Which of the following is the credibility exposure formula ne?

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Multiple Choice

Which of the following is the credibility exposure formula ne?

The idea is to determine how much exposure is needed to achieve a target level of relative precision when estimating the mean of S. Start from the fact that the relative standard error of the sample mean is CV(S) divided by the square root of the number of observations n, where CV(S) = σ(S)/μ(S). In symbols, relative error ≈ CV(S) / sqrt(n). If you want the relative margin of error to match a specified bound tied to a z-value, you set CV(S)/sqrt(n) equal to k divided by z_{1+p/2}. Solving for n gives n = (z_{1+p/2} / k)^2 * [CV(S)]^2. This is precisely ne = ((z_{1+p/2}) / k)^2 * (CV of S)^2. Using CV makes the formula scale-free, capturing dispersion relative to the mean rather than relying on the mean or the variance alone. The other forms would tie exposure to the mean or to the variance directly, which doesn’t reflect how much data you need relative to the size of the mean when aiming for a specified relative precision. Example: if CV(S) = 0.2, z_{1+p/2} = 1.96 (for ~95% CI), and desired relative precision k = 0.05, then ne ≈ (1.96/0.05)^2 * 0.04 ≈ 61. So about 61 exposure units are needed to meet that precision target.

The idea is to determine how much exposure is needed to achieve a target level of relative precision when estimating the mean of S. Start from the fact that the relative standard error of the sample mean is CV(S) divided by the square root of the number of observations n, where CV(S) = σ(S)/μ(S). In symbols, relative error ≈ CV(S) / sqrt(n).

If you want the relative margin of error to match a specified bound tied to a z-value, you set CV(S)/sqrt(n) equal to k divided by z_{1+p/2}. Solving for n gives n = (z_{1+p/2} / k)^2 * [CV(S)]^2. This is precisely ne = ((z_{1+p/2}) / k)^2 * (CV of S)^2.

Using CV makes the formula scale-free, capturing dispersion relative to the mean rather than relying on the mean or the variance alone. The other forms would tie exposure to the mean or to the variance directly, which doesn’t reflect how much data you need relative to the size of the mean when aiming for a specified relative precision.

Example: if CV(S) = 0.2, z_{1+p/2} = 1.96 (for ~95% CI), and desired relative precision k = 0.05, then ne ≈ (1.96/0.05)^2 * 0.04 ≈ 61. So about 61 exposure units are needed to meet that precision target.

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