Recovering Panjer Parameters From Probabilities

Why Recovering the Parameters Takes More Than Two Ratios

A sequence of count probabilities can sometimes be tested for a Panjer recursion by examining ratios of neighboring probabilities. If the ratios have the required form, two of them determine candidate values of $a$ and $b$. But a two-ratio fit is only a candidate: it does not establish that the remaining probabilities follow the same recursion. The initial probability relation and any finite-support endpoint also need to be checked.

This distinction matters in exam problems. A distribution may agree with a proposed ratio pattern at several positive counts while failing at count one, or it may fit two selected ratios but not a third. The method below makes these checks explicit. It is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: Let $p_k=P(N=k)$, for nonnegative integer counts $k$. A standard $(a,b,1)$ Panjer recursion has the form $$ p_k=\left(a+\frac{b}{k}\right)p_{k-1},\qquad k\geq 1. $$ When $p_{k-1}>0$, define the probability ratio $r_k=p_k/p_{k-1}$. Then the recursion is equivalent at that count to $r_k=a+b/k$. In particular, it includes $p_1=(a+b)p_0$.

The ratio is undefined when its denominator is zero, so in that case check the probability recursion itself. Also, a sequence of ratios alone does not ensure that the probabilities are nonnegative or sum to one. Those are separate requirements for a probability distribution.

Solving for the Candidate Parameters

Suppose ratios are available at two distinct positive counts $j$ and $k$, with $p_{j-1}>0$ and $p_{k-1}>0$. Under the proposed recursion, $$ r_j=a+\frac{b}{j},\qquad r_k=a+\frac{b}{k}. $$ Subtracting eliminates $a$, and then either equation gives $a$:

r_j-r_k=b\left(\frac{1}{j}-\frac{1}{k}\right), \qquad b=\frac{jk}{k-j}(r_j-r_k), \qquad a=r_j-\frac{b}{j}.

The indices must be distinct; otherwise the equations do not provide two different values of $1/k$. For consecutive ratios, the formula simplifies. If the observed ratios are $r_k$ and $r_{k+1}$, then

b=k(k+1)(r_k-r_{k+1}),\qquad a=r_k-\frac{b}{k}.

This is a useful calculation shortcut, not a verification shortcut. Once $a$ and $b$ are obtained, use them to predict every other ratio that is defined and check those predictions against the data.

Why the two-ratio formula works: The term $a$ is the same in both equations, so subtraction removes it. The remaining difference is $b$ times the known difference $1/j-1/k$. Since $j\ne k$, that difference is nonzero, and division isolates $b$. Substitution into either original equation then determines $a$. Two distinct ratios therefore determine at most one candidate pair, but do not by themselves prove that the full probability sequence follows the recursion.

A Verification Procedure

A reliable check proceeds from the candidate fit to the actual probability sequence. In particular, do not start at $k=2$ if the question asks whether the sequence belongs to the standard $(a,b,1)$ class: the defining recursion begins at $k=1$.

Choose usable ratios.
Use two ratios at distinct positive count indices whose denominators are positive. Record the indices as well as the ratios.

Solve for $a$ and $b$.
Apply the two-ratio equations and retain exact fractions when the input permits them.

Check all other ratios.
For each defined ratio $r_m$, compare it with $a+b/m$. A mismatch rejects the candidate for the stated sequence.

Check the initial count and endpoints.
Verify $p_1=(a+b)p_0$. For finite support, check the recursion at the last positive count and at the first zero beyond it.

Check probability validity.
Confirm the stated probabilities are nonnegative and sum to one. A recursion fit alone does not establish either condition.

For finite support ending at $m$, the recursion at $m+1$ must produce $p_{m+1}=0$. If $p_m>0$, this requires $a+b/(m+1)=0$. Once both adjacent probabilities are zero, later recursion equations also give zero, provided the sequence is stated to remain zero. This endpoint check is one way to detect an attempted infinite-support fit to a finite sequence.

Worked Example: Recovering Parameters and Checking a Finite Endpoint

A hypothetical count model assigns probabilities $p_0=27/64$, $p_1=27/64$, $p_2=9/64$, $p_3=1/64$, and $p_k=0$ for $k\geq4$. Recover candidate parameters from $r_2$ and $r_3$, then verify the standard recursion across the support.

Method. Use the two positive ratios to solve for $a,b$, then check the remaining ratio at count one and the finite endpoint. This avoids treating a fit to two ratios as a complete classification.

First calculate the ratios: $$ r_2=\frac{p_2}{p_1}=\frac{9/64}{27/64}=\frac{1}{3}, \qquad r_3=\frac{p_3}{p_2}=\frac{1/64}{9/64}=\frac{1}{9}. $$

Using counts $j=2$ and $k=3$, $$ b=\frac{2\cdot3}{3-2}\left(\frac{1}{3}-\frac{1}{9}\right) =6\cdot\frac{2}{9}=\frac{4}{3}. $$ Then $$ a=r_2-\frac{b}{2}=\frac{1}{3}-\frac{2}{3}=-\frac{1}{3}. $$

At count one, the recursion factor is $a+b=-1/3+4/3=1$, so it predicts $p_1=1\cdot p_0=27/64$, as given. At count two, the factor is $-1/3+(4/3)/2=1/3$, giving $p_2=(1/3)(27/64)=9/64$. At count three, it is $-1/3+(4/3)/3=1/9$, giving $p_3=(1/9)(9/64)=1/64$.

At count four, the factor is $-1/3+(4/3)/4=0$, so $p_4=0$, as stated. For $k\geq5$, $p_{k-1}=0$, so the recursion produces $p_k=0$. Finally, the probabilities sum to $$ \frac{27+27+9+1}{64}=1. $$ The candidate is verified across the support. The common mistake avoided here is stopping after solving from $r_2,r_3$ without checking either $k=1$ or the endpoint $k=4$.

Worked Example: Rejecting a Two-Ratio Fit

A hypothetical frequency distribution has probabilities $p_0=0.10$, $p_1=0.20$, $p_2=0.20$, $p_3=0.12$, $p_4=0.12$, and $p_5=0.26$, with zero probability above five. Determine whether the ratios support one standard $(a,b,1)$ recursion.

Method. Fit $a,b$ using $r_2,r_3$, then compare the fitted model with unused ratios. The probabilities are nonnegative and sum to $1.00$, so the issue is whether the recursion fits.

The selected ratios are $$ r_2=\frac{0.20}{0.20}=1,\qquad r_3=\frac{0.12}{0.20}=0.6. $$ Thus $$ b=2\cdot3(1-0.6)=2.4,\qquad a=1-\frac{2.4}{2}=-0.2. $$

The candidate predicts $r_1=a+b=2.2$, but the observed value is $r_1=0.20/0.10=2$. It also predicts $r_4=-0.2+2.4/4=0.4$, while the observed ratio is $0.12/0.12=1$. Either disagreement is enough to reject the candidate. The common mistake avoided is assuming that two matching ratios prove the whole sequence follows the recursion.

The Initial Probability Is Part of the Test

A special boundary check is needed when $p_0=0$. In the standard recursion, $p_1=(a+b)p_0$, which forces $p_1=0$. Therefore, a distribution with $p_0=0$ and $p_1>0$ cannot satisfy the standard $(a,b,1)$ recursion, regardless of whether ratios from count two onward fit $a+b/k$. Such a sequence may arise in a model with a modified boundary, but that is not the standard recursion being tested here.

Worked Example: Positive-Count Ratios Do Not Fix a Failed Boundary

A hypothetical count distribution has $p_0=0$, $p_1=0.50$, $p_2=0.30$, and $p_3=0.20$, with zero probability for higher counts. Test whether its positive-count ratios can establish standard $(a,b,1)$ membership.

Method. Fit the ratios available from counts two and three, then test the defining recursion directly at count one. This avoids silently replacing the standard boundary condition with a restricted ratio rule.

The two defined ratios are $$ r_2=\frac{0.30}{0.50}=\frac{3}{5}, \qquad r_3=\frac{0.20}{0.30}=\frac{2}{3}. $$ They give $$ b=2\cdot3\left(\frac{3}{5}-\frac{2}{3}\right) =6\left(-\frac{1}{15}\right)=-\frac{2}{5}, \qquad a=\frac{3}{5}-\frac{-2/5}{2}=\frac{4}{5}. $$ These values do reproduce the two positive-count ratios: $a+b/2=4/5-1/5=3/5$, and $a+b/3=4/5-2/15=2/3$. But at count one they require $$ p_1=(a+b)p_0=\left(\frac{4}{5}-\frac{2}{5}\right)0=0, $$ contradicting the stated $p_1=0.50$. The sequence is not in the standard $(a,b,1)$ class. The common mistake avoided is calling a fit for counts two and above a standard recursion while ignoring $k=1$.

Common Exam Traps

Using the wrong index factor. For adjacent ratios $r_k,r_{k+1}$, the difference in reciprocals is $1/k-1/(k+1)=1/[k(k+1)]$. Therefore $b=k(k+1)(r_k-r_{k+1})$, not just $k(r_k-r_{k+1})$.
Reversing the ratio order. The defined ratio is $r_k=p_k/p_{k-1}$. Reversing the probabilities changes the value and no longer matches $a+b/k$.
Checking only the ratios used to fit. Those ratios determine a candidate pair by construction. Other ratios and the count-one recursion provide the meaningful tests.
Dividing by zero. If $p_{k-1}=0$, $r_k$ is undefined. Check $p_k=(a+b/k)p_{k-1}$ directly instead.
Ignoring rounded inputs. If probabilities are reported approximately, small discrepancies may reflect rounding. State whether the candidate fits the reported values exactly or only within an explicitly chosen tolerance; do not silently treat rounded ratios as exact.

Key takeaway: Two ratios at distinct count indices determine candidate values of $a$ and $b$. A standard $(a,b,1)$ classification requires the recursion $p_k=(a+b/k)p_{k-1}$ to hold starting at $k=1$ and throughout the stated support, with valid probability values.

Check Your Understanding

For each problem, solve for candidates where possible and check the stated conditions rather than relying on the initial fit alone.

A count distribution has $p_0=0.25$, $p_1=0.45$, $p_2=0.18$, $p_3=0.045$, and $p_4=0.009$. Use $r_2,r_3$ to find $a,b$, then decide whether the sequence satisfies the standard recursion. (Answer: $a=-0.05,\ b=0.90$; it fails, since the candidate predicts $r_1=0.85$, not $0.45/0.25=1.8$.)
Suppose $r_1=1.4$ and $r_3=0.6$. Recover $a,b$, and give the ratios predicted at counts two and four. (Answer: $a=0.2,\ b=1.2$; predicted $r_2=0.8$ and $r_4=0.5$.)
A finite-support model has $p_0=64/125$, $p_1=48/125$, $p_2=12/125$, $p_3=1/125$, and $p_k=0$ for $k\geq4$. Recover $a,b$ from $r_1,r_2$, then check $r_3$ and the endpoint. (Answer: $a=-1/4,\ b=1$; $r_3=1/12$, and the factor at count four is zero, so the stated endpoint is consistent.)
A distribution has $p_0=0,\ p_1=0.4,\ p_2=0.24,\ p_3=0.12$, and zero probability thereafter. Ratios from counts two and three can be fitted by some $a,b$. Can it satisfy the standard $(a,b,1)$ recursion? (Answer: No; $p_1=(a+b)p_0=0$ would be required, contradicting $p_1=0.4$.)
Reported ratios are $r_2=0.70,\ r_3=0.50,\ r_4=0.40$. Fit $a,b$ from the first two and test the third. (Answer: $b=1.20,\ a=0.10$; the candidate predicts $r_4=0.40$, so all three reported ratios are consistent, though the initial probability relation must still be checked for a full classification.)

What You'll Learn