Discretizing Severity for Aggregate Loss Calculations

Why Discretize Severity?

Aggregate loss calculations often begin with a continuous claim severity, while a calculation method may require severity probabilities on a discrete grid. Discretization replaces each possible loss by a nearby grid amount. This makes the severity model usable in discrete calculations, but it also changes the distribution of aggregate losses. The direction and size of that change depend on the rounding rule.

Let $X$ be a nonnegative claim severity and let $h>0$ be the grid spacing. The grid amounts are $0,h,2h,\ldots$. We will use three standard choices: round down (lower rounding), round up (upper rounding), and round to the nearest grid point (midpoint rounding). Unless stated otherwise, assume $X$ is continuous, so the probability of landing exactly on a grid boundary is zero.

Definition: For grid spacing $h$, define the lower-rounded severity $Y^L=h\lfloor X/h\rfloor$, the upper-rounded severity $Y^U=h\lceil X/h\rceil$, and the midpoint-rounded severity $Y^M=h\lfloor X/h+1/2\rfloor$. For a continuous severity, the choice at a midpoint tie does not affect probabilities. Lower rounding assigns $X$ to the grid point below it; upper rounding assigns it to the grid point above it; midpoint rounding assigns it to the nearest grid point.

The superscripts $L$, $U$, and $M$ identify rounding rules, not powers. In particular, $Y^L$ is a rounded payment amount, not a loss modified by a policy deductible. Each rule produces a discrete random variable, even when $X$ is continuous.

Finding the Discrete Severity Probabilities

It is useful to work in grid units first. For lower rounding, the probability assigned to $kh$ is the probability that the original severity lies in the interval from $kh$ up to, but not including, $(k+1)h$. For upper rounding, the interval assigned to $kh$ lies just below that grid point. For midpoint rounding, the interval is centered at the grid point.

\begin{aligned} P(Y^L=kh)&=P(kh\le X<(k+1)h), && k=0,1,2,\ldots,\\ P(Y^U=kh)&=P((k-1)h<X\le kh), && k=1,2,\ldots,\\ P(Y^M=0)&=P(0\le X<h/2),\\ P(Y^M=kh)&=P((k-\tfrac12)h\le X<(k+\tfrac12)h), && k=1,2,\ldots. \end{aligned}

For a nonnegative continuous severity, $P(Y^U=0)=0$. If the severity has a point mass at zero or at a grid boundary, endpoint probabilities must be handled according to the chosen rounding convention; the displayed interval formulas assume there is no such mass. Probabilities in the discrete severity model must sum to one across all grid points, including any tail grid points.

Worked Example: Exponential Severity on a Unit Grid

A hypothetical claim severity has an exponential distribution with rate $1$ per dollar, so $P(X>x)=e^{-x}$ for $x\ge0$. Use grid spacing $h=1$. Find the probabilities assigned to $0,1,2$ by each rounding rule.

Method. Apply the interval formulas to the exponential survival function. Write $q=e^{-1}=0.3678794$, so $e^{-2}=q^2=0.1353353$, $e^{-1/2}=0.6065307$, and $e^{-3/2}=0.2231302$.

For lower rounding, $P(Y^L=0)=P(0\le X<1)=1-e^{-1}=0.6321206$. Next, $P(Y^L=1)=P(1\le X<2)=e^{-1}-e^{-2}=0.2325442$, and $P(Y^L=2)=P(2\le X<3)=e^{-2}-e^{-3}=0.0855482$.

For upper rounding, the zero probability is $0$, because $X>0$ with probability one. The first two positive grid probabilities are $P(Y^U=1)=P(0<X\le1)=1-e^{-1}=0.6321206$ and $P(Y^U=2)=P(1<X\le2)=e^{-1}-e^{-2}=0.2325442$.

For midpoint rounding, $P(Y^M=0)=P(X<0.5)=1-e^{-0.5}=0.3934693$. The next two probabilities are $P(Y^M=1)=P(0.5\le X<1.5)=e^{-0.5}-e^{-1.5}=0.3834005$ and $P(Y^M=2)=P(1.5\le X<2.5)=e^{-1.5}-e^{-2.5}=0.1410452$.

Check. The probabilities shown need not sum to one because only the first three grid points are listed. For example, the lower-rounded tail probability at grid points $3$ and above is $P(X\ge3)=e^{-3}=0.0497871$; adding it to the three displayed lower probabilities gives $1.0000000$, apart from rounding.

How Rounding Changes an Aggregate Distribution

Suppose the aggregate loss is $S=\sum_{j=1}^{N}X_j$, where $N$ is the claim count and $X_j$ are claim severities. Apply the same rounding rule and grid to each claim, and write $S^L=\sum_{j=1}^{N}Y_j^L$, $S^U=\sum_{j=1}^{N}Y_j^U$, and $S^M=\sum_{j=1}^{N}Y_j^M$. The claim count and severity assumptions must be retained: rounding changes the severities, not the count model. When severities are independent with a common distribution and independent of $N$, the rounded claims remain independent with the corresponding discrete severity distribution.

The key comparison follows claim by claim. For every nonnegative severity value, $Y^L\le X\le Y^U$. Adding these inequalities over the claims gives $S^L\le S\le S^U$. Therefore, for any threshold $t$, the lower-rounded aggregate CDF is at least the continuous aggregate CDF, which in turn is at least the upper-rounded aggregate CDF.

S^L\le S\le S^U \quad\Longrightarrow\quad P(S^L\le t)\ge P(S\le t)\ge P(S^U\le t).

This is a pathwise argument: for each realized set of claims, the rounded aggregate is ordered relative to the actual aggregate. Midpoint rounding has no corresponding ordering. Some claims round upward and others downward, so $S^M$ can be either larger or smaller than $S$. As a result, midpoint rounding is not generally an upper or lower bound on aggregate probabilities.

Derivation: Conditional on $N=n$, a rounded aggregate is a sum of $n$ discrete severities. If $p_k=P(Y=kh)$, then for a threshold $mh$, its conditional probability is the sum of the probabilities of all $n$-tuples of grid indices whose total is at most $m$. Each tuple has probability equal to the product of its severity probabilities when claims are independent. Averaging those conditional probabilities over the claim-count probabilities gives $$ P(S^D\le mh)=\sum_{n=0}^{\infty}P(N=n)\! \sum_{\substack{k_1,\ldots,k_n\ge0\\k_1+\cdots+k_n\le m}} p_{k_1}\cdots p_{k_n}, $$ where $D$ denotes the chosen rounding rule. For $n=0$, the empty sum is zero, so the $N=0$ term contributes to $S^D=0$. This expression explains why severity rounding probabilities must be combined with the claim-count model rather than interpreted as aggregate probabilities by themselves.

Worked Aggregate Comparisons

Worked Example: Comparing All Three Rules at a Grid Threshold

A hypothetical portfolio has claim count $N$ with $P(N=0)=0.25$, $P(N=1)=0.50$, and $P(N=2)=0.25$. Given $N$, severities are independent exponential random variables with rate $1$ per dollar. Use $h=1$ and calculate the probability that the rounded aggregate is at most $2$, under each rounding rule. Also compare with the continuous aggregate probability.

Method. Condition on the three possible claim counts. For $N=0$, the aggregate is zero. For $N=1$, sum the severity probabilities at grid points $0,1,2$. For $N=2$, add the probabilities of all pairs of grid indices whose sum is at most $2$.

For lower rounding, the one-claim probability is $P(Y^L\le2)=P(X<3)=1-e^{-3}=0.9502129$. The two-claim event includes the pairs $(0,0),(0,1),(1,0),(0,2),(2,0),(1,1)$. With $p_0=0.6321206$, $p_1=0.2325442$, and $p_2=0.0855482$, its probability is $p_0^2+2p_0p_1+2p_0p_2+p_1^2=0.855799$ (approximately). Thus $P(S^L\le2)=0.25+0.50(0.9502129)+0.25(0.855799)=0.9391$, to four decimals.

For upper rounding, one claim is at most $2$ precisely when $X\le2$, giving $1-e^{-2}=0.8646647$. For two claims, each rounded claim is at least $1$; their rounded total is at most $2$ only when both round to $1$. The relevant probability is $(1-e^{-1})^2=0.3995764$. Therefore $P(S^U\le2)=0.25+0.50(0.8646647)+0.25(0.3995764)=0.7822$, to four decimals.

For midpoint rounding, the one-claim probability is $P(X<2.5)=1-e^{-2.5}=0.9179150$. Use $p_0=0.3934693$, $p_1=0.3834005$, and $p_2=0.1410452$, as calculated above. The two-claim probability is $p_0^2+2p_0p_1+2p_0p_2+p_1^2=0.714521$ (approximately). Hence $P(S^M\le2)=0.25+0.50(0.9179150)+0.25(0.714521)=0.8876$, to four decimals.

For comparison, the one-claim continuous probability is $P(X\le2)=1-e^{-2}=0.8646647$. The sum of two independent rate-$1$ exponentials has CDF $1-e^{-2}(1+2)=0.5939942$ at $2$. The continuous mixture probability is therefore $0.25+0.50(0.8646647)+0.25(0.5939942)=0.8308$, to four decimals.

Check. The results satisfy $0.9391\ge0.8308\ge0.7822$, as required for lower and upper rounding. The midpoint result, $0.8876$, is not a bound. The common mistake avoided here is treating a one-claim rounded CDF as the aggregate CDF; the $N=2$ term requires summing all qualifying pairs.

Worked Example: Two Rounded Exponential Claims at a Small Threshold

Two independent hypothetical severities are exponential with rate $1$ per dollar. Round each down to the nearest dollar. Find the probability that their rounded total is at most $1$, then compare it with the probability that their actual total is at most $1$.

Method. For the rounded total to be at most $1$, the grid-index pairs must be $(0,0),(0,1)$, or $(1,0)$. With $q=e^{-1}=0.3678794$, the lower-rounded probabilities are $p_0=1-q=0.6321206$ and $p_1=q(1-q)=0.2325442$. Thus $P(S^L\le1)=p_0^2+2p_0p_1$.

Substitution gives $(0.6321206)^2+2(0.6321206)(0.2325442)=0.693568$ (approximately), or $0.6936$ to four decimals. The expression can also be checked as $(1-q)^2(1+2q)$, which gives the same result.

The actual sum has the two-claim exponential-sum CDF $P(S\le x)=1-e^{-x}(1+x)$ for $x\ge0$. At $x=1$, this is $1-2e^{-1}=0.2642411$, or $0.2642$ to four decimals. The lower-rounded probability is larger, consistent with $S^L\le S$. The main trap is to include only the case where both claims round to zero; the two ordered cases with one claim rounding to $1$ also meet the threshold.

Worked Example: Midpoint Rounding Need Not Be a Bound

Two independent hypothetical severities are uniform on $[0,4]$, and the claim count is fixed at $N=2$. Round each severity to the nearest dollar using $h=1$. Find the probability that the rounded aggregate is at most $4$, and compare it with the actual aggregate probability.

Method. The midpoint bins have widths $0.5$ at grid points $0$ and $4$, and widths $1$ at grid points $1,2,3$. Since the severity density is $1/4$, the rounded probabilities are $p_0=0.125$, $p_1=p_2=p_3=0.250$, and $p_4=0.125$.

For the rounded total to be at most $4$, if the first grid index is $0$, the second can be $0$ through $4$, contributing $0.125(1)=0.125$. If the first index is $1$, the second can be $0$ through $3$, contributing $0.250(0.875)=0.21875$. For first indices $2,3,4$, the respective contributions are $0.250(0.625)=0.15625$, $0.250(0.375)=0.09375$, and $0.125(0.125)=0.015625$. The total is $0.609375$, or $0.6094$ to four decimals.

For the actual severities, the joint density is constant over the square $[0,4]\times[0,4]$. The region $x_1+x_2\le4$ is a right triangle with half the square’s area, so $P(S\le4)=0.5000$. Midpoint rounding gives a larger probability here, but this example does not make midpoint rounding an upper bound in general; other models and thresholds can give a smaller probability. The common mistake is to assume that nearest-grid rounding must preserve the direction of an aggregate inequality.

Practical Checks and Exam Traps

A good discretization calculation starts by identifying the grid and the event being evaluated. A threshold of $mh$ for the rounded aggregate is a threshold on the sum of grid indices, not a claim-by-claim threshold of $mh$. For example, at aggregate threshold $2h$, a pair with indices $(1,1)$ qualifies, but a pair with indices $(0,3)$ does not. This index interpretation prevents missing or adding combinations.

Do not confuse the rules. Lower rounding uses intervals $[kh,(k+1)h)$; upper rounding assigns the interval immediately below $kh$ to $kh$; midpoint rounding uses intervals centered on grid points.
Check the zero grid probability. For a strictly positive continuous severity, upper rounding has no mass at zero. Lower and midpoint rounding generally do.
Keep the frequency model in the calculation. The $N=0$ probability contributes directly to a nonnegative aggregate CDF at any threshold $t\ge0$, while $N=1,2,\ldots$ require their own conditional probabilities.
Use the right interpretation of a CDF. “At most $mh$” includes every discrete aggregate grid point from zero through $mh$, not only the point $mh$.
Do not claim a midpoint bound. The pathwise inequalities apply to lower and upper rounding, not to midpoint rounding.

Discretization is an approximation choice, not a change in the underlying claim-count model. Its effect can be substantial when the grid is coarse relative to typical claim sizes or when the requested aggregate threshold is near several rounded claim amounts. Lower and upper results provide useful bracketing checks; midpoint results require direct interpretation rather than an assumed direction of error. These distinctions are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

Use the stated severity and frequency assumptions. Give probabilities to four decimals where needed.

A severity is uniform on $[0,6]$. With $h=2$, what probability does lower rounding assign to the grid amount $2$? (Answer: $1/3$.)
A severity is exponential with rate $0.5$ per dollar. With $h=2$, what probability does upper rounding assign to the grid amount $4$? (Answer: $e^{-1}-e^{-2}\approx0.2325$.)
A severity is exponential with rate $1$ per dollar. With $h=1$, what probability does midpoint rounding assign to zero? (Answer: $1-e^{-0.5}\approx0.3935$.)
Two independent exponential severities have rate $1$ per dollar. Round both down with $h=1$. Find the probability their rounded total is at most $1$. (Answer: $p_0^2+2p_0p_1\approx0.6936$, where $p_0=1-e^{-1}$ and $p_1=e^{-1}(1-e^{-1})$.)
A hypothetical claim count has probabilities $P(N=0)=0.4$, $P(N=1)=0.4$, and $P(N=2)=0.2$. Given $N$, severities are independent uniform variables on $[0,2]$. Round down with $h=1$. Find the probability that the rounded aggregate is at most $1$. (Answer: $0.9500$.)

What You'll Learn