Compound Poisson Risk Models and Finite-Horizon Reserves

From Claims to Reserves

An insurer collects premiums over time and pays claims when they arrive. A reserve model tracks the resulting balance: it starts with initial capital, increases at a fixed premium rate, and decreases by the amounts of claims. Over a fixed horizon, this gives a direct way to calculate the reserve’s mean, variability and distribution. These calculations describe the reserve at the horizon; by themselves, they do not establish that the insurer avoids insolvency at every earlier time.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Let $(N_t)_{t\ge0}$ be a Poisson process with rate $\lambda>0$, and let $Y_1,Y_2,\ldots$ be independent and identically distributed nonnegative claim sizes, independent of the Poisson process. Assume $m_1=\mathbb{E}[Y_1]<\infty$ and $m_2=\mathbb{E}[Y_1^2]<\infty$. Define the aggregate claims by $C_t=\sum_{k=1}^{N_t}Y_k$, where the sum is zero when $N_t=0$. For initial capital $u\ge0$ and premium rate $c\ge0$, the reserve is $R_t=u+ct-C_t$. The reserve process has state space $\mathbb{R}$, with time index $t\ge0$. We use the filtration generated by the claim arrivals and the claim sizes observed up to each time $t$.

Definition: In the classical compound Poisson risk model, claims arrive according to a Poisson process of constant rate $\lambda$, and arrival $k$ incurs an independent claim size $Y_k$. The finite-horizon reserve at time $T$ is $R_T=u+cT-C_T$. The expected claim amount per unit time is $\lambda m_1$.

The distinction between claim count and claim size is central. The random total $C_T$ varies both because the number of claims varies and because their sizes vary. Its variance therefore includes the variability caused by the random count, not just the variability of an individual claim.

Mean, Variance, and Expected Profitability

Theorem: For $T\ge0$, under the finite-second-moment assumptions above, $$ \mathbb{E}[R_T]=u+(c-\lambda m_1)T,\qquad \operatorname{Var}(R_T)=\lambda Tm_2. $$ In particular, for $T>0$, the expected reserve increases over the horizon exactly when $c>\lambda m_1$, is unchanged when $c=\lambda m_1$, and decreases when $c<\lambda m_1$.

Proof. Condition on the claim count $N_T=n$. Independence and identical distribution of the claim sizes give $\mathbb{E}[C_T\mid N_T=n]=nm_1$ and $\operatorname{Var}(C_T\mid N_T=n)=n(m_2-m_1^2)$. Since a Poisson random variable with mean $\lambda T$ has mean and variance $\lambda T$, conditioning first on $N_T$ gives $\mathbb{E}[C_T]=\mathbb{E}[N_T]m_1=\lambda Tm_1$. The conditional variance identity gives $$ \operatorname{Var}(C_T) =\mathbb{E}[\operatorname{Var}(C_T\mid N_T)] +\operatorname{Var}(\mathbb{E}[C_T\mid N_T]) =\lambda T(m_2-m_1^2)+\lambda Tm_1^2 =\lambda Tm_2. $$ Because $R_T=u+cT-C_T$, its expectation is $u+cT-\mathbb{E}[C_T]$, and subtracting a random variable does not change its variance. This proves the two formulas. For $T>0$, $\mathbb{E}[R_T]-u=(c-\lambda m_1)T$, whose sign is the sign of $c-\lambda m_1$, proving the final claim. $\square$

The inequality $c>\lambda m_1$ is called the net profit condition: expected premium income per unit time exceeds expected claim outgo per unit time. It is a condition on the mean, not a guarantee of positive reserve. Even when it holds, a sufficiently large claim can make the reserve negative. Conversely, a reserve can remain positive over a particular finite horizon even when the expected reserve is decreasing.

Worked Example: Checking the Net Profit Condition

Suppose claims arrive at rate $\lambda=2$ per year. Each claim is $1$ or $3$ units, each with probability $\tfrac12$. Take $u=10$, premium rate $c=3$ per year, and horizon $T=\tfrac12$ year. Find the expected reserve and its variance, and assess the net profit condition.

The claim moments are $$ m_1=\tfrac12(1)+\tfrac12(3)=2,\qquad m_2=\tfrac12(1^2)+\tfrac12(3^2)=\tfrac12+\tfrac92=5. $$ The premium earned is $cT=3\cdot\tfrac12=\tfrac32$, and the expected aggregate claims are $\lambda Tm_1=2\cdot\tfrac12\cdot2=2$. Thus $$ \mathbb{E}[R_T]=10+\tfrac32-2=\tfrac{19}{2},\qquad \operatorname{Var}(R_T)=2\cdot\tfrac12\cdot5=5. $$ The net profit comparison is $c=3$ versus $\lambda m_1=2\cdot2=4$, so the condition fails. Over this half-year, the expected reserve change is $(3-4)\cdot\tfrac12=-\tfrac12$, which checks that $\mathbb{E}[R_T]=10-\tfrac12=\tfrac{19}{2}$.

The Finite-Horizon Reserve Distribution

The reserve distribution can be described by conditioning on the number of claims. Write $F^{*n}$ for the distribution of a sum of $n$ independent claim sizes with the same law as $Y_1$; set $F^{*0}$ to be a point mass at zero. Since $N_T$ has a Poisson distribution with mean $\lambda T$, the reserve law is a mixture of the distributions of $u+cT-\sum_{k=1}^nY_k$, weighted by the Poisson probabilities.

Proposition: For any set $B$ for which the probability is defined, $$ \mathbb{P}(R_T\in B) =\sum_{n=0}^{\infty}e^{-\lambda T}\frac{(\lambda T)^n}{n!} \mathbb{P}\!\left(u+cT-\sum_{k=1}^nY_k\in B\right). $$ For any real $a$, $\mathbb{P}(R_T\ge a)=\mathbb{P}(C_T\le u+cT-a)$. The weak inequality matters when the aggregate claims can have atoms.

Proof. The events $\{N_T=n\}$, for integers $n\ge0$, are disjoint and have total probability one. On $\{N_T=n\}$, the reserve is $u+cT-\sum_{k=1}^nY_k$. Independence of the claim sizes and $N_T$ means that the conditional probability of this reserve belonging to $B$, given $N_T=n$, is the probability inside the sum. Multiplying by $\mathbb{P}(N_T=n)=e^{-\lambda T}(\lambda T)^n/n!$ and summing over $n$ proves the mixture formula. Finally, $R_T\ge a$ is equivalent, by $R_T=u+cT-C_T$, to $C_T\le u+cT-a$. This proves the stated reserve probability, including its endpoint convention. $\square$

Worked Example: Fixed Claim Amounts

Let claims arrive at rate $\lambda=\tfrac12$ per year, with every claim equal to $2$. Let $u=5$, $c=2$, and $T=2$. Find the distribution of the terminal reserve and the probability it is nonnegative.

Here $\lambda T=\tfrac12\cdot2=1$, the premium earned is $cT=2\cdot2=4$, and $C_T=2N_T$. Therefore $R_T=5+4-2N_T=9-2N_T$. For each integer $k\ge0$, $$ \mathbb{P}(R_T=9-2k)=\mathbb{P}(N_T=k)=\frac{e^{-1}}{k!}. $$ The reserve is nonnegative exactly when $9-2N_T\ge0$, or $N_T\le\tfrac92$. Since $N_T$ is integer-valued, this is $N_T\le4$, so $$ \mathbb{P}(R_T\ge0)=e^{-1}\left(1+1+\tfrac12+\tfrac16+\tfrac1{24}\right)=\frac{65}{24e}. $$ For a moment check, $m_1=2$ and $m_2=4$, so the formulas give $\mathbb{E}[R_T]=5+(2-1)\cdot2=7$ and $\operatorname{Var}(R_T)=\tfrac12\cdot2\cdot4=4$. From $R_T=9-2N_T$, the Poisson mean and variance give $9-2\mathbb{E}[N_T]=9-2=7$ and $4\operatorname{Var}(N_T)=4$, agreeing with both results.

Worked Example: A Reserve Probability with Random Claim Sizes

Let $\lambda=1$, $T=1$, $u=5$, and $c=2$. Each claim is $1$ or $2$, each with probability $\tfrac12$. Find $\mathbb{P}(R_T\ge4)$.

Here $\lambda T=1$ and $u+cT=5+2=7$, so $R_T\ge4$ is equivalent to $C_T\le3$. The probabilities for aggregate claims $0,1,2,3$ can be found by conditioning on the claim count. For zero claims, $\mathbb{P}(C_T=0)=e^{-1}$. For one claim, the count probability is $e^{-1}$, and the claim is $1$ with probability $\tfrac12$, giving $\mathbb{P}(C_T=1)=\tfrac12e^{-1}$. For two claims, $C_T=2$ occurs either with one claim of size $2$, with probability $e^{-1}\cdot\tfrac12$, or with two claims both of size $1$, with probability $\tfrac{e^{-1}}{2!}\cdot\tfrac14=\tfrac18e^{-1}$. Thus $\mathbb{P}(C_T=2)=\tfrac58e^{-1}$.

For $C_T=3$, two claims must have sizes $1$ and $2$, in either order, or three claims must all have size $1$. The first possibility has probability $\tfrac{e^{-1}}{2!}\cdot\tfrac12=\tfrac14e^{-1}$; the second has probability $\tfrac{e^{-1}}{3!}\cdot\tfrac18=\tfrac1{48}e^{-1}$. Hence $\mathbb{P}(C_T=3)=\tfrac{13}{48}e^{-1}$, and $$ \mathbb{P}(R_T\ge4) =\mathbb{P}(C_T\le3) =e^{-1}\left(1+\tfrac12+\tfrac58+\tfrac{13}{48}\right) =\frac{115}{48e}. $$ As a check on the strictness, the event includes $C_T=3$, because that gives $R_T=4$; replacing $C_T\le3$ by $C_T<3$ would omit that positive-probability case. Also, $m_1=\tfrac12(1)+\tfrac12(2)=\tfrac32$ and $m_2=\tfrac12(1)+\tfrac12(4)=\tfrac52$. The moment formulas give $\mathbb{E}[R_T]=7-\tfrac32=\tfrac{11}{2}$ and $\operatorname{Var}(R_T)=\tfrac52$, consistent with the model parameters.

Common Pitfall: Omitting Count Variability

A tempting but incorrect calculation is to use $\lambda T\operatorname{Var}(Y_1)$ for the variance of aggregate claims. In the first worked example, $\operatorname{Var}(Y_1)=\tfrac12(1-2)^2+\tfrac12(3-2)^2=1$, and $\lambda T=2\cdot\tfrac12=1$. That incorrect formula would give aggregate variance $1$. The correct calculation uses $m_2=5$, so $\operatorname{Var}(C_T)=\lambda Tm_2=1\cdot5=5$. The missing $4$ is the contribution of the random claim count: the conditional variance contributes $\lambda T(m_2-m_1^2)=1$, and variation in the conditional mean contributes $\lambda Tm_1^2=4$. Their sum is $5$.

A second distinction is between terminal reserve and ruin over the whole interval. The event $R_T\ge0$ concerns one time only. It does not rule out the possibility that an early claim makes the reserve negative and later premium income raises it again. The net profit condition concerns expected reserve growth, while a finite-horizon reserve probability concerns the distribution at a specified endpoint; neither statement alone is a pathwise guarantee against ruin.

Check Your Understanding

Use the model definitions and check any numerical probability against its event’s endpoint.

If $m_1$ and $m_2$ are finite, what are the mean and variance of $R_T$? Which claim-size moment appears in the variance?
For deterministic claim size $b>0$, express $R_T$ in terms of $N_T$, and write the condition for $R_T\ge0$.
Suppose $c=\lambda m_1$. What is $\mathbb{E}[R_T]$, and what does this equality say—and not say—about solvency?
For a reserve threshold $a$, why is $\mathbb{P}(R_T\ge a)$ expressed using $C_T\le u+cT-a$, rather than a strict inequality?
Explain why $\lambda T\operatorname{Var}(Y_1)$ omits part of the aggregate claim variance when the claim count is random.

Compound Poisson Risk Models and Finite-Horizon Reserves

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