Sensitivity of the Adjustment Coefficient

Why Sensitivity Matters

The adjustment coefficient $R$ is the positive root used in the classical ruin model to describe an exponential bound on ultimate ruin probability. Its value depends on the claim arrival rate, the premium income rate, and the claim severity distribution. Sensitivity analysis asks how that root changes when one of those inputs changes, with the treatment of the other inputs stated explicitly.

A crucial distinction is whether premium income is held fixed or is recalculated when frequency or severity changes. For example, increasing claim frequency while leaving the premium rate unchanged is not the same experiment as increasing frequency while preserving the same safety loading. The derivative can have a different value—and even a different interpretation—depending on that choice.

Model and definition: In the classical compound Poisson surplus model, claims arrive according to a Poisson process with rate $\lambda>0$ per unit time, claim amounts are independent and identically distributed nonnegative random variables $X$, and premium income accumulates at constant rate $c$. Write $\mu=E[X]$ and $M_X(r)=E[e^{rX}]$, where the moment generating function is finite in a neighborhood of the positive root. The adjustment coefficient $R>0$ satisfies $$ \lambda\bigl(M_X(R)-1\bigr)=cR. $$ The usual net profit condition is $c>\lambda\mu$. A positive root is assumed to exist; the sensitivity formulas below apply locally when the root is finite and the equation is differentiable at it.

Implicit Differentiation

Define $H(R;\lambda,c,\eta)=\lambda(M_X(R;\eta)-1)-cR$, where $\eta$ denotes any severity-distribution parameter. At an adjustment coefficient, $H(R;\lambda,c,\eta)=0$. If $H_R\ne0$, implicit differentiation gives

\frac{\partial R}{\partial q}=-\frac{H_q}{H_R}, \qquad H_R=\lambda M_X'(R;\eta)-c,

for a parameter $q$, with the other model inputs held fixed according to the chosen parameterization. This is a local derivative: for a small change $\Delta q$, the corresponding first-order approximation is $\Delta R\approx(\partial R/\partial q)\Delta q$. It does not by itself give the exact root after a large change.

Why the denominator is positive: Under the usual nondegenerate claim model, $H(0)=0$, $H'(0)=\lambda\mu-c<0$, and $H$ is convex because $H''(r)=\lambda E[X^2e^{rX}]>0$ wherever this expectation exists. At a finite positive root, the curve has risen back to zero after initially falling, so its slope there is positive: $H_R=\lambda M_X'(R)-c>0$. Thus, the signs of the sensitivities are determined by the corresponding numerator derivatives, with the minus sign in the implicit-differentiation formula.

Fixed $c$ sensitivities: Holding the premium rate and severity distribution fixed, $$ \frac{\partial R}{\partial \lambda} =-\frac{M_X(R)-1}{\lambda M_X'(R)-c}<0, \qquad \frac{\partial R}{\partial c} =\frac{R}{\lambda M_X'(R)-c}>0. $$ The first derivative applies only while the model remains in the regime where a positive root exists, including the net profit condition.

The frequency derivative above is specifically for fixed $c$. If instead premiums are set by a fixed safety loading $\theta>0$, then $c=(1+\theta)\lambda\mu$, and the root equation becomes

M_X(R)-1-(1+\theta)\mu R=0.

Here $\lambda$ cancels. With severity and loading held fixed, the adjustment coefficient does not change with frequency, so $\partial R/\partial\lambda=0$ in this parameterization. This does not mean frequency has no effect on every ruin measure; it means only that this root is unchanged under this particular premium rule.

Worked Example: Frequency and Premium Rate Under Different Assumptions

A hypothetical insurer has Poisson claim frequency $\lambda=2$ per year and exponential claim amounts with mean $\mu=1$ thousand dollars. First, use premium income $c=3$ thousand dollars per year and find $R$, then calculate its sensitivities to $\lambda$ and $c$ when each is varied separately. Finally, calculate the frequency sensitivity if the premium rate is instead reset to maintain a safety loading of $\theta=0.5$. All monetary rates use thousands of dollars per year.

Method and root. For an exponential severity with mean $1$, $M_X(r)=1/(1-r)$ for $r<1$. The equation with fixed $c=3$ is $\,2(1/(1-R)-1)=3R$. For a positive root, divide by $R$: $2/(1-R)=3$, so $1-R=2/3$ and $R=1/3$ per thousand dollars.

At this root, $M_X(R)=1/(2/3)=3/2$ and $M_X'(R)=1/(1-R)^2=9/4$. Therefore the denominator in both sensitivity formulas is $D=\lambda M_X'(R)-c=2(9/4)-3=9/2-3=3/2$.

Fixed-premium frequency sensitivity. The numerator is $M_X(R)-1=3/2-1=1/2$, giving $\partial R/\partial\lambda=-(1/2)/(3/2)=-1/3$. Locally, adding one claim per year while keeping $c=3$ fixed reduces the adjustment coefficient by about $1/3$ per thousand dollars.

Fixed-frequency premium sensitivity. The derivative is $\partial R/\partial c=R/D=(1/3)/(3/2)=2/9$. Thus, a small increase of $0.1$ thousand dollars per year in $c$ changes $R$ by approximately $(2/9)(0.1)=0.0222$ per thousand dollars.

Premium reset to fixed loading. If $\theta=0.5$, the exponential-claim root is $R=\theta/((1+\theta)\mu)=0.5/(1.5\cdot1)=1/3$, independent of $\lambda$. Hence its frequency derivative under this premium rule is $0$, not $-1/3$.

Check. With fixed premium, $c=3>\lambda\mu=2$, and increasing $\lambda$ uses up the safety margin, consistent with the negative derivative. With fixed loading, $c$ rises proportionally with $\lambda$, leaving the root equation unchanged.

Sensitivity to Severity Parameters

When the severity distribution changes, the premium rule matters again. If $c$ is fixed and the distribution depends on $\eta$, then $\partial R/\partial\eta=-\lambda(\partial M_X(R;\eta)/\partial\eta)/(\lambda M_X'(R;\eta)-c)$. Under a fixed safety loading, both the moment generating function and the mean can depend on $\eta$. Differentiating the loading-based root equation gives

\frac{\partial R}{\partial\eta} = -\frac{\partial M_X(R;\eta)/\partial\eta-(1+\theta)R\,\partial\mu/\partial\eta} {M_X'(R;\eta)-(1+\theta)\mu}.

The mean term in the numerator is essential: a severity change affects not only the claim distribution but also the premium rate when premiums are recalculated using the same loading. The denominator is the derivative of the loading-based equation with respect to $R$, and is positive at a regular positive root.

Worked Example: Changing the Mean of Exponential Claims

A hypothetical portfolio has exponential claim amounts with mean $\mu=2$ thousand dollars and safety loading $\theta=0.25$. Premium income is reset as $c=(1+\theta)\lambda\mu$ when the mean changes. Find the adjustment coefficient and its derivative with respect to $\mu$ at the stated value. The claim frequency $\lambda$ is any fixed positive rate.

Method. For exponential claims, $M_X(R)=1/(1-\mu R)$. Substituting into the loading-based equation gives $1/(1-\mu R)-1=(1+\theta)\mu R$. For $R>0$, the left side is $\mu R/(1-\mu R)$. Divide both sides by $\mu R$ to obtain $1/(1-\mu R)=1+\theta$, hence $1-\mu R=1/(1+\theta)$ and $R=\theta/((1+\theta)\mu)$.

At $\theta=0.25$ and $\mu=2$, this gives $R=0.25/(1.25\cdot2)=0.1$ per thousand dollars. Differentiating the closed form with respect to $\mu$, while holding $\theta$ fixed, gives $\partial R/\partial\mu=-\theta/((1+\theta)\mu^2)$.

Substitution yields $\partial R/\partial\mu=-0.25/(1.25\cdot 2^2)=-0.25/5=-0.05$ per thousand dollars squared. A local increase of $0.1$ thousand dollars in the mean therefore corresponds to an approximate root change of $(-0.05)(0.1)=-0.005$ per thousand dollars.

Check. The coefficient $R$ is positive, and its derivative with respect to the mean is negative. A larger exponential mean makes the claim distribution heavier on the loss scale; the positive root falls, as expected.

Discrete Severity: Differentiate the Probability and the Mean

For a discrete severity model, a parameter can shift probability between claim sizes. Both the moment generating function and the mean must then be differentiated. The next example shows this calculation explicitly and illustrates why the direction of a probability parameter must be read carefully.

Worked Example: Probability of a Large Claim

A hypothetical portfolio has claim amount $1$ or $3$ thousand dollars, with $P(X=3)=p$ and $P(X=1)=1-p$. Claims arrive at rate $\lambda=1$ per year. Premium income follows a fixed safety loading with $1+\theta=2/\ln 2$. At $p=0.5$, calculate the adjustment coefficient and its sensitivity to $p$.

Root and mean. The mean is $\mu(p)=(1-p)1+3p=1+2p$, so at $p=0.5$, $\mu=2$. The moment generating function is $M_X(R;p)=(1-p)e^R+pe^{3R}$. Test $R=\ln2$: $e^R=2$, $e^{3R}=8$, and $M_X(R;0.5)=0.5(2)+0.5(8)=5$. Thus $M_X-1=4$. The loading equation's right side is $(1+\theta)\mu R=(2/\ln2)(2)(\ln2)=4$, so $R=\ln2$ is the positive root.

Differentiate the equation. Write $H(R,p)=M_X(R;p)-1-(1+\theta)\mu(p)R=0$. Its derivative with respect to the root is $H_R=M_X'(R;p)-(1+\theta)\mu$. At $R=\ln2$ and $p=0.5$, $M_X'=0.5e^R+1.5e^{3R}=0.5(2)+1.5(8)=1+12=13$. Therefore $H_R=13-(2/\ln2)(2)=13-4/\ln2$, approximately $7.2292$.

The derivative with respect to $p$ is $H_p=e^{3R}-e^R-(1+\theta)\mu'(p)R$. Since $\mu'(p)=2$, the value is $H_p=8-2-(2/\ln2)(2)(\ln2)=6-4=2$. Implicit differentiation now gives $\partial R/\partial p=-H_p/H_R=-2/(13-4/\ln2)$, approximately $-0.2767$ per unit change in $p$.

Check. Increasing $p$ shifts probability from a claim of $1$ to a claim of $3$, raising the mean. With the loading rule, premiums also increase, but the calculated derivative remains negative at this parameter value. The positive denominator and $H_p=2$ give the stated negative sign.

Common Mistakes and Interpretation

Changing the premium rule without changing the derivative. The fixed-$c$ frequency derivative is not valid when $c$ changes with $\lambda$. Under fixed $\theta$ and fixed severity, $\lambda$ cancels from the root equation.
Forgetting the premium response to severity. Under fixed loading, differentiating only $M_X$ omits the term $(1+\theta)R\,\partial\mu/\partial\eta$.
Reversing a discrete probability parameter. If $p=P(X=3)$, then $\partial M_X/\partial p=e^{3R}-e^R$. If $p=P(X=1)$, the sign is reversed. State what event $p$ represents before differentiating.
Treating a local derivative as an exact change. A first-order approximation is suitable for small input changes. For a substantial change, solve the new root equation rather than extrapolating the derivative.

In the discrete example, a common arithmetic and sign trap is to report $H_p=-2$ after writing $H_p=8-2-4$. The displayed calculation equals $2$, not $-2$; consequently the sensitivity is $-2/H_R$, not $2/H_R$. Checking the derivative's direction against the severity change is a useful additional safeguard.

Key takeaway: Differentiate the equation that actually defines the root, not a formula from a different premium convention. State which inputs are held fixed, include every parameter-dependent term, and use the sign of $H_R$ to interpret the result.

Check Your Understanding

For each problem, identify the premium convention before applying a sensitivity formula.

Claims are exponential with mean $\mu=4$ thousand dollars and safety loading $\theta=0.25$. Find $R$ and $\partial R/\partial\mu$ when premium is reset to maintain the loading. (Answer: $R=0.05$ per thousand dollars; $\partial R/\partial\mu=-0.0125$ per thousand dollars squared.)
Claims are exponential with mean $1$ thousand dollars, $\lambda=3$ per year, and fixed $c=4$ thousand dollars per year. Find $R$ and $\partial R/\partial c$, holding $\lambda$ fixed. (Answer: $R=0.25$ per thousand dollars; $\partial R/\partial c=0.1875$ per thousand dollars squared.)
For the two-point severity model in the worked example, evaluate $\partial R/\partial p$ at $p=0.5$, with $p=P(X=3)$ and the same fixed loading. (Answer: $-2/(13-4/\ln2)\approx-0.2767$ per unit change in $p$.)
Suppose severity and safety loading are fixed, and premium income is $c=(1+\theta)\lambda\mu$. What is $\partial R/\partial\lambda$ under this premium rule? (Answer: $0$, because $\lambda$ cancels from the root equation.)

What You'll Learn