Choosing Loss-Grid Width for Aggregate Recursion

Why Grid Width Is a Modeling Choice

Panjer recursion computes probabilities on a lattice: aggregate losses are represented as integer multiples of a chosen monetary unit. If the unit is $h$ dollars, the recursion works with loss values $0,h,2h,\ldots$. A smaller $h$ gives a finer representation, but it also requires more recursion steps to reach a fixed monetary threshold. Choosing $h$ is therefore a balance, not a rule to make the grid as fine as possible.

The decision has three parts. First, the grid should resolve the threshold of interest adequately. Second, severity discretization should be fine enough for the required accuracy. Third, the number of lattice points must be computationally manageable. The grid itself does not make a continuous severity exact; it replaces it with a discrete approximation.

Definition: For a nonnegative loss $X$ and grid width $h>0$, the lower discretization is $X_h=h\lfloor X/h\rfloor$. Its lattice probabilities are $g_j=P(jh\leq X<(j+1)h)$, for integers $j\geq 0$. Thus $X_h=jh$ whenever $X$ falls in the $j$th interval. An atom exactly at $jh$ is assigned to that grid point.

This convention ensures $0\leq X-X_h<h$ for every finite $X$. It also makes $X_h\leq X$, so for an aggregate formed from the same claim count and the same individual losses, the discretized aggregate cannot exceed the original aggregate. Lower discretization is useful for interpreting the direction of the resulting probability bias, but it does not by itself guarantee that the bias is small.

Panjer Recursion on the Chosen Grid

Suppose $N$ is an independent claim count in the $(a,b,0)$ class, meaning its probabilities $p_n=P(N=n)$ satisfy $p_n=(a+b/n)p_{n-1}$ for $n\geq1$. Let the claim severities be independent copies of $X_h$, independent of $N$, and let $S_h$ be their aggregate. Write $q_k=P(S_h=kh)$. When $1-a g_0\neq0$, the Panjer recursion is

q_0=\sum_{n=0}^{\infty}p_n g_0^n,\qquad q_k=\frac{1}{1-a g_0}\sum_{j=1}^{k}\left(a+\frac{bj}{k}\right)g_jq_{k-j},\quad k\geq1.

For a Poisson claim count with parameter $\lambda$, $a=0$, $b=\lambda$, and $q_0=\exp\{\lambda(g_0-1)\}$. This recursion applies to the discretized, nonnegative lattice severity distribution; a continuous severity model must first be converted to grid probabilities $g_j$. To report a result through monetary threshold $B$, the recursion is needed through $m=\lfloor B/h\rfloor$. The represented threshold is $mh$, which is less than or equal to $B$ and differs from it by less than $h$.

A smaller grid reduces this threshold-rounding distance, but does not automatically reduce probability error by a known amount. Probability can change sharply near a threshold, especially when the aggregate distribution has substantial mass around it. The grid width should therefore be selected with the threshold and the severity model in mind, then checked by refinement where accuracy matters.

What Lower Discretization Does to the Mean

For compound Poisson frequency with $E[X]<\infty$, the original aggregate mean is $\lambda E[X]$, and the discretized aggregate mean is $\lambda E[X_h]$. The difference follows directly from the lower-grid construction:

0\leq E[S]-E[S_h] =\lambda E[X-X_h] \leq \lambda h,\qquad \lambda\geq0.

If $\lambda>0$, the upper inequality is strict, since $X-X_h<h$ for every finite loss and hence $E[X-X_h]<h$. If $\lambda=0$, both aggregate means are zero and the difference is zero, so the non-strict bound is the valid statement for all $\lambda\geq0$.

Set the threshold.
Identify the largest aggregate amount $B$ through which probabilities are required.

Translate resolution into a grid constraint.
If the threshold representation must be within $\epsilon$ dollars, require $h\leq\epsilon$.

Check error and work.
Use a suitable severity-specific error calculation or bound, and calculate $m=\lfloor B/h\rfloor$. For a direct implementation of the recursion, smaller $h$ can substantially increase the number of operations.

Choose and validate.
Select a feasible width, then compare with a finer grid if the probability result is sensitive or an accuracy requirement is tight.

Worked Grid-Selection Examples

Worked Example: Satisfying Resolution and a Recursion Limit

A hypothetical portfolio has a Poisson claim count with $\lambda=3$ and individual losses uniformly distributed from $0$ to $\$100$. A calculation is required through $B=\$1{,}000$. The threshold representation must be within $\$25$, and the implementation permits at most 80 positive recursion indices. Choose the largest feasible grid width. For comparison, calculate the severity and aggregate mean loss from lower discretization.

Method. Convert the resolution and index limits into constraints on $h$. Then use the uniform severity model to calculate the mean loss exactly. This avoids treating a smaller grid as automatically preferable.

The resolution requirement gives $h\leq\$25$. The recursion limit requires

m=\left\lfloor\frac{1000}{h}\right\rfloor\leq80.

A width of $\$12.50$ would give $m=80$; the largest width satisfying the index limit is $\$12.50$. But the resolution requirement also allows widths up to $\$25$. Combining the two constraints carefully, $h$ must be at least $\$12.50$ and at most $\$25$. The largest feasible width is therefore $h=\$25$, giving $m=1000/25=40$.

With $X$ uniform on $[0,100]$, the four lower-grid values $0,25,50,75$ each have probability $0.25$. Thus

$$ E[X_h]=0(0.25)+25(0.25)+50(0.25)+75(0.25)=37.50. $$

The original mean is $E[X]=100/2=\$50$, so the per-claim mean reduction is $\$50-\$37.50=\$12.50$. The original compound Poisson aggregate mean is $3(50)=\$150$; the discretized mean is $3(37.50)=\$112.50$. The aggregate mean reduction is $\$37.50$. The general bound is $\lambda h=3(25)=\$75$, which correctly exceeds the calculated reduction.

As a work comparison, $h=\$12.50$ requires $m=80$ indices rather than 40. A straightforward recursion whose work grows approximately as $m^2$ can require about $(80/40)^2=4$ times the work. The selected $\$25$ grid meets both stated constraints with fewer indices. The check is that $h\leq25$ and $m=40\leq80$.

Worked Example: Enforcing an Aggregate Mean-Error Tolerance

A hypothetical compound Poisson model has $\lambda=4$, with severity uniform on $[0,100]$ dollars. The allowed downward error in the aggregate mean from lower discretization is at most $\$40$. The recursion must extend through $\$800$. Among grid widths that divide $\$100$, choose the largest width satisfying the mean-error tolerance, and find the number of positive indices.

Method. Calculate the mean discretization error for this uniform model rather than relying only on the general upper bound. This avoids confusing a sufficient bound with the exact error for the specified severity.

For a width $h$ that divides 100, the lower-grid levels are $0,h,\ldots,100-h$, each with probability $h/100$. Their average is $E[X_h]=(100-h)/2$. Since $E[X]=50$, the per-claim mean reduction is $h/2$, and the aggregate mean reduction is

4\left(50-\frac{100-h}{2}\right)=2h.

The tolerance requires $2h\leq40$, or $h\leq20$. The largest eligible width is therefore $h=\$20$. Its grid levels are $0,20,40,60,80$, each with probability $0.20$, so $E[X_h]=40$. The original aggregate mean is $4(50)=\$200$, and the discretized aggregate mean is $4(40)=\$160$, giving exactly $\$40$ of downward error.

The number of positive recursion indices through $\$800$ is $m=800/20=40$. The general bound is $\lambda h=4(20)=\$80$; it is valid but not sharp here. The calculated error satisfies the tolerance exactly, and the check $2h=40$ confirms that any larger eligible width would fail it.

Worked Example: Seeing the CDF Effect of a Coarse Grid

A hypothetical portfolio has a Poisson claim count with $\lambda=1$. Each claim is independently $\$20$ or $\$40$, with probability $1/2$ each. Compare the exact probability that aggregate loss is at most $\$40$ with the probability computed after lower discretization using $h=\$40$.

Method. Count the possible claim-count cases for the exact model, then identify the discretized severity probabilities and apply the Poisson thinning property. This makes clear how a coarse grid can alter a threshold probability, even when the grid-point calculation is internally exact.

For the original model, if $N=0$, aggregate loss is zero. If $N=1$, it is at most $\$40$. If $N=2$, it is at most $\$40$ only when both claims are $\$20$, an event with conditional probability $(1/2)^2=1/4$. If $N\geq3$, the aggregate is at least $\$60$. Since $N\sim\operatorname{Poisson}(1)$, the exact probability is

P(S\leq40)=e^{-1}+\left(e^{-1}\right) +\left(\frac{e^{-1}}{2!}\right)\left(\frac14\right) =e^{-1}(2.125)=0.7817.

With $h=40$, a $\$20$ claim maps to zero and a $\$40$ claim maps to $\$40$. Thus $g_0=g_1=0.5$. The number of claims mapped to the positive grid value is Poisson with mean $1(0.5)=0.5$. The discretized aggregate is at most $\$40$ if that number is zero or one, giving

P(S_h\leq40)=e^{-0.5}(1+0.5)=0.9098.

The discretized CDF exceeds the exact CDF by $0.9098-0.7817=0.1281$, or about 12.81 percentage points. This direction is consistent with lower discretization: $S_h\leq S$, so $P(S_h\leq B)\geq P(S\leq B)$. It is not a general error bound. The check is that the coarse grid has indeed moved the $\$20$ claims down to zero, making the discretized aggregate smaller.

Interpreting Accuracy and Computational Cost

For a fixed threshold $B$, the number of positive indices is approximately $B/h$. In a direct Panjer implementation, calculating each new $q_k$ involves a sum over earlier indices, so total work through $m$ is typically of order $m^2$, while storage can be of order $m$. Implementation details can change these costs, but the central trade-off remains: halving $h$ roughly doubles the index range and can roughly quadruple direct-recursion work.

The mean bound is useful when aggregate mean accuracy matters, but it does not control the error in every tail or CDF probability. In particular, a small mean difference does not prove a small error at a specific threshold. A careful calculation records the chosen severity discretization, the grid probability masses, the threshold index, and the result under at least one finer grid when precision is material.

Consideration	What the grid width controls	What it does not guarantee
Threshold representation	The threshold rounds down to a lattice point less than $h$ dollars away	A probability error of less than a fixed amount
Severity approximation	Each lower-discretized loss is less than $h$ below its original value	A small aggregate tail error based only on $h$
Computational work	The number of indices through $B$ is about $B/h$	That a fine grid is computationally feasible

Common Mistakes

Using the mean bound as a probability bound. The inequality for $E[S]-E[S_h]$ is about expected loss, not the difference between two threshold probabilities.
Forgetting the zero grid point. Under lower discretization, losses below $h$ are assigned to zero; this can materially affect the recursion.
Rounding the threshold up. Through threshold $B$, use $m=\lfloor B/h\rfloor$, not $\lceil B/h\rceil$. The latter includes a lattice point above the requested threshold.
Claiming a strict mean bound when $\lambda=0$. For compound Poisson frequency, $0\leq E[S]-E[S_h]\leq\lambda h$ holds for every $\lambda\geq0$; strict inequality at the upper end requires $\lambda>0$.
Choosing $h$ from computation alone. A grid that is cheap to compute may be too coarse to represent the threshold or severity distribution adequately.

Key takeaway: Select $h$ by combining a threshold-resolution constraint, a severity-error assessment, and a recursion-size limit. For lower discretization, $X_h\leq X$ and the compound Poisson mean reduction is at most $\lambda h$, with a strict bound when $\lambda>0$. Validate important probability results by refining the grid. This approach is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

Work through each problem using lower discretization and state the grid index or error calculation requested.

A compound Poisson model has $\lambda=2$ and $X$ uniform on $[0,60]$. For $h=15$, calculate $E[X_h]$ and the downward error in the aggregate mean. (Answer: $E[X_h]=\$22.50$; aggregate mean error $=\$15$.)
A recursion is needed through $B=\$1{,}800$, with threshold representation within $\$30$ and no more than 60 positive indices. What is the largest feasible width? (Answer: $h=\$30$, with $m=60$.)
Let $\lambda=3$, $X$ be uniform on $[0,120]$, and $h=\$30$. Calculate the original and discretized aggregate means and their difference. (Answer: $\$180$, $\$135$, and a $\$45$ downward difference.)
A Poisson($1$) claim count has severity $\$10$ or $\$30$, each with probability $1/2$. Under lower discretization with $h=\$20$, calculate $P(S_h\leq\$20)$. (Answer: $e^{-0.5}(1+0.5)=0.9098$.)
For the same threshold $B$, a direct recursion uses $m=60$ indices with one grid and $m=120$ with a finer grid. Approximately how does the work compare if it grows as $m^2$? (Answer: the finer grid requires about $(120/60)^2=4$ times the work.)

What You'll Learn