Absolute and Relative Precision Requirements

Why Precision Requirements Need a Probability Statement

A statement such as “the claim count should be accurate to within 25 claims” describes a tolerance, but it does not say how often that tolerance is expected to be met. A probability requirement supplies the missing standard: for example, the count must be within 25 of its target with probability at least 0.90. The distinction matters in limited-fluctuation credibility, where a precision standard is used to describe how reliable an observed result or estimate must be.

Precision can be measured in the original units, such as dollars or number of claims, or as a proportion of the target. These are different requirements. An error of $10,000 may be modest for a $2 million loss estimate but large for a $20,000 estimate. The first is an absolute error; the second comparison naturally suggests a relative error.

Definition: Let $T$ be a random quantity used to estimate or represent a target value $\theta$. An absolute-error requirement with tolerance $a>0$ and probability level $p$ is $P(|T-\theta|\le a)\ge p$. For a positive target $\theta>0$, a relative-error requirement with tolerance $r>0$ is $P(|T-\theta|/\theta\le r)\ge p$, equivalently $P(|T-\theta|\le r\theta)\ge p$. The tolerance describes the permitted error; $p$ describes the required probability of meeting it.

For a count, $a$ is measured in claims. For a frequency, it is measured in frequency units, such as claims per exposure. For a loss estimate, it is measured in dollars. A relative tolerance $r=0.10$, for example, means an error no greater than 10% of the specified target. The target in the denominator is $\theta$, not the random result $T$. Relative error is undefined when $\theta=0$, so a relative criterion needs a nonzero target.

Writing the Criterion Before Calculating

The safest first step is to write the event that counts as an acceptable result. For an absolute tolerance, the acceptable values of $T$ lie between $\theta-a$ and $\theta+a$. For a relative tolerance with a positive target, they lie between $\theta(1-r)$ and $\theta(1+r)$. This setup makes the center, width, units, and tail convention explicit.

\begin{aligned} \text{Absolute:}\quad &P(\theta-a\le T\le\theta+a)\ge p,\\ \text{Relative:}\quad &P\bigl(\theta(1-r)\le T\le\theta(1+r)\bigr)\ge p, \qquad \theta>0. \end{aligned}

These are two-sided criteria: results too far below the target and results too far above it both count as failures. If a problem instead specifies a one-sided condition, such as “no more than 5% below the target,” the event must be written with one tail only. A two-sided formula should not be substituted for a one-sided requirement.

If $T$ has a discrete distribution, use its actual probability model when an exact answer is required. For a claim count $N$, the absolute criterion is the sum of the probabilities of all integer counts satisfying $\theta-a\le N\le\theta+a$. A normal approximation may be used when appropriate; for an integer-valued count, a continuity correction adjusts the interval endpoints by one-half. Whether the resulting approximation is adequate depends on the distribution and its parameters.

Normal-Model Standardization

Suppose, for the purpose of evaluating a criterion, that $T$ is normally distributed with mean $\theta$ and standard deviation $\sigma>0$. Standardization uses $Z=(T-\theta)/\sigma$, which has the standard normal distribution. The distance from the target to either tolerance boundary, measured in standard deviations, determines the probability of meeting a symmetric requirement.

P(|T-\theta|\le a) = P\left(-\frac{a}{\sigma}\le Z\le\frac{a}{\sigma}\right) = \Phi\left(\frac{a}{\sigma}\right) - \Phi\left(-\frac{a}{\sigma}\right).

Here $\Phi$ is the standard normal cumulative distribution function. The formula follows directly by subtracting $\theta$ from each side of the interval and dividing by the positive quantity $\sigma$. By symmetry of the standard normal distribution, it can also be written $2\Phi(a/\sigma)-1$. For a relative tolerance, replace $a$ by $r\theta$, giving $2\Phi(r\theta/\sigma)-1$. The normal-model condition meets the required probability when this value is at least $p$.

This calculation depends on the stated normal model and the assumption that the distribution is centered at the target. If $E[T]\ne\theta$, the acceptable interval is not symmetric around the distribution mean, and the two endpoints must be standardized separately. The normal formula is not an exact probability for a discrete count; use the discrete model or identify the result as a normal approximation.

Specify the target and the tolerance.
State $\theta$, and write the absolute tolerance $a$ or relative tolerance $r$. Keep their units clear.

Write the acceptable event.
For a two-sided requirement, use $|T-\theta|\le a$ or $|T-\theta|\le r\theta$.

Use the specified probability model.
Calculate the event probability directly, or standardize under a justified normal model. Apply a continuity correction if approximating a discrete count.

Compare with the required level.
The criterion is satisfied only if the probability is at least $p$.

Worked Applications

Worked Example: Absolute Precision for a Claim Count

A hypothetical portfolio has an expected annual claim count of 400. For this calculation, approximate the count $N$ by a normal distribution with mean 400 and variance 400. Assess the absolute requirement that the count be within 30 claims of its expected value. Use a continuity correction and decide whether the probability is at least 0.90.

Method and event. The absolute tolerance is $a=30$, and the target is $\theta=400$. Since $N$ is integer-valued, $|N-400|\le30$ means $370\le N\le430$. The continuity-corrected interval is $369.5\le N\le430.5$.

Standardize. The standard deviation is $\sigma=\sqrt{400}=20$. The standardized endpoints are $(369.5-400)/20=-1.525$ and $(430.5-400)/20=1.525$.

Thus the normal approximation to the probability is $\Phi(1.525)-\Phi(-1.525)$, approximately 0.8727. This is below 0.90, so the requirement is not met under this approximation.

Check and trap avoided. The interval includes both endpoints, 370 and 430, which is why the continuity-corrected endpoints are 369.5 and 430.5. Omitting the correction would give a different approximation. The answer is an approximation, not an exact probability under a specified count distribution.

Worked Example: Relative Precision for a Claim Frequency

A hypothetical insurer estimates a claim frequency $T$ with target $\theta=0.060$ claims per exposure. Suppose $T$ is normally distributed with mean 0.060 and standard deviation 0.004. Find the probability that the estimate is within 10% of its target, and assess a required probability of 0.90.

Method and tolerance. The relative tolerance is $r=0.10$, so the absolute error permitted is $r\theta=0.10(0.060)=0.006$ claims per exposure. The acceptable interval is $0.060-0.006\le T\le0.060+0.006$, or $0.054\le T\le0.066$.

Standardize. The lower and upper standardized endpoints are $(0.054-0.060)/0.004=-1.5$ and $(0.066-0.060)/0.004=1.5$. Therefore the probability is $\Phi(1.5)-\Phi(-1.5)$, approximately 0.8664.

Since 0.8664 is less than 0.90, this estimate does not meet the stated probability requirement. The calculation is in frequency units throughout; the permitted error 0.006 is 10% of 0.060, not 10% of the standard deviation.

Check and trap avoided. The interval is centered on the target frequency. Treating 10% as an absolute tolerance of 0.10 claims per exposure would produce an interval far wider than the one specified.

Worked Example: Comparing Absolute and Relative Loss Precision

For a hypothetical aggregate-loss estimate $T$, assume $T\sim N(2{,}000{,}000,(100{,}000)^2)$. Compare (a) an absolute tolerance of $150,000 and (b) a relative tolerance of 8%. For each criterion, calculate the probability of meeting the tolerance and determine whether it reaches 0.90.

(a) Absolute criterion. The target is $\theta=\$2{,}000{,}000$, the tolerance is $a=\$150{,}000$, and the standard deviation is $\sigma=\$100{,}000$. The standardized tolerance is $a/\sigma=1.5$, so the probability is $\Phi(1.5)-\Phi(-1.5)$, approximately 0.8664. This is below 0.90.

(b) Relative criterion. The permitted error is $r\theta=0.08(\$2{,}000{,}000)=\$160{,}000$. The standardized tolerance is $160{,}000/100{,}000=1.6$, so the probability is $\Phi(1.6)-\Phi(-1.6)$, approximately 0.8904. This also falls short of 0.90.

Check and trap avoided. An 8% tolerance here is $160,000, not $8,000 and not 8% of the standard deviation. The relative criterion is slightly wider than the $150,000 absolute criterion, so its probability is higher, as the calculated values confirm. Neither probability reaches the specified standard.

Common Errors and Practical Checks

A precision calculation should be checked both algebraically and in context. The standardized tolerance $a/\sigma$, or $r\theta/\sigma$, must be dimensionless. If it still has units of dollars, claims, or claims per exposure, the standardization is incomplete. A larger tolerance or a smaller standard deviation should increase the probability of meeting a two-sided criterion; a result that does the opposite is a warning to recheck the work.

Confusing tolerance with probability. “Within 10 claims” specifies $a=10$; it does not specify $p=0.90$ or any other probability level.
Using a relative tolerance as an absolute amount. A relative tolerance must be multiplied by the target: $r\theta$.
Using the wrong denominator. Relative error is measured against the stated target $\theta$, not the observed estimate $T$.
Counting only one tail. A two-sided requirement includes results above and below the target. A one-sided condition needs a different event.
Treating an approximation as exact. A normal approximation to an integer count should be identified as such, with a continuity correction when appropriate.
Mixing units. Convert all quantities to compatible units before calculating. For example, a dollar tolerance cannot be divided by a standard deviation stated in thousands of dollars without conversion.

Key takeaway: Translate the wording into an event first. Absolute precision uses $a$; relative precision uses $r\theta$. Then calculate the probability under the stated model and compare it with the required level $p$.

Check Your Understanding

For each problem, write or evaluate the probability event before deciding whether the criterion is met.

A hypothetical claim-count estimate is approximated by a normal distribution with mean 900 and variance 900. What is the approximate probability that it is within 45 of its mean, using the continuous normal model without a continuity correction? (Answer: approximately 0.8664.)
A frequency estimate $T\sim N(0.025,(0.0025)^2)$ must be within an absolute tolerance of 0.005. Find the probability of meeting the requirement. (Answer: approximately 0.9545.)
A positive loss estimate $T\sim N(800,120^2)$ has target 800 dollars and a relative tolerance of 15%. Find the probability of meeting the criterion. (Answer: approximately 0.6827.)
A reserve estimate $T\sim N(5{,}000{,}000,(250{,}000)^2)$ must be within $400,000 of its target. Does it meet a required probability of 0.95? (Answer: No; the probability is approximately 0.8904.)
A frequency estimate $T\sim N(0.040,(0.004)^2)$ must be within 20% of its target. Find the probability of meeting the criterion and compare it with 0.95. (Answer: approximately 0.9545; it meets the requirement.)

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