Aggregate Probability Errors From Severity Discretization

Why Severity Discretization Changes Aggregate Probabilities

Aggregate recursions often require severity to take values on an integer grid. When the underlying loss amount does not lie on that grid, it must be assigned to a grid point. This is a computational representation of severity, not a change to the original loss model—and it can change the probability that aggregate loss is below a chosen threshold.

The effect depends on three things: the grid width, the rule used to assign a severity to the grid, and the threshold at which aggregate probabilities are compared. Rounding each claim down can create zero-valued grid losses and usually shifts aggregate losses downward. Rounding up shifts them upward. Rounding to the nearest grid point has no general direction of bias.

Definition: For a grid width $h>0$, a grid value is an integer multiple of $h$. Under floor discretization, a severity $X$ is represented by $X_h=h\lfloor X/h\rfloor$; under ceiling discretization, it is represented by $h\lceil X/h\rceil$. Under nearest-grid discretization, it is represented by the closest multiple of $h$, with a stated tie rule if ties are possible. The aggregate loss computed from these represented severities is a discretized aggregate loss, not necessarily the original aggregate loss.

To study the effect precisely, let $S$ be the original aggregate loss and $S_h$ the aggregate after discretizing each claim. If every individual severity is rounded down, then $S_h\leq S$ for every outcome. If every severity is rounded up, then $S\leq S_h$. Consequently, at any fixed threshold $x$,

P(S_h^{\mathrm{up}}\leq x) \leq P(S\leq x) \leq P(S_h^{\mathrm{down}}\leq x).

These bounds follow from event inclusion: under rounding up, $S_h^{\mathrm{up}}\leq x$ implies $S\leq x$; under rounding down, $S\leq x$ implies $S_h^{\mathrm{down}}\leq x$. Nearest-grid rounding does not supply either bound in general, because it can round some claims up and others down.

Recursion on a Severity Grid

Consider a compound Poisson model. The number of claims $N$ has a Poisson distribution with parameter $\lambda$, and claim severities are independent and identically distributed, independent of $N$, and represented on a grid of width $h$. Write $f_j=P(X_h=jh)$ for $j=0,1,\ldots$, and write $g_k=P(S_h=kh)$ for the aggregate grid probabilities. A zero-valued discretized severity is allowed; it can occur, for example, when a positive severity is rounded down.

Compound Poisson recursion: With the assumptions above, $$ g_0=e^{\lambda(f_0-1)},\qquad g_k=\frac{\lambda}{k}\sum_{j=1}^{k}j f_j g_{k-j},\quad k\geq 1. $$ The term $f_0$ affects $g_0$, but it does not introduce an extra factor in the recursion for $g_k$.

A short derivation explains why the formula applies. Let $F(z)=\sum_{j\geq0}f_jz^j$ be the probability generating function of one discretized severity, and $G(z)=\sum_{k\geq0}g_kz^k$ that of the aggregate grid index. For Poisson frequency, conditioning on $N$ gives $G(z)=\exp\{\lambda(F(z)-1)\}$. Differentiating yields $G'(z)=\lambda F'(z)G(z)$. Comparing coefficients of $z^{k-1}$ on both sides gives the recursion; the constant term of $G$ gives $g_0$.

G'(z)=\lambda F'(z)G(z) \quad\Longrightarrow\quad kg_k=\lambda\sum_{j=1}^{k}j f_j g_{k-j}.

The formula is for compound Poisson frequency and grid-valued severity probabilities. It is not a distribution-free formula for arbitrary claim counts. Once the grid probabilities have been determined from the selected rounding rule, the recursion calculates the aggregate probabilities on that grid. For a threshold $x$, the grid event $S_h\leq x$ includes indices $k$ satisfying $kh\leq x$, or equivalently $k\leq\lfloor x/h\rfloor$.

Worked Comparisons

Worked Example: A Coarse Floor Grid Changes the Claim-Count Cutoff

Suppose $N\sim\operatorname{Poisson}(1.5)$, and every claim has exact severity $1.6$ dollars. Compare the exact probability $P(S\leq 3.5)$ with the probability after rounding each severity down to the nearest dollar. Show the recursion calculation for the rounded model.

Method. First translate the threshold into the number of claims allowed under each severity representation. Then calculate the corresponding Poisson probabilities. The frequency parameter is $\lambda=1.5$.

For the exact severity, $S=1.6N$. The threshold allows at most two claims, since $2(1.6)=3.2\leq3.5$, while $3(1.6)=4.8>3.5$. Therefore

P(S\leq3.5)=P(N\leq2) =e^{-1.5}\left(1+1.5+\frac{1.5^2}{2}\right) =e^{-1.5}(3.625) =0.8088.

For the rounded model, every severity becomes $1$ dollar, so $f_1=1$, $f_0=0$, and $g_0=e^{-1.5}=0.2231302$. The recursion gives $g_k=(1.5/k)g_{k-1}$. Thus $g_1=1.5(0.2231302)=0.3346952$, $g_2=(1.5/2)(0.3346952)=0.2510214$, and $g_3=(1.5/3)(0.2510214)=0.1255107$. Since $3$ rounded dollars are within the $3.5$-dollar threshold, the rounded event includes $N\leq3$:

P(S_h\leq3.5)=g_0+g_1+g_2+g_3 =0.9343575 =0.9344.

The rounded probability is larger, as expected for floor discretization. The difference is $0.9343575-0.8088468=0.1255107$, or $0.1255$ to four decimal places. The comparison is checked by the ordering $S_h\leq S$.

Worked Example: Comparing Two Floor-Grid Widths at One Threshold

A hypothetical portfolio has $N\sim\operatorname{Poisson}(1.2)$. Each claim is $0.6$ dollars or $1.6$ dollars, each with probability $0.5$. Compare the recursively calculated probabilities of aggregate loss at most $1.4$ dollars using floor grids of width $h=1$ and $h=0.5$.

Method. For each grid, form the severity probabilities $f_j$, apply the compound Poisson recursion, and sum only grid outcomes at or below the threshold. The original two severities are not both positive grid losses under either floor rule.

For $h=1$, the rounded severities are $0$ and $1$, so $f_0=f_1=0.5$. Then $g_0=e^{1.2(0.5-1)}=e^{-0.6}=0.5488116$, and $g_1=(1.2/1)(1)(0.5)g_0=0.6g_0=0.3292870$. The threshold allows grid indices 0 and 1, so

P(S_1\leq1.4)=g_0+g_1 =0.8780986 =0.8781.

For $h=0.5$, the rounded severities are $0.5$ and $1.5$, corresponding to indices 1 and 3. Thus $f_1=f_3=0.5$, and $g_0=e^{1.2(0-1)}=e^{-1.2}=0.3011942$. The recursion gives $g_1=1.2(0.5)g_0=0.1807165$, and $g_2=(1.2/2)(1)(0.5)g_1=0.0542150$. The $1.5$-dollar grid severity exceeds the threshold, so the included indices are 0, 1, and 2:

P(S_{0.5}\leq1.4)=g_0+g_1+g_2 =0.5361257 =0.5361.

As a check, the exact event also requires no $1.6$-dollar claims and at most two $0.6$-dollar claims. The two claim-type counts are independent Poisson variables with means $0.6$ and $0.6$, so the exact probability is $e^{-1.2}(1+0.6+0.6^2/2)=0.5361$. The half-dollar floor grid happens to preserve this event at this threshold, whereas the one-dollar grid rounds the $0.6$-dollar claim to zero and substantially increases the probability.

Worked Example: Floor and Nearest Rounding Can Err in Opposite Directions

Suppose $N\sim\operatorname{Poisson}(1)$, and each claim is $0.7$ or $1.7$ dollars, with probability $0.5$ for each. At threshold $1.5$ dollars, compare the exact probability with floor and nearest-dollar discretization.

Method. Determine which claim-count combinations meet the threshold for each representation, then use the recursion or equivalent Poisson probabilities. The tie rule for nearest rounding is immaterial because neither severity is halfway between grid points.

In the exact model, the number of each claim type is independently Poisson with mean $0.5$. The event requires no $1.7$-dollar claims and at most two $0.7$-dollar claims, because $2(0.7)=1.4\leq1.5$ but $3(0.7)=2.1>1.5$. Hence

P(S\leq1.5)=e^{-0.5}e^{-0.5} \left(1+0.5+\frac{0.5^2}{2}\right) =e^{-1}(1.625) =0.5978.

Under floor rounding, the severities become 0 and 1 dollars. The positive grid-loss count is Poisson with mean $0.5$, so the threshold permits at most one such loss: $P(S_1\leq1.5)=e^{-0.5}(1+0.5)=0.9098$.

Under nearest-dollar rounding, the severities become 1 and 2 dollars, with $f_1=f_2=0.5$. The recursion gives $g_0=e^{-1}=0.3678794$ and $g_1=1(1)(0.5)g_0=0.1839397$. Only indices 0 and 1 meet the threshold, so $P(S_1\leq1.5)=g_0+g_1=0.5518$. The floor result is above the exact probability and the nearest-grid result is below it; nearest rounding is not a one-sided bound.

Interpreting Discretization Error Correctly

A difference between two aggregate probabilities can have several causes: rounding changes the probability assigned to each severity grid point, those changed severities affect the aggregate distribution, and the threshold may include a different set of grid indices. A threshold that lies between grid points is especially important. The event $S_h\leq x$ includes the largest grid amount not exceeding $x$, not the next grid amount.

Specify the original model.
Record the frequency distribution and parameters, the severity distribution, and whether the stated threshold applies to exact or discretized losses.

Define the grid rule.
State the grid width and whether each severity is rounded down, up, or to the nearest point. Determine all resulting $f_j$, including $f_0$ when relevant.

Calculate aggregate grid probabilities.
Use the recursion appropriate to the stated frequency model. For compound Poisson frequency, include $g_0=e^{\lambda(f_0-1)}$ and use the coefficient recursion for $k\geq1$.

Apply the threshold after recursion.
Sum $g_k$ over all $k$ with $kh\leq x$, then compare with the exact probability or another grid result at the same monetary threshold.

This order matters. Changing the grid while accidentally changing the threshold changes two things at once, so the resulting difference cannot be attributed solely to severity discretization. When comparing grid widths, keep the original monetary threshold fixed and use each grid's own index condition $k\leq\lfloor x/h\rfloor$.

Common Exam Trap

A frequent error is to keep the original claim-count cutoff after rounding severity. In the first worked example, the exact model allowed at most two claims because each exact claim cost $1.6$. After floor rounding, each grid claim costs $1$, so the rounded aggregate event at $3.5$ dollars allows three claims.

The incorrect calculation is to use $P(N\leq2)=0.8088$ for the rounded model as well. That calculation silently retains the exact-severity claim-count cutoff. The correction is to test the rounded aggregate amount: $3(1)=3\leq3.5$, while $4(1)=4>3.5$, so the rounded model requires $P(N\leq3)=0.9344$. Always convert the threshold using the severity representation currently being evaluated.

Key takeaway: Grid width and rounding rule can materially change aggregate probabilities, especially when floor rounding creates a zero-loss atom or moves the threshold's claim-count cutoff. State the discretization explicitly, recurse with the resulting severity probabilities, and compare all models at the same monetary threshold.

Check Your Understanding

For each problem, keep the stated monetary threshold fixed and account for the grid rule before summing aggregate probabilities.

A Poisson frequency has $\lambda=0.8$, and every exact claim costs $1.3$ dollars. Round claims down to whole dollars. Find the exact and discretized probabilities of aggregate loss at most $2.4$ dollars. (Answer: exact $0.8088$; discretized $0.9526$)
A compound Poisson model has $\lambda=0.8$, with grid severity probabilities $f_0=0.25$, $f_1=0.50$, and $f_2=0.25$, on a unit grid. Use recursion to find $P(S_h\leq2)$. (Answer: $0.9220$)
Let $N\sim\operatorname{Poisson}(1)$, with claim severity $0.6$ or $1.6$ dollars, each with probability $0.5$. Round down to a half-dollar grid. Find the probability that the discretized aggregate is at most $1.1$ dollars. (Answer: $0.5978$)
Let $N\sim\operatorname{Poisson}(1.4)$, and let every exact claim cost $2.2$ dollars. Round down to whole dollars. At threshold $6.5$ dollars, find both the exact and discretized aggregate probabilities. (Answer: exact $0.8335$; discretized $0.9463$)
At a fixed threshold $x$, a floor-grid calculation gives $P(S_h^{\mathrm{down}}\leq x)=0.72$, and a ceiling-grid calculation gives $P(S_h^{\mathrm{up}}\leq x)=0.61$. State the implied bound on the exact probability, assuming both calculations use the same underlying severity model and grid width. (Answer: $0.61\leq P(S\leq x)\leq0.72$)

What You'll Learn