Why Use a Generating Function for the (a,b,0) Class?
The Panjer recursion describes the ratio between successive probabilities in a frequency distribution. A probability generating function (PGF) packages those probabilities into one function. For a distribution in the (a,b,0) class, the recursion translates into a differential equation for that function. The equation gives a compact route to the PGF and, when the parameters define a valid distribution, to its mean.
This connection is useful because it provides a check on probabilities obtained from the recursion and a way to calculate moments without summing every probability separately. The key is to distinguish the differential equation, which follows from the recursion, from the additional parameter conditions needed for a probability distribution to exist. The results here are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.
The notation \((a,b,0)\) refers to this probability recursion, not to a choice of \(p_0\) by itself. The recursion determines probabilities relative to \(p_0\); normalization, \(\sum_{k=0}^{\infty}p_k=1\), determines \(p_0\). In finite-support cases, the recursion must terminate with probability zero rather than assign a negative probability to a later value of \(N\).
From the Probability Recursion to a Differential Equation
Multiply the recursion by \(k\), so that \(kp_k=(ak+b)p_{k-1}\). Now multiply by \(z^k\) and sum over \(k\geq1\). Initially, take \(|z|<1\), where the PGF power series and its termwise derivative are valid. The left-hand side is \(zG'(z)\). On the right-hand side, changing the summation index to \(j=k-1\) gives $$ \sum_{k=1}^{\infty}(ak+b)p_{k-1}z^k =az\sum_{j=0}^{\infty}(j+1)p_jz^j+bz\sum_{j=0}^{\infty}p_jz^j. $$
Since \(\sum_{j=0}^{\infty}p_jz^j=G(z)\) and \(\sum_{j=0}^{\infty}jp_jz^j=zG'(z)\), the right-hand side is \(az(zG'(z)+G(z))+bzG(z)\). Equating both sides and rearranging yields the differential equation.
The factor \(z\) in the first equation can be divided out to obtain the second equation for \(z\ne0\). The resulting equation also holds at \(z=0\) by continuity. This derivation explains why the coefficient \(a\) multiplies \(z\) in the differential equation: one factor of \(z\) appears when the recursion is shifted from \(p_k\) to \(p_{k-1}\).
Solving and Normalizing the Equation
When \(a\ne0\), separate variables wherever \(G(z)>0\): $$ \frac{G'(z)}{G(z)}=\frac{a+b}{1-az}. $$ Integrating and using \(G(0)=p_0\) gives \(G(z)=p_0(1-az)^{-(a+b)/a}\). If the model is a probability distribution with finite mean, then \(G(1)=1\). For admissible parameters with \(a<1\), this determines \(p_0\).
The real power in this expression is interpreted on the part of the real interval from 0 to 1 where \(1-az>0\). The parameter conditions below ensure this for the cases in which the expression is used. For \(a=0\), dividing by \(a\) is not valid. The differential equation instead becomes \(G'(z)=bG(z)\), whose solution is \(G(z)=p_0e^{bz}\); normalization gives \(p_0=e^{-b}\).
For a nondegenerate distribution, the admissible parameters have one of the following forms. A degenerate distribution with \(P(N=0)=1\) is also possible when the recursion makes all positive-count probabilities zero.
- \(a>0\): \(a<1\) and \(b>-a\). These conditions give nonnegative probabilities and a normalizable unbounded distribution.
- \(a=0\): \(b>0\). The PGF is the normalized exponential solution above.
- \(a<0\): there is an integer \(m\geq1\) such that \(b=-(m+1)a\). The recursion terminates at \(p_{m+1}=0\), so the support is \(0,1,\ldots,m\).
The admissibility check is not optional. For \(a>0\), the ratio \(p_k/p_{k-1}\) approaches \(a\) as \(k\) increases, so \(a\geq1\) cannot give a summable, nondegenerate probability sequence. For \(a<0\), the ratio eventually becomes negative unless it reaches zero exactly at an integer boundary. When \(b=-(m+1)a\), the ratio at \(k=m+1\) is zero, as required for termination.
Deriving the Mean
For a nonnegative integer-valued random variable, \(G'(1)=E[N]\) when the mean is finite. Every admissible nondegenerate case above has a finite mean. Set \(z=1\) in the differential equation and use \(G(1)=1\):
The denominator \(1-a\) arises from evaluating the differential equation at \(z=1\); it is not part of the probability recursion’s numerator. This formula applies after admissibility and normalization have been checked. If a proposed pair \((a,b)\) does not define a probability distribution, substituting it into the mean expression does not make it valid.
Worked Examples
Worked Example: Find the PGF and Mean for an Unbounded Model
A hypothetical frequency model has recursion \(p_k=(\frac13+\frac{2}{3k})p_{k-1}\) for \(k\geq1\). Find its normalized PGF and mean, and check the first two positive-count probabilities. State whether the parameters are admissible.
Method and admissibility. Here \(a=\frac13\) and \(b=\frac23\). Since \(0<a<1\) and \(b>-a\), the parameters are admissible for a nondegenerate unbounded model.
The exponent is $$ \frac{a+b}{a}=\frac{1}{1/3}=3. $$ Thus normalization gives $$ p_0=(1-a)^3=(2/3)^3=\frac{8}{27}, \qquad G(z)=\frac{8}{27}(1-z/3)^{-3}=\frac{8}{(3-z)^3}. $$ At \(z=1\), \(G(1)=8/2^3=1\), which checks the normalization.
The recursion gives $$ p_1=(a+b)p_0=1\cdot\frac{8}{27}=\frac{8}{27}, \qquad p_2=\left(\frac13+\frac{2}{3\cdot2}\right)p_1 =\frac23\cdot\frac{8}{27}=\frac{16}{81}. $$ The mean from the differential equation is $$ E[N]=\frac{a+b}{1-a} =\frac{1}{2/3}=\frac32=1.5. $$ Check. The PGF is one at \(z=1\), and the mean is positive, as expected for a nondegenerate frequency. The method avoids treating the recursion’s first ratio as if it were the mean.
Worked Example: A Finite-Support Model
A hypothetical count has \(a=-\frac14\) and \(b=1\). Determine whether the recursion terminates at a finite count, find its normalized probabilities, and calculate the mean.
Method and admissibility. Since \(b=-(m+1)a\), the required support endpoint satisfies $$ 1=-(m+1)(-1/4)=\frac{m+1}{4}, \qquad m+1=4,\qquad m=3. $$ The parameters are admissible for support \(0,1,2,3\); the ratio at \(k=4\) is \(-\frac14+\frac14=0\).
Starting with an unnormalized weight of 1 at zero, apply the recursion: $$ w_0=1,\quad w_1=\frac34,\quad w_2=\left(-\frac14+\frac12\right)\frac34=\frac{3}{16},\quad w_3=\left(-\frac14+\frac13\right)\frac{3}{16}=\frac{1}{64}. $$ Their sum is $$ 1+\frac34+\frac{3}{16}+\frac{1}{64} =\frac{64+48+12+1}{64} =\frac{125}{64}. $$ Dividing each weight by this sum gives $$ p_0=\frac{64}{125},\quad p_1=\frac{48}{125},\quad p_2=\frac{12}{125},\quad p_3=\frac{1}{125}. $$ They sum to \(125/125=1\), and the next probability is zero. The mean is $$ E[N]=0\cdot\frac{64}{125} +1\cdot\frac{48}{125} +2\cdot\frac{12}{125} +3\cdot\frac{1}{125} =\frac{75}{125}=\frac35. $$ The formula gives the same result: $$ \frac{a+b}{1-a} =\frac{3/4}{5/4}=\frac35. $$ Check. The direct weighted sum agrees with the formula. The method avoids continuing the recursion beyond its zero-probability boundary.
Worked Example: Handle the Boundary Case \(a=0\)
A hypothetical frequency satisfies \(p_k=(2/k)p_{k-1}\) for \(k\geq1\). Find its PGF, \(p_0\), and mean. Explain why the nonzero-\(a\) PGF expression should not be used directly.
Method and admissibility. Here \(a=0\) and \(b=2>0\), so this is an admissible boundary case. Use \(G'(z)=bG(z)\), not an expression with division by \(a\): $$ G(z)=p_0e^{2z}. $$ Normalization requires \(G(1)=1\), hence \(p_0e^2=1\) and \(p_0=e^{-2}\). Therefore \(G(z)=e^{2(z-1)}\).
The mean is \(G'(1)=2e^{2(1-1)}=2\). Equivalently, the general mean expression gives \((a+b)/(1-a)=(0+2)/1=2\). Check. The PGF is one at \(z=1\), and the mean is positive. The method avoids dividing by zero when solving the differential equation.
Common Exam Trap: Omitting the Denominator
A frequent error is to substitute \(z=1\) into the differential equation but forget the coefficient \(1-az\). For example, take admissible parameters \(a=\frac14\) and \(b=\frac12\). The incorrect calculation \(E[N]=a+b=\frac34\) omits this coefficient.
The correct substitution is $$ (1-a)G'(1)=(a+b)G(1), \qquad \frac34G'(1)=\frac34\cdot1, \qquad E[N]=G'(1)=1. $$ The denominator \(1-a=\frac34\) is essential. A useful check is to differentiate the normalized PGF, or to verify the result against probabilities from the recursion when enough terms are available.
Using the Result Reliably
The PGF equation links a local probability rule to global distributional information. It can confirm the normalization constant, expose parameter choices that cannot define a probability sequence, and yield the mean with a single substitution. In frequency modelling, these checks help distinguish a valid model from a formal recurrence that produces invalid probabilities.
A reliable sequence of steps is:
Write \(p_k=(a+b/k)p_{k-1}\) and read off \(a\) and \(b\).
Verify the relevant parameter conditions, including finite termination if \(a<0\).
For \(a\ne0\), use the power form; for \(a=0\), solve \(G'=bG\) separately.
Use \(G(1)=1\), then evaluate the differential equation at \(z=1\) to obtain \(G'(1)\).
The main boundary checks are therefore simple but important: do not divide by \(a\) when \(a=0\); do not continue a finite-support recursion past its zero probability; and do not use the mean formula to validate parameters that are inadmissible. Keeping those checks separate from the algebra makes the PGF method both efficient and dependable.
Check Your Understanding
For each problem, first decide whether the parameters are admissible before using the PGF or mean formula.
- A frequency has \(a=\frac12\) and \(b=\frac12\). Find \(p_0\) and \(E[N]\) from the normalized PGF. (Answer: \(p_0=\frac14,\ E[N]=2\))
- A frequency has \(a=0\) and \(b=3\). State its normalized PGF and mean, using the boundary-case equation. (Answer: \(G(z)=e^{3(z-1)},\ E[N]=3\))
- A proposed recursion has \(a=-\frac13\) and \(b=\frac43\). Find the finite support endpoint and mean. (Answer: support \(0,1,2,3\), and \(E[N]=\frac34\))
- A proposed nondegenerate model has \(a=\frac54\) and \(b=\frac14\). Is it admissible as a normalized unbounded frequency distribution? Explain using the ratio or PGF condition. (Answer: No; \(a>1\), so the successive probability ratio tends to a value greater than 1 and the probabilities cannot form a summable sequence.)
- For an admissible model with \(a=\frac13\) and \(b=\frac13\), calculate \(E[N]\), showing the substitution from the differential equation at \(z=1\). (Answer: \(E[N]=\frac{(1/3+1/3)}{1-1/3}=1\))