One-Sided and Two-Sided Credibility Criteria

Why the Number of Tails Matters

A credibility standard specifies how close observed experience should be to its expected value, and how often that closeness is required. The direction of the error matters. A requirement concerned only with an excess above expectation is a one-sided requirement; a requirement that controls both excesses and shortfalls is two-sided. At the same stated confidence level, the two-sided requirement generally needs a larger normal quantile, and therefore a larger claim count.

The essential task is to translate the wording into the probability event before choosing a quantile. “At most 10% above the expected count” describes one direction. “Within 10% of the expected count” describes both directions. Treating those statements as interchangeable can materially understate the required count.

Definition: Let \(N\) be a claim count with expected value \(\lambda\), and let \(r\) be the permitted relative error, where \(r>0\). A one-sided upper requirement at confidence level \(p\) controls \(P(N\leq (1+r)\lambda)\). A one-sided lower requirement controls \(P(N\geq (1-r)\lambda)\). A two-sided requirement controls \(P((1-r)\lambda\leq N\leq(1+r)\lambda)\). The confidence level \(p\) is the probability assigned to the stated event.

“Confidence level” alone does not identify the criterion: the direction, or whether both directions are controlled, must also be clear. For a two-sided requirement with total probability outside the interval \(\alpha=1-p\), the conventional normal approximation puts \(\alpha/2\) in each tail. For a one-sided requirement, the entire excluded probability \(1-p\) is in the specified tail.

Choosing the Correct Normal Quantile

Write \(\Phi(z)\) for the cumulative distribution function of a standard normal random variable. The one-sided quantile \(z_p\) satisfies \(\Phi(z_p)=p\). The two-sided quantile for central probability \(p\) satisfies \(\Phi(z)=1-(1-p)/2=(1+p)/2\). Thus, the quantile for a two-sided \(p\) requirement is \(z_{(1+p)/2}\), not \(z_p\).

For example, a one-sided 90% requirement uses \(z_{0.90}=1.2816\). A two-sided 90% requirement uses \(z_{0.95}=1.6449\), since 5% is placed in each tail. The confidence level is the same, but the two-sided criterion is more demanding because it protects against errors in both directions.

Result: Under a normal approximation, the relevant quantile is \(z_p\) for a one-sided confidence level \(p\), and \(z_{(1+p)/2}\) for a two-sided confidence level \(p\). A more demanding quantile leads to a larger required count when the tolerance \(r\) is unchanged.

Deriving the Count Requirement

For the limited-fluctuation count standard, suppose \(N\) follows a Poisson distribution with mean \(\lambda\). Then \(E[N]=\lambda\) and \(\operatorname{Var}(N)=\lambda\). For a sufficiently large mean, use the normal approximation \((N-\lambda)/\sqrt{\lambda}\approx N(0,1)\). This is an approximation, not an exact statement about a discrete count.

\frac{(1+r)\lambda-\lambda}{\sqrt{\lambda}}=r\sqrt{\lambda}, \qquad \frac{(1-r)\lambda-\lambda}{\sqrt{\lambda}}=-r\sqrt{\lambda}.

For an upper one-sided standard, the approximated probability of meeting the requirement is \(\Phi(r\sqrt{\lambda})\). For a lower one-sided standard, symmetry of the standard normal distribution gives the same probability. For the two-sided standard, the approximated probability is the central area between \(-r\sqrt{\lambda}\) and \(r\sqrt{\lambda}\). Requiring these probabilities to be at least \(p\) gives the corresponding count formulas.

\begin{aligned} \text{One-sided:}\quad & \Phi(r\sqrt{\lambda})\geq p &\Longrightarrow\quad & \lambda\geq \left(\frac{z_p}{r}\right)^2,\\ \text{Two-sided:}\quad & 2\Phi(r\sqrt{\lambda})-1\geq p &\Longrightarrow\quad & \lambda\geq \left(\frac{z_{(1+p)/2}}{r}\right)^2. \end{aligned}

The calculation produces a minimum required expected count. If the applicable count must be a whole number, round the result up to the next integer; rounding down would fail the inequality. The formulas presume the Poisson mean-variance relationship and the normal approximation. They should not be applied unchanged to a count model with a different variance or when an exact discrete probability is specifically required.

Translate the wording.
Decide whether the condition controls one specified direction or both directions, and identify the relative tolerance \(r\).

Identify the confidence probability.
Use the stated probability \(p\), distinguishing it from the tail probability.

Choose the quantile.
Use \(z_p\) for one side and \(z_{(1+p)/2}\) for two sides.

Calculate and round up.
Evaluate the appropriate squared-quantile formula, then check that the rounded count satisfies the unrounded threshold.

Worked Applications

Worked Example: Comparing 90% One-Sided and Two-Sided Standards

A hypothetical portfolio has a Poisson claim count with expected count \(\lambda\). A standard requires the observed count to be within 20% of its mean. Find the approximate minimum expected count for a 90% one-sided upper requirement and for a 90% two-sided requirement.

Method. Apply the normal-approximation limited-fluctuation formula separately, because the first condition controls only an excess and the second controls both excesses and shortfalls. The relative tolerance is \(r=0.20\).

One-sided upper requirement. The quantile is \(z_{0.90}=1.2816\). Therefore,

\lambda\geq\left(\frac{1.2816}{0.20}\right)^2 =(6.408)^2 =41.062464.

The minimum whole-number count is \(\lceil41.062464\rceil=42\). The check is \(42\geq41.062464\), while 41 is below the threshold.

Two-sided requirement. The central 90% interval leaves 5% in each tail, so the quantile is \(z_{0.95}=1.6449\). Thus,

\lambda\geq\left(\frac{1.6449}{0.20}\right)^2 =(8.2245)^2 =67.64240025.

The minimum whole-number count is \(\lceil67.64240025\rceil=68\). The check is \(68\geq67.64240025\), while 67 is below the threshold. The two-sided standard needs more expected claims because it controls deviations on both sides. The common mistake this comparison avoids is using \(1.2816\) for the two-sided 90% condition.

Worked Example: A 95% Requirement With a 25% Tolerance

For a hypothetical Poisson claim count, calculate the minimum expected count for (a) a 95% one-sided lower requirement and (b) a 95% two-sided requirement, both using a 25% relative tolerance.

Method. A lower requirement is still one-sided: it controls a shortfall, not both tails. Set \(r=0.25\). The one-sided 95% quantile is \(z_{0.95}=1.6449\); the two-sided 95% quantile is \(z_{0.975}=1.96\).

For the one-sided lower requirement,

\lambda\geq\left(\frac{1.6449}{0.25}\right)^2 =(6.5796)^2 =\frac{2.70569601}{0.0625} =43.29113616.

Round up to 44. The check is \(44\geq43.29113616\), and 43 is below the threshold.

For the two-sided requirement,

\lambda\geq\left(\frac{1.96}{0.25}\right)^2 =(7.84)^2 =\frac{3.8416}{0.0625} =61.4656.

Round up to 62. The check is \(62\geq61.4656\), and 61 is below the threshold. The two-sided count is larger because the same 95% probability must cover both sides of the mean. The mistake avoided here is treating “lower” as if it automatically meant “two-sided.”

Worked Example: Reading an Upper-Only Error Condition

A hypothetical portfolio’s Poisson claim count must be no more than 15% above its expected value with probability at least 90%. Find the approximate minimum expected count. Also calculate the count that would be required if the wording instead required the count to be within 15% on either side with 90% probability.

Method. “No more than 15% above” is an upper one-sided condition, so use \(z_{0.90}=1.2816\). Here \(r=0.15\):

\lambda\geq\left(\frac{1.2816}{0.15}\right)^2 =(8.544)^2 =\frac{1.64249856}{0.0225} =72.999936.

Round up to 73. The check is \(73\geq72.999936\), while 72 is below the threshold.

If the condition is instead within 15% on either side, use the 90% two-sided quantile \(z_{0.95}=1.6449\):

\lambda\geq\left(\frac{1.6449}{0.15}\right)^2 =\frac{2.70569601}{0.0225} =120.253156.

Round up to 121. The check is \(121\geq120.253156\), while 120 is below the threshold. This example highlights the wording trap: the first standard gives no guarantee about a shortfall, whereas “within 15%” imposes a condition on both tails.

Common Exam Traps and Practical Checks

Using the confidence level as the quantile subscript for both cases. A 90% two-sided condition uses \(z_{0.95}\), not \(z_{0.90}\), because half of the excluded 10% lies in each tail.
Ignoring the direction. “At least,” “no less than,” or “shortfall” usually signals a lower-tail condition; “at most,” “no more than,” or “excess” usually signals an upper-tail condition. Read the event precisely.
Rounding to the nearest integer. The inequality specifies a minimum. Always round a non-integer result upward.
Forgetting the model assumption. These formulas use a Poisson count and a normal approximation. A different count model, a small mean, or an exact-probability requirement calls for a different check.
Assuming equal confidence means equal sample size. At the same \(p\) and \(r\), the two-sided quantile is larger, so its required expected count is larger.

Key takeaway: First decide whether the probability statement is one-sided or two-sided. Then use \(z_p\) or \(z_{(1+p)/2}\), respectively. Under the Poisson normal approximation, divide that quantile by the relative tolerance, square the result, and round the required count upward.

The distinction is useful whenever a credibility standard is written in words rather than as a probability event. A single phrase such as “within the expected value” changes the tail allocation, quantile, and required count. Making the event explicit before calculating is a reliable way to avoid an otherwise easy-to-miss understatement.

Check Your Understanding

For each question, assume the Poisson normal approximation and round a required expected count up to a whole number.

A one-sided 90% standard permits a relative error of 10%. What is the minimum expected count? (Answer: 165)
A two-sided 90% standard permits a relative error of 10%. What is the minimum expected count? (Answer: 271)
A one-sided 95% lower-tail standard permits a relative error of 20%. What is the minimum expected count? (Answer: 68)
A two-sided 95% standard permits a relative error of 20%. What is the minimum expected count? (Answer: 97)
A one-sided 99% upper-tail standard permits a relative error of 25%. Using \(z_{0.99}=2.3263\), what is the minimum expected count? (Answer: 87)

What You'll Learn