Why the Rounding Rule Matters
Aggregate-loss recursions commonly use a discrete severity distribution whose possible claim payments lie on a regular grid. When actual severities do not lie on that grid, they must be assigned to grid points or represented by probabilities at nearby points. That choice changes the severity distribution. It can change its mean, its variance, or both, and those differences carry through to an aggregate calculation.
A scheme is not automatically “better” because it is more convenient. Lower rounding avoids overstating individual losses, but tends to reduce the mean. Upper rounding tends to do the opposite. Nearest rounding may have a small mean error in some cases and a substantial one in others. Linear interpolation can preserve the mean exactly, but generally adds variance. The appropriate choice depends on what the calculation is intended to approximate.
For a value \(x\) between adjacent grid points, write \(kh\le x<(k+1)h\). At a grid point, the usual schemes leave the value unchanged. For nearest rounding, a tie rule must be specified; in this tutorial, an exact tie is assigned to the higher grid point.
Common Rounding Schemes
| Scheme | Rule for \(kh\le x<(k+1)h\) | Typical effect on the mean |
|---|---|---|
| Lower rounding | Replace \(x\) by \(kh\) | Does not increase the mean; strictly decreases it for values between grid points |
| Upper rounding | Replace \(x\) by \((k+1)h\) | Does not decrease the mean; strictly increases it for values between grid points |
| Nearest rounding | Replace \(x\) by the closer of \(kh\) and \((k+1)h\) | Can increase or decrease the mean |
| Linear interpolation | Assign probabilities to both adjacent points so their weighted mean is \(x\) | Preserves the mean |
The mean effect follows directly from the individual mapping: lower rounding replaces each loss by a value no greater than it, while upper rounding replaces each loss by a value no less than it. Nearest rounding has no such one-sided property. For linear interpolation, the probabilities are chosen to make the conditional expected grid value equal to the original loss.
Why Linear Interpolation Preserves the Mean—and Changes the Variance
Set \(p=(x-kh)/h\). Under linear interpolation, the rounded value is \(kh\) with probability \(1-p\) and \((k+1)h\) with probability \(p\). Its conditional expectation is
Averaging over the possible values of \(X\) therefore gives \(E[Y]=E[X]\). This is the reason the method preserves the mean; it is not an accidental property of a particular example.
The conditional variance of this two-point assignment is
If \(X\) itself is random and the interpolation is understood as a randomized assignment conditional on \(X\), the law of total variance gives
Thus linear interpolation preserves the mean and adds a nonnegative amount to the variance. The added amount is zero for losses already on grid points and positive for losses strictly between grid points. This is a property of this randomized, mean-preserving scheme; it should not be attributed to every rounding rule.
Worked Example: Comparing Nearest Rounding with the Original Moments
A hypothetical claim severity, measured in thousands of dollars, is \(1\), \(4\), or \(9\), with respective probabilities \(0.2\), \(0.5\), and \(0.3\). Use a grid spacing of \(h=5\) and nearest rounding. Find the original and rounded means and variances.
Method. Map each value to its nearest multiple of 5, then compute the first and second raw moments of both distributions. This avoids assuming that nearest rounding preserves the mean or variance.
The mappings are \(1\mapsto0\), \(4\mapsto5\), and \(9\mapsto10\). The rounded distribution therefore has probabilities \(0.2\) at 0, \(0.5\) at 5, and \(0.3\) at 10.
For the original severity,
Its second raw moment and variance are
For the rounded severity,
Nearest rounding increases the mean by \(5.5-4.9=0.6\) thousand dollars and increases the variance by \(12.25-8.49=3.76\) thousand-dollars squared. Check: the rounded probabilities sum to \(0.2+0.5+0.3=1\), and the rounded mean \(5.5\) lies between the grid minimum and maximum, 0 and 10.
Worked Example: Lower, Upper, and Mean-Preserving Rounding
A hypothetical severity is 2 thousand dollars with probability \(0.4\), and 7 thousand dollars with probability \(0.6\). On a grid with \(h=5\), calculate the mean and variance under lower rounding, upper rounding, and linear interpolation. Compare them with the original moments.
Method. Construct the severity distribution under each rule and calculate \(E[Y]\) and \(E[Y^2]-(E[Y])^2\). This avoids confusing a mean-preserving assignment with deterministic rounding.
The original mean is \(2(0.4)+7(0.6)=0.8+4.2=5\). The second raw moment is \(2^2(0.4)+7^2(0.6)=1.6+29.4=31\). Hence the original variance is \(31-5^2=6\), in thousand-dollars squared.
Under lower rounding, 2 maps to 0 and 7 maps to 5. Thus \(Y\) is 0 with probability \(0.4\) and 5 with probability \(0.6\). Its mean is \(0(0.4)+5(0.6)=3\), and its second raw moment is \(0+25(0.6)=15\). The variance is \(15-3^2=6\).
Under upper rounding, 2 maps to 5 and 7 maps to 10. The mean is \(5(0.4)+10(0.6)=2+6=8\), and the second raw moment is \(25(0.4)+100(0.6)=10+60=70\). The variance is \(70-8^2=6\).
For linear interpolation, 2 is assigned to 0 with probability \((5-2)/5=0.6\) and to 5 with probability \(2/5=0.4\). The value 7 is assigned to 5 with probability \((10-7)/5=0.6\) and to 10 with probability \(2/5=0.4\). Combining these conditional assignments gives probabilities \(0.4(0.6)=0.24\) at 0, \(0.4(0.4)+0.6(0.6)=0.52\) at 5, and \(0.6(0.4)=0.24\) at 10.
The interpolated mean is \(0(0.24)+5(0.52)+10(0.24)=2.6+2.4=5\). Its second raw moment is \(25(0.52)+100(0.24)=13+24=37\), so its variance is \(37-5^2=12\). The mean is preserved, while the variance increases from 6 to 12. Check: the interpolated probabilities sum to \(0.24+0.52+0.24=1\), and the mean agrees with the original mean of 5.
Choosing a Scheme for an Aggregate Calculation
A severity approximation is used inside a frequency-severity model, so a severity moment change may affect the aggregate distribution as well. For a compound Poisson model, let \(N\sim\operatorname{Poisson}(\lambda)\), assume claim severities are independent and identically distributed as \(Y\), and assume \(N\) is independent of the severities. For aggregate loss \(S=\sum_{j=1}^{N}Y_j\), the standard moment identities are \(E[S]=\lambda E[Y]\) and \(\operatorname{Var}(S)=\lambda E[Y^2]\). These identities show why an approximation that matches the severity mean need not match aggregate variance: the second raw moment matters too.
Worked Example: Selecting a Scheme for a Compound Poisson Calculation
A hypothetical portfolio has a Poisson claim count with parameter \(\lambda=2\). Each claim is 2 or 7 thousand dollars, with probabilities \(0.4\) and \(0.6\). A recursion requires severity points on the grid \(0,5,10,\ldots\), and the stated objective is to preserve the expected aggregate loss. Compare the lower-rounding and linear-interpolation aggregate means and variances.
Method. First find the severity moments under each scheme, then apply the compound Poisson moment identities. Since the objective is expected aggregate loss, this method avoids choosing lower rounding merely because its individual payments never exceed the original losses.
The original severity has \(E[X]=5\) and \(E[X^2]=31\), as calculated from \(2(0.4)+7(0.6)=5\) and \(2^2(0.4)+7^2(0.6)=31\). Its aggregate mean is \(2(5)=10\) thousand dollars, and its aggregate variance is \(2(31)=62\) thousand-dollars squared.
Under lower rounding, the severity is 0 with probability \(0.4\) and 5 with probability \(0.6\). Thus \(E[Y]=3\) and \(E[Y^2]=15\). The approximate aggregate mean is \(2(3)=6\) thousand dollars, and the variance is \(2(15)=30\) thousand-dollars squared.
Under linear interpolation, the severity probabilities are \(0.24\) at 0, \(0.52\) at 5, and \(0.24\) at 10. The moments are \(E[Y]=5\) and \(E[Y^2]=37\). The approximate aggregate mean is \(2(5)=10\) thousand dollars, and its variance is \(2(37)=74\) thousand-dollars squared.
Linear interpolation is the appropriate choice for the stated mean-preservation objective: its aggregate mean matches the original 10 thousand dollars, whereas lower rounding gives 6 thousand dollars. It does not preserve aggregate variance; here it gives 74 rather than 62 thousand-dollars squared. Check: both Poisson aggregate means equal \(\lambda\) times the corresponding severity mean, and both aggregate variances equal \(\lambda\) times the corresponding severity second raw moment.
Common Exam Traps and Practical Guidance
Rounding is not a harmless change of units. A grid spacing that is large relative to the variation in individual claim amounts can create meaningful moment differences. Before using a discretized severity in a recursion, identify what is being approximated and calculate the moments needed to evaluate the effect.
- Assuming nearest rounding preserves the mean. It does not. The direction and size of the error depend on where the probability mass lies relative to grid points.
- Confusing lower rounding with variance reduction. Lower rounding guarantees \(Y\le X\), so it cannot increase the mean. It does not guarantee that \(\operatorname{Var}(Y)\le\operatorname{Var}(X)\); variance depends on the full distribution.
- Stopping after checking the mean. Linear interpolation preserves the mean but can increase the variance. If aggregate dispersion matters, also calculate the second raw moment.
- Dropping a grid value of zero without checking the model. A discretization may assign probability to zero. That probability is part of the severity distribution and must be handled consistently in the calculation.
- Mixing units. If severities are stated in thousands of dollars, means are in thousands of dollars and variances are in thousand-dollars squared. Convert only after the calculation if a different reporting unit is required.
Check Your Understanding
For each problem, identify the rounding rule or moment being tested before calculating.
- A hypothetical severity is 3 or 8 thousand dollars, with probabilities \(0.25\) and \(0.75\). On a grid with \(h=10\), find its lower-rounded and upper-rounded means. (Answer: lower-rounded mean \(0\) thousand dollars; upper-rounded mean \(10\) thousand dollars.)
- A severity is 1, 3, or 6 thousand dollars with probabilities \(0.2\), \(0.5\), and \(0.3\). Use nearest rounding to a grid with \(h=4\). Find the original and rounded means. (Answer: original mean \(3.5\) thousand dollars; rounded mean \(4.4\) thousand dollars.)
- A severity is 3 or 8 thousand dollars with probabilities \(0.4\) and \(0.6\). Using linear interpolation on a grid with \(h=5\), find the probabilities at 0, 5, and 10, and the resulting variance. (Answer: probabilities \(0.16, 0.48, 0.36\); variance \(12\) thousand-dollars squared.)
- A compound Poisson model has \(\lambda=3\) and uses the linearly interpolated severity distribution from Question 3. Find the aggregate mean and variance. (Answer: mean \(18\) thousand dollars; variance \(144\) thousand-dollars squared.)
- A hypothetical severity is equally likely to be 2 or 9 thousand dollars. On a grid with \(h=5\), nearest rounding maps the values to 0 and 10. Find the original and rounded variances. (Answer: original variance \(12.25\) thousand-dollars squared; rounded variance \(25\) thousand-dollars squared.)