Discretization Choices for Continuous Severity

Why Discretization Choices Matter

Aggregate-loss recursions operate on discrete severities, but a claim-payment amount can follow a continuous distribution. To use a recursion, an actuary assigns each possible payment to a point on a grid. The grid spacing and assignment rule determine the resulting probability masses, and those choices can affect the aggregate probabilities and moments.

The severity to discretize is the payment amount that enters the aggregate model. If a contract has already changed a gross loss through a deductible, limit, or coinsurance, discretize the resulting payment severity, not the unmodified gross loss. That distinction matters because a contract can create payment atoms or a positive-payment threshold even when gross severity is continuous.

Definition: Let $X$ be a nonnegative continuous payment severity and let $h>0$ be a lattice spacing. A discretized severity $Y$ takes values $0,h,2h,\ldots$. Its lattice probability at $jh$ is $p_j=P(Y=jh)$. A discretization rule specifies which interval of values of $X$ is assigned to each lattice point.

The distribution function $F_X(x)=P(X\le x)$ provides the interval probabilities. For a continuous severity, endpoint choices do not affect the probabilities because individual endpoints have probability zero. The following common rules assign each interval differently.

Rule	Assignment	Typical effect on the mean
Lower (round down)	Assign $X$ to $h\lfloor X/h\rfloor$	Does not exceed $X$, so it tends to lower the mean
Upper (round up)	Assign $X$ to $h\lceil X/h\rceil$	Does not fall below $X$, so it tends to raise the mean
Nearest (midpoint)	Assign $X$ to the nearest lattice point	Balances rounding within each interval, but need not preserve the mean exactly

For example, under lower rounding, $p_j=P(jh\le X<(j+1)h)$ for $j\ge 0$. Under nearest rounding, $p_0=P(X<h/2)$, and $p_j=P((j-\tfrac12)h\le X<(j+\tfrac12)h)$ for $j\ge1$. The lattice probabilities must retain the probability of all possible severities, including a zero bucket or an upper-tail bucket if the chosen rule creates one.

How Interval Probabilities Enter the Aggregate Recursion

Once the probabilities $p_j$ are specified, a discrete aggregate model can be evaluated. For a Poisson claim count $N$ with mean $\lambda$, and independent severities $Y_1,Y_2,\ldots$ with lattice probabilities $p_j$, define $S=\sum_{r=1}^{N}Y_r$. Write $g_k=P(S=kh)$. The Poisson aggregate recursion is

g_0=\exp\{\lambda(p_0-1)\},\qquad g_k=\frac{\lambda}{k}\sum_{j=1}^{k}j p_j g_{k-j},\quad k\ge1.

This formula uses lattice units: $k$ is an index, while the corresponding aggregate amount is $kh$. A rounded severity can have $p_0>0$ even if the original continuous payment is strictly positive. In that case, $g_0$ is the probability of zero aggregate payment under the discretized model, not necessarily the probability of no claims.

Justification: The probability generating function of one discretized severity is $P_Y(z)=\sum_{j\ge0}p_jz^j$. For a Poisson count, the aggregate generating function is $G(z)=\exp\{\lambda(P_Y(z)-1)\}$. Differentiating gives $G'(z)=\lambda P_Y'(z)G(z)$. The coefficient of $z^{k-1}$ on the left is $kg_k$; on the right it is $\lambda\sum_{j=1}^k j p_jg_{k-j}$. Equating coefficients yields the recursion. Setting $z=0$ in $G$ gives $g_0=\exp\{\lambda(p_0-1)\}$. The derivation assumes a Poisson count independent of identically distributed, independent severities; a different frequency model requires its corresponding recursion.

Worked Examples

Worked Example: Comparing Three Lattices for a Uniform Payment

A hypothetical payment severity $X$ is uniform on $[0,4]$ dollars. Use a grid spacing of $h=\$1$ and calculate the mean under lower, upper, and nearest rounding. Compare each result with the continuous mean.

Method. For $X\sim\operatorname{Uniform}(0,4)$, the continuous mean is $(0+4)/2=\$2$, and each interval of length $\$1$ has probability $1/4$.

Lower rounding. The intervals $[0,1),[1,2),[2,3),[3,4]$ are assigned to $0,1,2,3$, respectively. Thus $p_0=p_1=p_2=p_3=1/4$, and

E[Y]=0\left(\frac14\right)+1\left(\frac14\right)+2\left(\frac14\right)+3\left(\frac14\right)=\$1.50.

Upper rounding. Those same four intervals are assigned to $1,2,3,4$. Each point has probability $1/4$, so

E[Y]=1\left(\frac14\right)+2\left(\frac14\right)+3\left(\frac14\right)+4\left(\frac14\right)=\$2.50.

Nearest rounding. Values below $0.50$ round to 0, and values from $3.50$ through 4 round to 4. Therefore $p_0=p_4=1/8$, while $p_1=p_2=p_3=1/4$. The mean is

E[Y]=0\left(\frac18\right)+1\left(\frac14\right)+2\left(\frac14\right)+3\left(\frac14\right)+4\left(\frac18\right)=\$2.00.

The nearest rule preserves the mean in this particular example, while lower and upper rounding shift it by $-\$0.50$ and $+\$0.50$. The mean match is not guaranteed for other distributions or grid widths. The common mistake is to assume that nearest rounding always preserves the mean; here it does so because the uniform distribution is symmetric across the grid intervals.

Check. Each set of lattice probabilities sums to 1, and the lower and upper approximations lie on opposite sides of the continuous mean.

Worked Example: Midpoint Probabilities in a Poisson Aggregate Model

A hypothetical portfolio has a Poisson number of claims with mean $\lambda=2$. Each continuous claim payment is exponential with mean $\$2$, so $F_X(x)=1-e^{-x/2}$ for $x\ge0$. Use nearest rounding with $h=\$1$ to calculate the probabilities of aggregate payments of $\$0$ and $\$1$.

Step 1: Calculate the severity masses needed. Nearest rounding puts $X<0.50$ at 0 and $0.50\le X<1.50$ at 1. Hence

p_0=1-e^{-0.25}=0.2212,\qquad p_1=F_X(1.50)-F_X(0.50)=e^{-0.25}-e^{-0.75}=0.3064.

Step 2: Calculate the zero aggregate probability. Apply the Poisson recursion’s initial value:

g_0=e^{2(p_0-1)} =e^{2(0.221199-1)} =e^{-1.557602} =0.2106.

Step 3: Calculate the probability of one lattice unit. For $k=1$, only the $j=1$ term appears:

$$ g_1=2p_1g_0 =2(0.306434)(0.210642) =0.1291. $$

Because $h=\$1$, $g_1$ is the probability of an aggregate payment of $\$1$. The discretized $p_0$ creates a zero-payment severity category, so $g_0$ includes outcomes with claims that round to zero as well as outcomes with no claims. A common mistake is to interpret $g_0$ as just the probability of no claims, $e^{-2}=0.1353$. That is the no-claim probability, not the zero aggregate-payment probability after rounding.

Check. The severity masses used are interval probabilities from the stated exponential distribution, and $g_1$ uses the $k=1$ recursion with the same dollar grid.

Worked Example: Keeping the Upper Tail When the Calculation Has a Cutoff

A hypothetical claim payment $X$ is exponential with mean $\$1$. Use nearest rounding with $h=\$1$, but store the distribution using only the lattice points 0, 1, 2, and 3. Assign every payment that rounds to 3 or higher to the point 3. Find the resulting severity probabilities and mean, then explain what they imply for aggregate calculations.

Step 1: Calculate the probabilities below the tail bucket. Since $F_X(x)=1-e^{-x}$, nearest rounding gives

p_0=F_X(0.5)=1-e^{-0.5}=0.393469,

p_1=F_X(1.5)-F_X(0.5)=e^{-0.5}-e^{-1.5}=0.383401,

p_2=F_X(2.5)-F_X(1.5)=e^{-1.5}-e^{-2.5}=0.141045.

Step 2: Put the remaining probability in the tail bucket. All values $X\ge2.5$ are assigned to 3, so $p_3=P(X\ge2.5)=e^{-2.5}=0.082085$. The mass check is

$$ p_0+p_1+p_2+p_3 =0.393469+0.383401+0.141045+0.082085 =1.000000. $$

Step 3: Calculate the discretized mean.

E[Y]=0p_0+1p_1+2p_2+3p_3 =0.383401+0.282090+0.246255 =\$0.911746\approx \$0.91.

The continuous mean is $\$1$, so assigning the entire upper tail to 3 underestimates the mean. The approximation is nevertheless a valid probability distribution because its masses sum to 1. If a Poisson aggregate calculation with $\lambda=1$ were needed only through aggregate index $k=2$, the $p_3$ mass would not enter $g_0,g_1,$ or $g_2$. It still matters for larger aggregate amounts and for the mean.

Common mistake and correction. A tempting but incorrect step is to discard $p_3$ and divide $p_0,p_1,p_2$ by their sum, $0.917915$, to make the retained probabilities sum to 1. That would describe the severity conditional on $X<2.5$, not the original payment model. Keep $p_3$ as a tail bucket when approximating the full distribution, or state explicitly that the model is conditional on falling below the cutoff.

Check. The tail bucket has probability $P(X\ge2.5)$, and its assigned value is 3, not the conditional mean of the unbounded tail. That assignment explains the downward mean bias.

Choosing a Rule and Treating the Tail

A smaller $h$ usually gives a finer representation of the severity distribution, but it also requires more lattice points and more recursion calculations. A larger $h$ can be computationally convenient, while making rounding effects more consequential. The appropriate choice depends on the aggregate quantities required and on how much error the application can tolerate.

The tail convention must match the calculation. If the aim is to compute aggregate probabilities only through $Kh$, severity masses at lattice points above $Kh$ cannot contribute to those probabilities: a single such severity already exceeds the target amount. Do not renormalize the probabilities up to $K$; doing so changes the model. If the aim includes aggregate moments or probabilities at larger amounts, retain and represent the tail rather than silently dropping it.

When a contract creates an atom—for example, a payment capped at a policy limit—the atom should be handled as an actual probability mass, not treated as though the severity remained continuous at that point. For an already discrete or mixed payment severity, calculate each interval probability with the correct endpoint and atom treatment. The convenient endpoint simplification for continuous distributions does not apply to atoms.

State the units. If $h=\$100$, an aggregate index $k=7$ represents $\$700$, not $\$7$.
Track the zero mass. Rounding can create $p_0>0$, which changes the probability of zero aggregate payment.
Check the total mass. The complete severity lattice, including any tail bucket, should sum to 1.
Check the direction of bias. Lower rounding cannot increase any individual severity; upper rounding cannot decrease one. Nearest rounding has no universal direction of mean bias.
Keep the contract order clear. Discretize the payment distribution that the aggregate model is intended to sum.

Key takeaway: A severity lattice is a modeling approximation, not just a computational format. Specify the grid, interval assignment, zero-payment treatment, and tail convention; then verify the probability mass and interpret aggregate results in the grid’s original units.

Check Your Understanding

For each problem, use the stated discretization rule and keep probability masses on the original model unless the question explicitly conditions on a restricted range.

A payment is uniform on $[0,3]$ dollars. With $h=\$1$ and nearest rounding, give $p_0,p_1,p_2,p_3$ and the discretized mean. (Answer: $p_0=p_3=1/6,\ p_1=p_2=1/3;\ E[Y]=\$1.50$.)
A payment is exponential with mean $\$1$. With $h=\$1$ and lower rounding, calculate $p_0$ and $p_1$. (Answer: $p_0=1-e^{-1}=0.6321,\ p_1=e^{-1}-e^{-2}=0.2325$.)
A Poisson claim count has mean 3. The discretized severity has $p_0=0.20$ and $p_1=0.50$, with the remaining mass at lattice points greater than 1. Using the Poisson recursion, find $g_0$ and $g_1$. (Answer: $g_0=e^{-2.4}=0.0907,\ g_1=3(0.50)g_0=0.1361$.)
A payment is exponential with mean $\$2$. Use nearest rounding with $h=\$1$, assigning all $X\ge2.5$ to 3. Calculate the probabilities at 0, 1, 2, and 3 and the discretized mean. (Answer: $p_0=0.2212,\ p_1=0.3064,\ p_2=0.1859,\ p_3=0.2865;\ E[Y]=\$1.5377$.)
A Poisson claim count has mean 2. Each payment is exponential with mean $\$1$, rounded to the nearest dollar with $h=\$1$. Find the probability of zero aggregate payment. (Answer: $p_0=1-e^{-0.5}=0.3935$, so $g_0=e^{2(p_0-1)}=0.2973$.)

Rule	Assignment	Typical effect on the mean
Lower (round down)	Assign \(X\) to \(h\lfloor X/h\rfloor\)	Does not exceed \(X\), so it tends to lower the mean
Upper (round up)	Assign \(X\) to \(h\lceil X/h\rceil\)	Does not fall below \(X\), so it tends to raise the mean
Nearest (midpoint)	Assign \(X\) to the nearest lattice point	Balances rounding within each interval, but need not preserve the mean exactly

What You'll Learn