claim. For discrete severities, conditioning on the number of claims gives a direct formula for every aggregate-loss probability. A recursion derived from that formula can compute probabilities efficiently, one loss value at a time, without repeatedly forming all the convolutions. Truncation bounds then quantify what is omitted when only finitely many claim counts or loss values are computed.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((N_t)_{t\ge0}\) be a Poisson process with rate \(\lambda\ge0\). Let \(Y_1,Y_2,\ldots\) be independent and identically distributed nonnegative integer-valued claim severities, independent of the Poisson process. Write \(g_j=\mathbb{P}(Y_1=j)\), so \(g_j\ge0\) and \(\sum_{j=0}^{\infty}g_j=1\). For a fixed horizon \(t\ge0\), set \(\theta=\lambda t\) and define the aggregate loss \(S_t=\sum_{r=1}^{N_t}Y_r\), with an empty sum equal to zero. Write \(f_k=\mathbb{P}(S_t=k)\) for integers \(k\ge0\).
Since \(N_t\) has a Poisson distribution with mean \(\theta\), conditioning on its value yields a mixture of severity-sum distributions. The case \(n=0\) contributes all its mass at zero; when \(\theta=0\), the count is zero almost surely and the aggregate loss is zero.
This is the conditioning formula. It is conceptually direct, but computing all the convolutions separately can be cumbersome. The following result gives a sequential calculation.
Proof. Given \(N_t=n\), independence of the severities gives \(\mathbb{E}[z^{S_t}\mid N_t=n]=G(z)^n\). Averaging over the Poisson count therefore gives \(F(z)=e^{-\theta}\sum_{n\ge0}\theta^nG(z)^n/n!=\exp(\theta(G(z)-1))\). The constant coefficient of this power series is \(\exp(\theta(g_0-1))\), proving the formula for \(f_0\). Differentiating for \(|z|<1\) gives \(F'(z)=\theta G'(z)F(z)\). The coefficient of \(z^{k-1}\) in \(F'(z)\) is \(k f_k\). In the product \(G'(z)F(z)\), that coefficient is \(\sum_{j=1}^{k}j g_j f_{k-j}\), because the coefficient of \(z^{j-1}\) in \(G'(z)\) is \(j g_j\). Equating coefficients proves the recursion for every \(k\ge1\). The derivation also covers \(\theta=0\): the displayed exponential gives \(f_0=1\), and the recursion gives \(k f_k=0\), hence \(f_k=0\) for \(k\ge1\). \(\square\)
The recursion only uses severity probabilities up to the loss value being computed. In particular, \(g_0\) affects \(f_0\), but not the sum in the recursion, which starts at \(j=1\). A zero-severity claim adds nothing to the aggregate loss; the formula accounts for such claims without treating them as positive loss jumps.
Worked Example: Two Possible Positive Severities
Let \(\theta=1\), and suppose a claim has severity \(1\) or \(2\), each with probability \(\tfrac12\). Thus \(g_1=\tfrac12\), \(g_2=\tfrac12\), and all other \(g_j=0\). Compute \(f_0,f_1,f_2,f_3\).
The severity probabilities sum to \(\tfrac12+\tfrac12=1\). Here \(g_0=0\), so \(f_0=\exp(1(0-1))=e^{-1}\). For \(k=1\), the recursion gives \(f_1=1\cdot g_1 f_0=\tfrac12e^{-1}\). For \(k=2\), it gives \(2f_2=g_1f_1+2g_2f_0=\tfrac12\cdot\tfrac12e^{-1}+2\cdot\tfrac12e^{-1}=\tfrac14e^{-1}+e^{-1}=\tfrac54e^{-1}\), hence \(f_2=\tfrac58e^{-1}\). For \(k=3\), \(3f_3=g_1f_2+2g_2f_1=\tfrac12\cdot\tfrac58e^{-1}+1\cdot\tfrac12e^{-1}=\tfrac5{16}e^{-1}+\tfrac8{16}e^{-1}=\tfrac{13}{16}e^{-1}\), hence \(f_3=\tfrac{13}{48}e^{-1}\).
Check by conditioning on the count. For loss zero, only \(N_t=0\) contributes, giving \(e^{-1}\). For loss one, there must be one claim of severity one, giving \(e^{-1}\cdot\tfrac12\). For loss two, either one claim has severity two, contributing \(e^{-1}\cdot\tfrac12\cdot\tfrac12\), or two claims both have severity one, contributing \(e^{-1}\cdot\tfrac1{2!}\cdot\tfrac14\). Their sum is \((\tfrac14+\tfrac18)e^{-1}=\tfrac38e^{-1}\)—this is not the earlier \(f_2\), because the first contribution was misweighted if written this way: one claim of severity two has probability \(e^{-1}\cdot 1\cdot\tfrac12=\tfrac12e^{-1}\), while the two-claim contribution is \(\tfrac18e^{-1}\). Thus the correct total is \(\tfrac58e^{-1}\), as obtained by recursion. For loss three, two claims must have severities one and two in either order, contributing \(e^{-1}\cdot\tfrac1{2!}\cdot(\tfrac14+\tfrac14)=\tfrac14e^{-1}\), and three claims must all have severity one, contributing \(e^{-1}\cdot\tfrac1{3!}\cdot\tfrac18=\tfrac1{48}e^{-1}\). The total is \((\tfrac14+\tfrac1{48})e^{-1}=\tfrac{13}{48}e^{-1}\), confirming \(f_3\).
The probability of a loss of at least \(3\) is \(1-f_0-f_1-f_2=1-\tfrac{17}{8e}\). Also, \(f_0+f_1+f_2+f_3=(1+\tfrac12+\tfrac58+\tfrac{13}{48})/e=\tfrac{115}{48e}\). This is less than \(1\); the remaining probability is assigned to losses above \(3\).
Conditioning and Recursion as Complementary Methods
The conditioning formula explains the distribution as a mixture: first select a count \(n\), then sum \(n\) independent severities. The recursion is usually preferable when probabilities are needed successively up to some threshold \(K\), since each \(f_k\) uses only earlier aggregate probabilities. Both methods require the same assumptions: the count is Poisson with the stated mean, the severities are identically distributed and independent of the count, and their values lie in the nonnegative integers. If the claim severities have a continuous distribution, the point-probability recursion does not apply in this form.
Worked Example: A Tail Probability from Successive Recursion
Continue with \(\theta=1\) and severities \(1\) and \(2\), each with probability \(\tfrac12\). Find the probability that the aggregate loss is at least \(4\), using probabilities through \(f_4\).
The already computed values are \(f_0=e^{-1}\), \(f_1=\tfrac12e^{-1}\), \(f_2=\tfrac58e^{-1}\), and \(f_3=\tfrac{13}{48}e^{-1}\). At \(k=4\), \(4f_4=g_1f_3+2g_2f_2=\tfrac12\cdot\tfrac{13}{48}e^{-1}+1\cdot\tfrac58e^{-1}=\tfrac{13}{96}e^{-1}+\tfrac{60}{96}e^{-1}=\tfrac{73}{96}e^{-1}\). Thus \(f_4=\tfrac{73}{384}e^{-1}\).
Because \(S_t\) is integer-valued, \(S_t\ge4\) is the complement of \(S_t\in\{0,1,2,3\}\). The sum of the four probabilities is \((1+\tfrac12+\tfrac58+\tfrac{13}{48})e^{-1}=\tfrac{115}{48}e^{-1}\). Therefore, \(\mathbb{P}(S_t\ge4)=1-\tfrac{115}{48e}\). As a further check, the probability of a loss strictly greater than \(4\) is \(1-\sum_{k=0}^4f_k=1-\tfrac{993}{384e}=1-\tfrac{331}{128e}\), since \(\tfrac{115}{48}+\tfrac{73}{384}=\tfrac{920+73}{384}=\tfrac{993}{384}\). The distinction between “at least \(4\)” and “strictly greater than \(4\)” matters: the first includes \(f_4\), while the second does not.
Truncation Bounds
Two different truncations occur in practical calculations. One may stop the conditioning sum after a maximum claim count, or compute loss probabilities only up to a maximum loss. The first omits scenarios with many claims; the second omits the aggregate-loss tail. The bounds below do not require a special severity distribution, though a finite severity mean is needed for the loss-tail bound.
Proof. The terms in the conditioning formula are nonnegative. Removing all terms with \(n>M\) therefore leaves a value no larger than \(f_k\), and their total mass over any one value \(k\) is no larger than their total mass over all aggregate losses. That total mass is \(\mathbb{P}(N_t>M)\), proving the first inequality. For the second, the integer-valued variable \(\binom{N_t}{M+1}\) is at least \(1\) whenever \(N_t>M\), so \(\mathbb{P}(N_t>M)\le\mathbb{E}[\binom{N_t}{M+1}]\). For a Poisson variable, direct summation gives \[ \mathbb{E}\!\left[\binom{N_t}{M+1}\right] =e^{-\theta}\sum_{n=M+1}^{\infty}\binom{n}{M+1}\frac{\theta^n}{n!} =\frac{\theta^{M+1}}{(M+1)!}e^{-\theta}\sum_{r=0}^{\infty}\frac{\theta^r}{r!} =\frac{\theta^{M+1}}{(M+1)!}. \] For the loss bound, \(S_t\) is nonnegative and integer-valued, so \(S_t>K\) implies \(S_t\ge K+1\), and \(\mathbf{1}_{\{S_t>K\}}\le S_t/(K+1)\). Taking expectations gives \(\mathbb{P}(S_t>K)\le\mathbb{E}[S_t]/(K+1)\). Conditioning on \(N_t=n\) gives \(\mathbb{E}[S_t\mid N_t=n]=n\mathbb{E}[Y_1]\); averaging and using \(\mathbb{E}[N_t]=\theta\) yields \(\mathbb{E}[S_t]=\theta\mathbb{E}[Y_1]\). Substitution proves the claimed bound. \(\square\)
The count bound controls the error in any one computed probability, and also controls the total mass omitted by truncating the count mixture. The loss-tail bound instead controls the chance of exceeding a chosen loss threshold. These are not interchangeable: a small count-truncation error does not by itself say which loss values carry the omitted mass.
Worked Example: Bounding Both Sources of Omission
Again take \(\theta=1\) and severities \(1\) and \(2\) with equal probabilities. Truncate the count mixture at \(M=2\), and bound the probability of a loss strictly greater than \(3\).
For the count truncation, the omitted count event is \(N_t>2\). The bound is \(\mathbb{P}(N_t>2)\le 1^{3}/3!=\tfrac16\). Consequently, for each fixed \(k\), the omitted contribution \(f_k-f_k^{(2)}\) is between \(0\) and \(\tfrac16\). This is a rigorous bound even though it does not calculate the omitted terms separately.
For the loss threshold, \(\mathbb{E}[Y_1]=1\cdot\tfrac12+2\cdot\tfrac12=\tfrac32\). Hence \(\mathbb{E}[S_t]=1\cdot\tfrac32=\tfrac32\), and the integer loss-tail bound gives \(\mathbb{P}(S_t>3)\le\tfrac{3/2}{4}=\tfrac38\). From the preceding example, the exact value is \(\mathbb{P}(S_t>3)=1-\tfrac{115}{48e}\), since \(S_t>3\) is the complement of losses \(0,1,2,3\). The inequality just derived supplies a conservative bound; the exact expression supplies the actual tail probability.
Zero Severities and a Common Pitfall
When \(g_0>0\), the aggregate loss can be zero even if claims occurred. In fact, the recursion gives \(f_0=\exp(\theta(g_0-1))\): zero aggregate loss occurs when every claim has severity zero, including the possibility of no claims. This differs from the case \(g_0=0\), where aggregate loss zero requires \(N_t=0\) and \(f_0=e^{-\theta}\).
A common computational mistake is to combine the coefficients in an aggregate probability sum incorrectly. In the first example, \(1+\tfrac12+\tfrac58+\tfrac{13}{48}\) must be computed with denominator \(48\): \((48+24+30+13)/48=115/48\), not \(121/48\). Thus the mass at losses \(0\) through \(3\) is \(115/(48e)\). The correction matters because the complementary tail probability is obtained by subtracting that mass from \(1\). A second frequent error is to use \(\mathbb{P}(S_t\ge K)\) where the requested event is \(\mathbb{P}(S_t>K)\); for an integer-valued loss, those events differ by the point probability \(f_K\).
For computational work, a reliable order is to verify that \(\sum_jg_j=1\), calculate \(f_0\), apply the recursion in increasing order of \(k\), and check any requested tail by subtracting the probabilities with the correct endpoint convention. When a finite range is reported, it is useful to state a truncation bound as well as the computed mass.
Check Your Understanding
Use the definitions and formulas above. In each tail calculation, distinguish “at least” from “strictly greater than.”
- If \(g_0=0\) and \(\theta>0\), what event gives aggregate loss zero, and what is \(f_0\)?
- For \(\theta=2\), \(g_1=\tfrac14\), \(g_2=\tfrac34\), and all other \(g_j=0\), use the recursion to calculate \(f_0\), \(f_1\), and \(f_2\).
- For a count truncation at \(M=3\), write the general Poisson bound on the omitted count probability.
- If the severities are nonnegative integers with finite mean, derive a bound for \(\mathbb{P}(S_t>5)\) in terms of \(\theta\) and \(\mathbb{E}[Y_1]\).
- Explain why the probabilities \(f_0,\ldots,f_K\) must be subtracted from \(1\) to obtain \(\mathbb{P}(S_t>K)\), whereas \(f_0,\ldots,f_{K-1}\) are subtracted to obtain \(\mathbb{P}(S_t\ge K)\).