Panjer Recursion for Aggregate Operational Losses

Why Use Panjer Recursion?

An annual operational loss total combines a random number of events with random severities. Even when the frequency and severity models are manageable separately, the aggregate distribution can be difficult to calculate directly: each possible total may arise from many event counts and combinations of severities. Panjer recursion provides a systematic way to calculate aggregate probabilities on a discrete monetary grid without enumerating all those combinations.

The method is exact for a discretized severity distribution and a suitable frequency distribution. It is not automatically exact for the original continuous loss model: discretizing severity introduces approximation, and calculating only a finite number of aggregate probabilities leaves an uncalculated tail.

Definition: Let $N$ be the number of events in a period and let $Y_1,Y_2,\ldots$ be their severities or payments. The aggregate loss is $S=\sum_{r=1}^{N}Y_r$, with $S=0$ when $N=0$. In the compound model, $N$ is independent of the severities, and the severities are independent and identically distributed. For a grid width $h>0$, define $f_j=P(Y=jh)$ and $g_k=P(S=kh)$ for nonnegative integers $j,k$.

Discretizing Severity and Specifying Frequency

Choose a grid width $h$ in the same monetary units as the losses. A common discretization rounds each positive loss down to the nearest grid point: $Y=h\lfloor X/h\rfloor$, where $X$ is the original loss. Rounding down is simple but can understate severity; rounding up can be conservative but may overstate it. Rounding to the nearest grid point reduces some local errors but does not guarantee that aggregate losses are unbiased. The chosen rule should be explicit and applied consistently.

Rounding down can assign positive probability to $Y=0$, even if the original losses are all positive. Thus $f_0$ need not mean that an event never occurred: it can represent events whose discretized payment is zero. This distinction matters because several zero-payment events can occur without making the aggregate loss positive.

Panjer recursion applies to frequency distributions satisfying the following recurrence, called the $(a,b,0)$ class:

Definition: A frequency distribution with probabilities $p_n=P(N=n)$ is in the $(a,b,0)$ class if $p_n=(a+b/n)p_{n-1}$ for $n\geq 1$, for constants $a,b$ and a valid probability distribution. Examples include Poisson frequency, with $a=0$ and $b=\lambda$, and geometric frequency $p_n=(1-q)q^n$, with $a=q$ and $b=0$.

For example, under Poisson frequency with mean $\lambda$, $p_n=e^{-\lambda}\lambda^n/n!$, so $p_n=(\lambda/n)p_{n-1}$. The required frequency recurrence therefore holds with $a=0$ and $b=\lambda$.

The Recursion and Why It Holds

Let $G_N(z)=E[z^N]$ be the probability generating function of the frequency, and let $F(z)=\sum_{j=0}^{\infty}f_jz^j$ be that of discretized severity in grid units. Independence gives the aggregate generating function $G_S(z)=G_N(F(z))$. The frequency recurrence implies

$$ (1-az)G_N'(z)=(a+b)G_N(z). $$

To see this, write $G_N'(z)=\sum_{n\geq1}np_nz^{n-1}$ and substitute $np_n=(an+b)p_{n-1}$. The resulting sums are $a(zG_N'(z)+G_N(z))+bG_N(z)$; rearranging gives the displayed identity. Applying the chain rule to $G_S(z)=G_N(F(z))$ then yields

$$ (1-aF(z))G_S'(z)=(a+b)F'(z)G_S(z). $$

Equating coefficients of $z^{k-1}$, for $k\geq1$, gives the recursion below. The term involving $f_0$ multiplies $g_k$, so it must be moved to the left-hand side. The initial probability is obtained from $G_S(0)=G_N(f_0)$.

Theorem: For a compound model with frequency in the $(a,b,0)$ class and discretized severities $f_j$, set $g_0=G_N(f_0)$. If $1-af_0\ne0$, then for $k\geq1$,

g_k=\frac{1}{1-af_0}\sum_{j=1}^{k}\left(a+\frac{bj}{k}\right)f_jg_{k-j}.

For Poisson frequency, this simplifies to $g_0=\exp(\lambda(f_0-1))$ and $g_k=(\lambda/k)\sum_{j=1}^{k}j f_jg_{k-j}$. The sum only uses severity probabilities and aggregate probabilities up to grid point $k$, which makes calculation sequential. If $1-af_0=0$, this form cannot be used as written; check the model and parameterization rather than dividing by zero.

Worked Example: Geometric Frequency with Unit Severity

A hypothetical operational process has a geometric event count $P(N=n)=0.75(0.25)^n$, $n\geq0$. Each event costs exactly $500. Use a grid width of $h=$500$ to calculate $P(S=kh)$ for $k=0,1,2,3$, and the probability that the aggregate exceeds $1,500.

Method. The frequency recurrence has $a=0.25$, $b=0$. A severity is one grid unit with certainty, so $f_1=1$ and all other $f_j=0$, including $f_0=0$. The starting value is $g_0=G_N(0)=p_0=0.75$.

Since $b=0$, the recursion becomes $g_k=a g_{k-1}$ for this unit-severity model. Thus $g_1=0.25(0.75)=0.1875$, $g_2=0.25(0.1875)=0.046875$, and $g_3=0.25(0.046875)=0.01171875$.

The cumulative probability through $1,500 is $0.75+0.1875+0.046875+0.01171875=0.99609375$. Therefore $P(S>$1,500)=1-0.99609375=0.00390625$. The exact geometric tail is also $P(N\geq4)=(0.25)^4=0.00390625$, which checks the result. To four decimal places, the requested tail probability is 0.0039.

The expected event count is $q/(1-q)=0.25/0.75=1/3$, so the expected aggregate is $(1/3)($500)=$166.67$. This is consistent with most probability being concentrated at zero or one event. Common mistake avoided: treating $p_0=0.75$ as the probability that the aggregate is zero without checking severity. Here it is correct because every event has positive severity; if $f_0>0$, the aggregate-zero probability would be different.

Worked Example: Poisson Frequency with Two Severity Sizes

A hypothetical operational loss model has $N\sim\operatorname{Poisson}(2)$. Each event has severity $1,000 or $2,000, each with probability 0.5. With $h=$1,000$, calculate the aggregate probabilities at $0, $1,000, $2,000, and $3,000, then find the probability that the aggregate exceeds $3,000.

Method. The frequency is Poisson, so $a=0$, $b=\lambda=2$, and $f_0=0$, $f_1=0.5$, $f_2=0.5$. The initial probability is $g_0=e^{-2}=0.1353353$.

At $k=1$, $g_1=(2/1)(1)(0.5)g_0=0.1353353$. At $k=2$, $g_2=(2/2)[(1)(0.5)g_1+(2)(0.5)g_0]$ $=0.5g_1+g_0=0.2030029$.

At $k=3$, only severity grid points 1 and 2 contribute: $g_3=(2/3)[(1)(0.5)g_2+(2)(0.5)g_1]$ $=(2/3)[0.1015015+0.1353353]=0.1578912$.

To four decimal places, the probabilities at $0, $1,000, $2,000, and $3,000 are 0.1353, 0.1353, 0.2030, and 0.1579. Their cumulative probability is $0.6315647$, so $P(S>$3,000)=1-0.6315647=0.3684353$, or 0.3684 to four decimal places.

As a model check, the mean event severity is $0.5($1,000)+0.5($2,000)=$1,500$, and the mean aggregate is $2($1,500)=$3,000$. The mass shown through $3,000 is only part of the distribution; a tail probability of 0.3684 is plausible and is not a numerical failure. Common mistake avoided: summing only the probabilities displayed and treating that sum as one. The omitted mass belongs to aggregate values above the last calculated grid point.

Worked Example: A Zero-Probability Cell from Discretization

A hypothetical model has Poisson event frequency with mean $\lambda=1.2$. After rounding event payments down to a $1,000 grid, suppose the discretized severity probabilities are $f_0=0.25$, $f_1=0.50$, and $f_2=0.25$. Calculate the aggregate probabilities at $0, $1,000, and $2,000, and estimate the probability of an aggregate above $2,000.

Method. Here $a=0$, $b=1.2$, and the Poisson starting value includes the zero-payment cell: $g_0=\exp(1.2(0.25-1))=\exp(-0.9)=0.4065697$.

For $k=1$, $g_1=1.2(0.5)g_0=0.2439418$. For $k=2$, $g_2=(1.2/2)[(1)(0.5)g_1+(2)(0.25)g_0]$ $=0.6[0.5(0.2439418)+0.5(0.4065697)]=0.1951534$.

The probabilities at $0, $1,000, and $2,000 are 0.4066, 0.2439, and 0.1952 to four decimal places. Their cumulative probability is $0.8456649$, so the residual probability above $2,000 is $1-0.8456649=0.1543351$, or 0.1543 to four decimal places.

The probability $g_0$ includes both no events and any number of events whose discretized payments are zero. Indeed, the Poisson generating function gives $P(S=0)=G_N(f_0)$, not $P(N=0)$. Common mistake avoided: using $e^{-1.2}$ for $g_0$. That would count only years with no events and would omit years containing events assigned to the zero-payment cell.

Finite Grids, Boundary Treatment, and Accuracy

In an implementation, calculate $g_0,g_1,\ldots,g_K$ in order. At each step $k$, use the severity probabilities $f_1,\ldots,f_k$ and previously calculated aggregate probabilities $g_{k-1},\ldots,g_0$. This yields the exact probabilities on the chosen grid through $Kh$, provided the model inputs for those cells are correct. It does not calculate the composition of the aggregate tail above $Kh$.

A finite severity table should not silently discard probability above its last cell or move it into the zero cell. If the goal is to calculate aggregate probabilities only through $Kh$, severity probabilities above $Kh$ cannot contribute to those aggregate cells; they can be left outside the displayed range, with the range and residual probability clearly reported. If tail detail or an accurate aggregate mean is required, retain and model the severity tail rather than treating it as absent. A top-coded severity cell is an approximation and should be identified as such.

Choose and document the grid.
State the monetary width $h$ and whether losses are rounded down, up, or to the nearest grid point. Convert all monetary amounts to the same units.

Check the severity probabilities.
Confirm that each $f_j$ is nonnegative and that the full modeled severity probabilities, including any zero and tail cells, sum to one.

Initialize and recurse.
Compute $g_0=G_N(f_0)$, then apply the recursion sequentially for each required $k$. Retain enough precision during calculation and round reported probabilities only at the end.

Check the aggregate results.
Verify that calculated probabilities are nonnegative and that their partial sum is at most one. Report $1-\sum_{k=0}^{K}g_k$ as the probability above $Kh$, not as a missing probability to renormalize away.

A useful expectation check follows from independence. If $E[N]$ is finite, then $E[S]=E[N]E[Y]$, where $E[Y]=h\sum_j j f_j$. Compare this value with an independently calculated mean for the discretized model. The check can reveal a unit conversion error or a missing severity cell, although a correct mean alone does not validate the tail probabilities. For a finite aggregate table, the expectation computed only from $g_0,\ldots,g_K$ will generally omit the tail contribution and should not be compared as if it were the full mean.

Grid width is a modeling decision, not merely a computational setting. A finer grid can reduce discretization error but increases the number of recursive calculations. Rounding down can shift the expected severity lower; rounding up can shift it higher. When a result is sensitive to the grid, repeat the calculation with a finer grid or a different justified discretization and compare the results. The recursion itself does not remove discretization or model uncertainty.

Key takeaway: Panjer recursion produces aggregate probabilities for a discretized compound model. State the frequency class, grid rule, and treatment of zero and tail severity cells; calculate recursively without premature rounding; and report the uncalculated aggregate tail rather than forcing a finite table to sum to one.

Check Your Understanding

For each problem, apply the stated grid and frequency model. Probabilities in the answers are rounded to four decimal places.

A Poisson model has mean frequency 1, and every event has severity exactly one grid unit. Find $P(S/h=0)$, $P(S/h=1)$, $P(S/h=2)$, and $P(S/h>2)$. (Answer: 0.3679, 0.3679, 0.1839, and 0.0803, respectively.)
Event frequency is geometric with $P(N=n)=0.8(0.2)^n$, and each event costs one grid unit. Find the probability that aggregate loss exceeds two grid units. (Answer: 0.0080.)
A Poisson frequency has mean 0.4. Discretized severity is one grid unit with probability 0.75 and two units with probability 0.25. Use the recursion to find the probability that aggregate loss exceeds two grid units. (Answer: 0.0314.)
A Poisson frequency has mean 2, and severity probabilities after rounding down to a $100 grid are $f_0=0.10$, $f_1=0.60$, and $f_2=0.30$. Find the probability that aggregate discretized loss is zero. (Answer: 0.1653.)

What You'll Learn