Multivariate Compound Poisson Models

One Arrival Process, Several Lines

An insurance event can generate claims in more than one line at once: a single storm may cause both property and vehicle losses, for example. A model with a separate arrival process for each line cannot represent that shared timing directly. A multivariate compound Poisson model instead uses one Poisson arrival process and attaches a vector of claim amounts to each arrival. The dependence between coordinates of that vector, together with the shared count of arrivals, determines dependence between aggregate losses.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Let $(N_t)_{t\ge0}$ be a homogeneous Poisson process with rate $\lambda>0$, and let $S_k$ be its $k$th arrival time. Let $(Y_k)_{k\ge1}$ be independent and identically distributed random vectors in $\mathbb{R}^d$, independent of $N$. Write $Y_k=(Y_{k,1},\ldots,Y_{k,d})^\mathsf{T}$; coordinate $a$ is the claim at arrival $k$ for line $a$. Define the filtration $\mathcal{F}_t=\sigma(N_u,\mathbf{1}_{\{S_k\le u\}}Y_k:0\le u\le t,\ k\ge1)$, which records arrivals and their marks up to time $t$.

Definition: The multivariate compound Poisson process is the vector-valued process $$ \mathbf{S}_t=\sum_{k=1}^{N_t}Y_k,\qquad t\ge0, $$ where an empty sum is the zero vector. Its $a$th marginal aggregate is $S_t^{(a)}=\sum_{k=1}^{N_t}Y_{k,a}$. The same arrivals are used in every coordinate, while a coordinate of a jump vector is allowed to be zero.

We use the model with a common homogeneous arrival rate throughout. The jump vector can have any joint distribution; its coordinates need not be independent. The sum is always well-defined at a finite time because $N_t$ is finite almost surely. For mean and covariance formulas, assume $\mathbb{E}\|Y_1\|^2<\infty$. At $t=0$, the aggregate is zero and all its moments are zero.

Joint Distribution and Moment Formulas

Theorem (Transform and moments): Put $\theta=\lambda t$, and let $\varphi_Y(u)=\mathbb{E}[\exp(iu^\mathsf{T}Y_1)]$ for $u\in\mathbb{R}^d$. For $t\ge0$, $$ \mathbb{E}\exp(iu^\mathsf{T}\mathbf{S}_t) =\exp\{\lambda t(\varphi_Y(u)-1)\}. $$ If $\mathbb{E}\|Y_1\|^2<\infty$, then $$ \mathbb{E}\mathbf{S}_t=\lambda t\,\mathbb{E}Y_1,\qquad \operatorname{Cov}(\mathbf{S}_t)=\lambda t\,\mathbb{E}[Y_1Y_1^\mathsf{T}]. $$ In particular, for coordinates $a,b$, $$ \operatorname{Cov}(S_t^{(a)},S_t^{(b)}) =\lambda t\,\mathbb{E}[Y_{1,a}Y_{1,b}]. $$

Proof. Fix $t$ and write $N_t=n$. Conditional on this count, $\mathbf{S}_t$ is a sum of $n$ independent vectors with the distribution of $Y_1$. Thus its conditional characteristic function is $\varphi_Y(u)^n$. Since $N_t$ has the Poisson distribution with mean $\theta$, averaging over its values gives $$ \mathbb{E}\exp(iu^\mathsf{T}\mathbf{S}_t) =\sum_{n=0}^{\infty}e^{-\theta}\frac{\theta^n}{n!}\varphi_Y(u)^n =e^{-\theta}\exp(\theta\varphi_Y(u)) =\exp\{\theta(\varphi_Y(u)-1)\}. $$ This also holds at $t=0$, when $\theta=0$.

For the moment statements, write $m=\mathbb{E}Y_1$ and $\Sigma=\operatorname{Cov}(Y_1)$. Conditional on $N_t=n$, the sum has mean $nm$ and covariance $n\Sigma$: its mean is the sum of the $n$ means, and the cross-covariances between distinct independent jumps are zero. The Poisson probabilities give $\mathbb{E}N_t=\theta$ and $\operatorname{Var}(N_t)=\theta$; indeed, summing the Poisson mass function gives $\mathbb{E}[N_t(N_t-1)]=\theta^2$, so $\mathbb{E}N_t^2=\theta^2+\theta$. Consequently, $$ \mathbb{E}\mathbf{S}_t=\mathbb{E}[N_t m]=\theta m, \qquad \operatorname{Cov}(\mathbf{S}_t) =\mathbb{E}[N_t\Sigma]+\operatorname{Cov}(N_t m) =\theta\Sigma+\theta mm^\mathsf{T}. $$ The last equality uses the conditional covariance decomposition, obtained by separating variation around the conditional mean from variation of that mean. Since $\Sigma+mm^\mathsf{T}=\mathbb{E}[Y_1Y_1^\mathsf{T}]$, the stated matrix formula follows. Taking its $(a,b)$ entry proves the coordinate formula. $\square$

The cross-covariance uses the raw jump product $\mathbb{E}[Y_{1,a}Y_{1,b}]$, not just the covariance between the two coordinates of one jump. Equivalently, $$ \operatorname{Cov}(S_t^{(a)},S_t^{(b)}) =\lambda t\left(\operatorname{Cov}(Y_{1,a},Y_{1,b}) +\mathbb{E}[Y_{1,a}]\mathbb{E}[Y_{1,b}]\right). $$ The second term accounts for variation in the number of shared arrivals.

The same argument gives a cross-time formula. For $s,t\ge0$, increments of the marked process over disjoint time intervals are independent: the Poisson counts on those intervals are independent, and the independent marks assigned to each count are independent as well. If $s\le t$, write $\mathbf{S}_t=\mathbf{S}_s+(\mathbf{S}_t-\mathbf{S}_s)$. The second term is independent of $\mathbf{S}_s$, so its covariance with $S_s^{(a)}$ is zero. Applying the equal-time formula at time $s$ therefore yields:

Cross-time formula: If the jump vector has finite second moments, then $$ \operatorname{Cov}(S_s^{(a)},S_t^{(b)}) =\lambda\min(s,t)\,\mathbb{E}[Y_{1,a}Y_{1,b}]. $$

Worked Calculations

Worked Example: Joint Claim Types Can Offset Jump Dependence

Suppose $\lambda=2$, observe the process at $t=1$, and let a jump vector have probabilities $\tfrac12$ for $(1,0)^\mathsf{T}$, $\tfrac14$ for $(0,2)^\mathsf{T}$, and $\tfrac14$ for $(1,1)^\mathsf{T}$. Find both aggregate means, variances, and their covariance.

The listed probabilities sum to $\tfrac12+\tfrac14+\tfrac14=1$. The first jump-coordinate means are $\mathbb{E}Y_{1,1}=\tfrac12(1)+\tfrac14(0)+\tfrac14(1)=\tfrac34$ and $\mathbb{E}Y_{1,2}=\tfrac12(0)+\tfrac14(2)+\tfrac14(1)=\tfrac34$. The raw second moments are $\mathbb{E}Y_{1,1}^2=\tfrac12(1)+\tfrac14(0)+\tfrac14(1)=\tfrac34$, $\mathbb{E}Y_{1,2}^2=\tfrac12(0)+\tfrac14(4)+\tfrac14(1)=\tfrac54$, and $\mathbb{E}[Y_{1,1}Y_{1,2}]=\tfrac12(0)+\tfrac14(0)+\tfrac14(1)=\tfrac14$.

Here $\lambda t=2$. Thus the aggregate means are $2(\tfrac34)=\tfrac32$ for each line; the variances are $2(\tfrac34)=\tfrac32$ and $2(\tfrac54)=\tfrac52$; and the cross-covariance is $2(\tfrac14)=\tfrac12$. As a check using the decomposition, the jump covariance is $\tfrac14-(\tfrac34)(\tfrac34)=-\tfrac5{16}$, so the aggregate cross-covariance is $2(-\tfrac5{16}+\tfrac9{16})=\tfrac12$, as before. The individual jump coordinates are negatively associated in this example, but variation in the common arrival count contributes the positive term that makes the aggregate covariance positive.

Worked Example: Independent Marks Do Not Make the Aggregates Independent

Let $\lambda=2$, $t=1$, and let each jump be $(B_1,B_2)^\mathsf{T}$, where $B_1$ and $B_2$ are independent Bernoulli random variables with success probability $\tfrac12$. The four possible vectors $(0,0),(1,0),(0,1),(1,1)$ each have probability $\tfrac14$, so their probabilities sum to $1$.

For either coordinate, $\mathbb{E}B_a=\tfrac12$ and $\mathbb{E}B_a^2=\tfrac12$. Independence of the two marks gives $\mathbb{E}[B_1B_2]=(\tfrac12)(\tfrac12)=\tfrac14$. The formulas therefore give aggregate mean $2(\tfrac12)=1$ and variance $2(\tfrac12)=1$ in each line, and cross-covariance $2(\tfrac14)=\tfrac12$. The marginal aggregates are each Poisson with mean $1$: at every arrival, a given line receives either zero or one, with probability $\tfrac12$ each, and the marginal count has Poisson mean $2(\tfrac12)=1$.

Despite independent marks within each jump, the two aggregate totals are not independent: their finite covariance is $\tfrac12$, whereas independent variables with finite second moments have covariance zero. The common arrival process couples the totals. In particular, the common arrival count affects both coordinates, even though the two Bernoulli marks attached to any one arrival are independent.

Worked Example: Common Shocks and Line-Specific Claims

Consider two insurance lines with $\lambda=3$ and observation time $t=1$. A jump is a common claim $(2,3)^\mathsf{T}$, a line-1-only claim $(1,0)^\mathsf{T}$, or a line-2-only claim $(0,1)^\mathsf{T}$, each with probability $\tfrac13$. The probabilities sum to $1$. Compute the aggregate means, variances, and covariance.

The jump means are $\mathbb{E}Y_{1,1}=\tfrac13(2)+\tfrac13(1)+\tfrac13(0)=1$ and $\mathbb{E}Y_{1,2}=\tfrac13(3)+\tfrac13(0)+\tfrac13(1)=\tfrac43$. The raw second moments are $\mathbb{E}Y_{1,1}^2=\tfrac13(4)+\tfrac13(1)+\tfrac13(0)=\tfrac53$, $\mathbb{E}Y_{1,2}^2=\tfrac13(9)+\tfrac13(0)+\tfrac13(1)=\tfrac{10}3$, and $\mathbb{E}[Y_{1,1}Y_{1,2}]=\tfrac13(6)+\tfrac13(0)+\tfrac13(0)=2$.

Since $\lambda t=3$, the aggregate means are $3(1)=3$ and $3(\tfrac43)=4$; the variances are $3(\tfrac53)=5$ and $3(\tfrac{10}3)=10$; and the covariance is $3(2)=6$. The covariance check through jump-level terms gives $\operatorname{Cov}(Y_{1,1},Y_{1,2})=2-(1)(\tfrac43)=\tfrac23$; adding the count contribution gives $3(\tfrac23+\tfrac43)=6$. The common-shock category contributes directly to the raw cross-product, while the shared arrival count also contributes to dependence.

Interpreting Dependence and Avoiding a Common Error

For nonnegative claim amounts, the cross-covariance is nonnegative because every product $Y_{1,a}Y_{1,b}$ is nonnegative. It is strictly positive exactly when $\mathbb{E}[Y_{1,a}Y_{1,b}]>0$ and $\lambda t>0$. If the two lines can never both have a positive claim at the same arrival, then their raw jump product is zero and their aggregate covariance is zero. In some such models, the two aggregate processes are in fact independent: for example, if each arrival is assigned to exactly one of two lines and the assignment probabilities are fixed, Poisson splitting gives independent line-specific Poisson processes. Zero covariance alone, however, is not a general proof of independence.

Common mistake: replacing the raw jump product in the aggregate covariance formula with the jump covariance. In the independent-Bernoulli example, the jump covariance is $\operatorname{Cov}(B_1,B_2)=0$. The incorrect calculation $2\operatorname{Cov}(B_1,B_2)=0$ would therefore predict zero aggregate covariance. The correct calculation is $$ \operatorname{Cov}(S_1^{(1)},S_1^{(2)}) =2\mathbb{E}[B_1B_2] =2\left(\tfrac14\right) =\tfrac12. $$ The missing contribution in the incorrect calculation is $2\mathbb{E}[B_1]\mathbb{E}[B_2]=2(\tfrac12)(\tfrac12)=\tfrac12$, arising from the random number of shared arrivals.

For modeling, the joint jump law is therefore a substantive assumption, not just a convenient way to list marginal claim sizes. It specifies which lines can be hit together and the sizes of simultaneous claims. The common arrival count then creates an additional dependence contribution, even when the jump coordinates themselves are independent.

Check Your Understanding

Use the stated formulas, and distinguish jump-level quantities from aggregate quantities.

If $\mathbb{E}[Y_{1,1}Y_{1,2}]=0$, what is the aggregate cross-covariance at time $t$? Does that fact by itself prove that the two aggregates are independent?
For a two-line model with $\lambda=4$, $t=\tfrac12$, and $\mathbb{E}[Y_{1,1}Y_{1,2}]=\tfrac34$, calculate the cross-covariance.
Explain why independent coordinates within each jump can still give dependent aggregate processes.
In the common-shock example, identify which jump type contributes to $\mathbb{E}[Y_{1,1}Y_{1,2}]$, and recompute the aggregate covariance from that raw product.
State the cross-time covariance between lines $a$ and $b$ at times $s$ and $t$, assuming finite second moments.

Multivariate Compound Poisson Models

What You'll Learn