Deriving the Compound Poisson Adjustment Coefficient

Why the Adjustment Coefficient Matters

A surplus model combines premium income with a random stream of claim payments. The adjustment coefficient summarizes how the premium rate and the claim-size distribution interact in that model. It is defined through an equation involving the moment-generating function of a single claim, so calculating it requires careful attention to the claim frequency, premium units, and the positive solution of the equation.

This tutorial develops that equation for a compound Poisson surplus process and solves it for several claim distributions. The coefficient has units inverse to the units used for claim amounts: if claims are measured in dollars, the coefficient is per dollar. The coefficient itself is not a ruin probability.

Definition: Let $N(t)$ be a Poisson process with rate $\lambda$ claims per unit time, and let $X_1,X_2,\ldots$ be independent, identically distributed, nonnegative claim amounts, independent of $N(t)$. Write $C(t)=\sum_{j=1}^{N(t)}X_j$ for aggregate claims by time $t$. With initial surplus $u$ and premium income at constant rate $c$ per unit time, the surplus is $U(t)=u+ct-C(t)$. The adjustment coefficient $R$ is a positive solution, when one exists, of $\lambda(M_X(R)-1)=cR$, where $M_X(r)=E[e^{rX}]$ is the moment-generating function of one claim.

Deriving the Adjustment-Coefficient Equation

The equation follows from the moment-generating function of a compound Poisson sum. First condition on the number of claims. Given $N(t)=n$, the aggregate claims are the sum of $n$ independent claim amounts, so $E[e^{rC(t)}\mid N(t)=n]=M_X(r)^n$. Averaging over the Poisson claim count gives:

E[e^{rC(t)}] =\sum_{n=0}^{\infty} e^{-\lambda t}\frac{(\lambda t)^n}{n!}M_X(r)^n =\exp\{\lambda t(M_X(r)-1)\}.

Since $U(t)=u+ct-C(t)$, this identity gives the expectation of $e^{-rU(t)}$. The adjustment-coefficient equation makes that expectation constant in time. More strongly, independent increments make the resulting process a martingale when the equation holds.

Write the surplus.
$e^{-rU(t)}=e^{-ru}e^{-rct}e^{rC(t)}$.

Take expectations.
Using the compound Poisson moment-generating function gives $E[e^{-rU(t)}]=e^{-ru}\exp\{t[\lambda(M_X(r)-1)-cr]\}$.

Set the time exponent to zero.
For the expectation to be constant in $t$, require $\lambda(M_X(r)-1)-cr=0$. The positive solution is the adjustment coefficient $R$.

\boxed{\lambda(M_X(R)-1)=cR}

This derivation also shows why the premium rate is multiplied by $R$, while the claim term involves the moment-generating function of an individual claim. The Poisson frequency is already accounted for by the factor $\lambda$. The equation uses $c$, the premium income per unit time, not the premium charged per policy or the total premium collected over an arbitrary period.

Positive Loading and the Positive Root

A positive safety loading means premium income exceeds expected aggregate claims per unit time. In this model, expected aggregate claims per unit time are $\lambda E[X]$, so the net profit condition is $c>\lambda E[X]$. This condition is important, but it does not by itself guarantee that the moment-generating function exists far enough to yield a positive root.

Result: Define $\psi(r)=\lambda(M_X(r)-1)-cr$ wherever $M_X(r)$ is finite. If $E[X]$ is finite and $c>\lambda E[X]$, then $\psi(0)=0$ and $\psi'(0)=\lambda E[X]-c<0$. Where the relevant moments exist, $\psi''(r)=\lambda E[X^2e^{rX}]\geq 0$, with strict inequality for nonzero claims. Thus $\psi$ initially decreases, and it is strictly convex for a nonzero claim distribution. If it later rises through zero within the domain of $M_X$, there is exactly one positive root. That root is $R$. A positive root should not be asserted if the moment-generating function’s domain does not permit a crossing.

The equation always has the root $r=0$, because $M_X(0)=1$. That root is not the adjustment coefficient: the term “adjustment coefficient” refers to a positive solution. In practice, first check the net profit condition, then use the claim distribution to form $M_X(r)$, and finally solve for the positive root within the moment-generating function’s domain.

Worked Calculations

Worked Example: Exponential Claims

A hypothetical insurer has Poisson claim frequency $\lambda=4$ per year. Individual claim amounts $X$ are exponential with mean $\mu=\$500$, so $M_X(r)=(1-500r)^{-1}$ for $r<1/500$. Premium income is $c=\$2{,}500$ per year. Find the adjustment coefficient. Use dollars as the claim-size unit.

Method and loading check. The expected claims per year are $\lambda E[X]=4(\$500)=\$2{,}000$. Since $c=\$2{,}500>\$2{,}000$, the net profit condition holds; the safety loading relative to expected claims is $(2500-2000)/2000=25\%$.

Substitute the exponential moment-generating function into the coefficient equation:

4\left(\frac{1}{1-500R}-1\right)=2500R.

For a positive root, simplify the left side and divide both sides by $R$:

\frac{2000R}{1-500R}=2500R \quad\Longrightarrow\quad \frac{2000}{1-500R}=2500.

Therefore $1-500R=2000/2500=0.8$, so $500R=0.2$ and $R=0.0004$ per dollar. The result satisfies the exponential moment-generating function’s domain, $R<1/500=0.002$. Substitution gives $4(1/0.8-1)=1$ on the left and $2500(0.0004)=1$ on the right, checking the root. The common mistake avoided here is cancelling $R$ without first distinguishing the positive root from the root $R=0$.

Worked Example: Erlang Claims and the Moment-Generating-Function Domain

A hypothetical portfolio has Poisson frequency $\lambda=2$ per year. Each claim has an Erlang distribution with shape 2 and scale $\beta=\$500$, so $E[X]=\$1{,}000$ and $M_X(r)=(1-500r)^{-2}$ for $r<1/500$. The premium income rate is $c=\$2{,}500$ per year. Find $R$, reporting the answer to four significant figures.

Method and loading check. Expected claims per year are $2(\$1{,}000)=\$2{,}000$, so $c>\lambda E[X]$. Set $z=500R$, which must satisfy $0<z<1$ because of the moment-generating-function domain. The equation becomes:

2\left((1-z)^{-2}-1\right)=2500\frac{z}{500}=5z.

Since $(1-z)^{-2}-1=(2z-z^2)/(1-z)^2$, divide by $z>0$ and multiply by $(1-z)^2$:

\frac{2(2-z)}{(1-z)^2}=5 \quad\Longrightarrow\quad 2(2-z)=5(1-2z+z^2).

Rearranging gives $5z^2-8z+1=0$, with roots $z=(4\pm\sqrt{11})/5$. Since $\sqrt{11}\approx3.31662479$, the smaller root is $z\approx(4-3.31662479)/5=0.13667504$. The larger root is greater than 1 and lies outside the moment-generating-function domain, so it is invalid. Thus $R=z/500\approx0.0002733501$ per dollar, or $0.0002734$ per dollar to four significant figures. Substitution into the quadratic gives $5z^2-8z+1\approx0$, checking the selected root. The method avoids accepting an algebraic root that makes the moment-generating function undefined.

Worked Example: Fixed Claim Amount and Newton’s Method

A hypothetical insurer has a Poisson claim frequency of one claim per year. Every claim is exactly $\$1{,}000$, and premium income is $\$1{,}500$ per year. Find the adjustment coefficient to four significant figures using Newton’s method.

Method and loading check. Expected claims per year are $1(\$1{,}000)=\$1{,}000$, so the net profit condition holds. For a fixed claim amount, $M_X(R)=e^{1000R}$. Set $y=1000R$, a dimensionless quantity. Then the coefficient equation is $e^y-1=1.5y$, or $f(y)=e^y-1-1.5y=0$. The positive root is not available by isolating $y$ with ordinary algebra.

Use Newton’s iteration $y_{n+1}=y_n-f(y_n)/f'(y_n)$, where $f'(y)=e^y-1.5$. Starting at $y_0=0.750000$, calculate:

f(y_0)=e^{0.75}-1-1.5(0.75) =2.117000-1-1.125000=-0.008000, $$ $$ f'(y_0)=2.117000-1.5=0.617000, \qquad y_1=0.750000-\frac{-0.008000}{0.617000}\approx0.762966.

At $y_1$, $e^{y_1}\approx2.144628$, so $f(y_1)\approx2.144628-1-1.144449=0.000179$, and $f'(y_1)\approx0.644628$. The next step is $y_2\approx0.762966-0.000179/0.644628=0.762688$. Thus $R=y_2/1000\approx0.0007627$ per dollar to four significant figures. Checking, $e^{0.762688}-1\approx1.1440$, while $1.5(0.762688)\approx1.1440$; the two sides agree at the shown precision. The common mistake avoided is reporting the dimensionless root $y$ as though it were $R$: the conversion $R=y/1000$ is required because claims were $\$1{,}000$.

Common Exam Traps

Stopping at zero. Substituting $R=0$ satisfies the equation, but it does not give the positive adjustment coefficient. Find the positive root or establish that one does not exist under the stated assumptions.
Using the wrong claim moment-generating function. $M_X(r)$ is for one claim. The compound Poisson frequency enters separately as $\lambda$; do not put the aggregate annual claim distribution in place of $X$.
Mixing time units. If $\lambda$ is claims per year, then $c$ must be dollars per year. Converting one rate to a different time unit requires converting the other as well.
Ignoring the domain. A candidate root is usable only if $M_X(R)$ is finite. For exponential claims with mean $\mu$, this requires $R<1/\mu$.
Confusing positive loading with a guaranteed root. Positive loading gives a negative initial slope of $\psi$ after zero. The moment-generating function must still permit $\psi$ to cross zero at a positive argument.

Key takeaway: For a compound Poisson surplus process, derive the equation from the claim-count moment-generating function, check $c>\lambda E[X]$, and solve $\lambda(M_X(R)-1)=cR$ for a positive root inside the domain of $M_X$. Keep the time units consistent and verify the result in the original equation.

Check Your Understanding

For each problem, use claim amounts in dollars and report adjustment coefficients to four significant figures where a positive root exists.

A hypothetical insurer has exponential claims with mean $\$50$, frequency $\lambda=2$ per year, and premium income $c=\$120$ per year. Find $R$. (Answer: $0.003333$ per dollar.)
Claims are exponential with mean $\$25$, frequency $\lambda=4$ per year, and premium income $c=\$120$ per year. Check the net profit condition and find $R$. (Answer: Net profit condition holds; $R=0.006667$ per dollar.)
Claims have an Erlang distribution with shape 2 and scale $\$2$, frequency $\lambda=1$ per year, and premium income $c=\$6$ per year. Find the positive adjustment coefficient. (Answer: $0.1162$ per dollar.)
Every claim is $\$2{,}000$, the Poisson frequency is $0.5$ per year, and premium income is $c=2500(e^{0.4}-1)$ dollars per year. Find $R$. (Answer: $0.0002000$ per dollar.)
Claims have finite mean $\$40$, frequency $\lambda=3$ per year, and premium income $c=\$110$ per year. Can a positive adjustment coefficient satisfying the net profit condition be asserted? (Answer: No; $c=110<3(40)=120$, so the net profit condition fails.)

What You'll Learn