Compound Poisson Process: Moments and MGF

What is a Compound Poisson Process?

A compound Poisson process models a total amount, $S(t)$, by summing up a random number of random amounts over a period of time. It's a "process of processes." One stochastic process governs the number of events, and another governs the size of each event.

Let's begin with a video that introduces this concept using a familiar example.

As the video explains, we combine two key components:

1. Claim Frequency: The number of claims up to time $t$, denoted $N(t)$, follows a homogeneous Poisson process with rate $\lambda$.
2. Claim Severity: The amount of each individual claim, $X_i$, is a random variable. We assume that all claim severities $X_1,X_2,\dots$ are independent and identically distributed (i.i.d.) and are also independent of the claim frequency process $N(t)$.

The total aggregate claim amount by time $t$, denoted $S(t)$, is the sum of all individual claims that have occurred by that time. If $N(t)=0$, the sum is empty and $S(t)=0$.

For a formal definition, let's consult a standard actuarial study note.

Mathematically, the compound Poisson process is defined as:

Our objective is to find the statistical properties of this new random variable, $S(t)$.

Mean of a Compound Poisson Process

The first and most intuitive property is the expected value, or mean, of the total claims. We can derive this using the law of total expectation, also known as the iterated-expectation method. The core idea is to first find the expectation of $S(t)$ by conditioning on the number of claims, $N(t)=n$, and then find the expectation of that result over all possible values of $n$.

The formula is remarkably simple and intuitive.

As the note states, the mean of the aggregate claims is:

Since $N(t)$ is a Poisson process with rate $\lambda$, we know $E[N(t)]=\lambda t$. Therefore:

This makes perfect sense: the expected total loss is the expected number of claims multiplied by the average loss per claim.

For those interested in the step-by-step derivation, the following video segment walks through the application of the law of total expectation. Given your background, you should find the logic quite familiar.

Variance of a Compound Poisson Process

Calculating the variance is less intuitive and requires a bit more algebraic work. The derivation relies on the law of total variance:

Applying this to $S(t)$ and conditioning on $N(t)$ gives us the general formula for the variance of a compound process.

The key formulas from this derivation are:

1. General formula: $\text{Var}[S(t)]=E[N(t)]\text{Var}[X]+\text{Var}[N(t)](E[X]{)}^2$
2. Simplified for Poisson: Since for a Poisson process $E[N(t)]=\text{Var}[N(t)]=\lambda t$, we can substitute this in: $$\text{Var}[S(t)]=\lambda t\text{Var}[X]+\lambda t(E[X]{)}^2$$ Factoring out $\lambda t$ and recalling that $\text{Var}[X]+(E[X]{)}^2=E[X^2]$, we arrive at a very tidy and important result: $$\text{Var}[S(t)]=\lambda tE[X^2]$$

This elegant formula, known as Wald's identity for variance, is extremely useful in practice because it only requires the first and second moments of the claim severity distribution, not the full distribution itself.

The full derivation of the variance involves more algebra. The following video segment covers it in detail, which may be a useful review of applying the law of total variance.

Moment Generating Function (MGF) of a Compound Poisson Process

The MGF is a powerful tool for characterizing a distribution, and for a compound Poisson process, it also has a remarkably elegant form. The derivation again uses conditional expectation, this time applied to the definition of the MGF, $M_{S(t)}(u)=E[e^{uS(t)}]$.

The MGF of $S(t)$ is given by:

where $M_X(u)$ is the MGF of the individual claim severity distribution $X$.

Let's walk through the derivation, as it's a classic application of techniques you're familiar with.

1. Start with the definition and condition on the number of claims, $N(t)$: $$M_{S(t)}(u)=E[e^{uS(t)}]=E\left[E\left[e^{u{\sum }_{i=1}^{N(t)}{X}_i}|N(t)\right]\right]$$
2. Given $N(t)=n$, the sum becomes ${\sum }_{i=1}^n{X}_i$. Since the $X_i$ are i.i.d., the inner expectation is: $$E\left[e^{u{\sum }_{i=1}^n{X}_i}\right]=E\left[\prod _{i=1}^n{e}^{u{X}_i}\right]=\prod _{i=1}^{n}E[e^{u{X}_i}]=(M_X(u){)}^n$$
3. Substitute this back into the outer expectation, which is taken over the distribution of $N(t)$: $$M_{S(t)}(u)=E[(M_X(u){)}^{N(t)}]$$
4. This expression has the form $E[c^{N(t)}]$ with $c=M_X(u)$. This is the definition of the MGF of $N(t)$ evaluated at $v=\ln (c)$. Let $M_{N(t)}(v)$ be the MGF of the Poisson($\lambda t$) distribution. Then: $$M_{S(t)}(u)=M_{N(t)}(\ln (M_X(u)))$$
5. The MGF of a Poisson($\lambda t$) random variable is $M_{N(t)}(v)=e^{\lambda t(e^v-1)}$. Substituting $v=\ln (M_X(u))$ gives: $$M_{S(t)}(u)=\exp (\lambda t(e^{\ln (M_X(u))}-1))=\exp (\lambda t(M_X(u)-1))$$

This composition of MGFs is a key theoretical result for compound distributions.

Applying the Formulas: An Example

Let's solidify these concepts with a typical exam-style problem.

Problem: An insurance company experiences claims arriving at a Poisson rate of $\lambda =20$ per day. The amount of each claim is random, with the following discrete distribution:

• $500 with probability 0.6
• $2,000 with probability 0.3
• $10,000 with probability 0.1

Let $S(t)$ be the total claims amount over $t$ days. Calculate the mean and standard deviation of the aggregate claims over a 10-day period.

Solution:
First, we analyze the claim severity distribution $X$.

1. Expected Claim Severity $E[X]$: $$E[X]=(500)(0.6)+(2000)(0.3)+(10000)(0.1)=300+600+1000=1900$$
2. Expected Squared Claim Severity $E[X^2]$: $$E[X^2]=(500^2)(0.6)+(2000^2)(0.3)+(10000^2)(0.1)$$ $$=(250,000)(0.6)+(4,000,000)(0.3)+(100,000,000)(0.1)$$ $$=150,000+1,200,000+10,000,000=11,350,000$$

Now, we analyze the aggregate claims $S(10)$ over $t=10$ days. The Poisson rate is $\lambda =20$.
3. Expected Aggregate Claims $E[S(10)]$:
$E[S(10)]=\lambda tE[X]=(20)(10)(1900)=200\times 1900=380,000$
4. Variance of Aggregate Claims $\text{Var}[S(10)]$:
$\text{Var}[S(10)]=\lambda tE[X^2]=(20)(10)(11,350,000)=200\times 11,350,000=2,270,000,000$
5. Standard Deviation:
$\text{StdDev}[S(10)]=\sqrt{\text{Var}[S(10)]}=\sqrt{2,270,000,000}\approx 47,645$

So, over a 10-day period, the insurer expects to pay $380,000 in claims, with a standard deviation of approximately $47,645.

Conclusion

In this lesson, we have extended the Poisson process to model not just claim frequency but also the total claim amount. By combining a Poisson process for the number of claims with an i.i.d. distribution for their severity, we formed the compound Poisson process.

Key Takeaways:

• A Compound Poisson Process is defined as $S(t)={\sum }_{i=1}^{N(t)}{X}_i$.
• The mean of the aggregate claims is $E[S(t)]=\lambda tE[X]$. It is the expected number of claims times the expected severity of each claim.
• The variance of the aggregate claims is $\text{Var}[S(t)]=\lambda tE[X^2]$. This depends on the second moment of the severity distribution.
• The Moment Generating Function is $M_{S(t)}(u)=\exp (\lambda t(M_X(u)-1))$.

These three formulas are fundamental tools in the actuary's toolkit for risk modeling. Having mastered the moments of the aggregate claims distribution, our next step will be to work with the full probability distribution of $S(t)$. In the next lesson, we will learn how to calculate probabilities for the total claim amount (e.g., $P(S(t)>k)$) and find conditional expectations, which are crucial for pricing more complex insurance contracts like those with deductibles or limits.