Use the gamma distribution shape-rate convention: with . The conditional Poisson distribution has mean and variance both equal to , so the Bühlmann–Straub model parameters are
For and , the Bühlmann–Straub credibility factor is . Thus
and the required expected-count estimate is
The total exposure determines how much information the observed count carries; the number of years alone is not the appropriate denominator.
Write the year- count as a sum of its individual counts, with . The specified conditional independence within that year gives
Dividing by and scaling the conditional variance by therefore gives
This is the Bühlmann–Straub model: larger exposures reduce the process noise in a year's average.
Define the population parameters and total observed exposure by
Here is the expected process variance and the variance of hypothetical means. Assume finite second moments. The law of total variance gives , and conditional independence between years gives for : their only shared variation is the latent conditional mean.
To derive the Bühlmann–Straub credibility estimate, minimize mean squared error among affine estimates of . For coefficients and , the optimal intercept is . Centering at and conditioning on shows that the resulting error is
The cross term vanishes because has conditional mean zero; the conditional noise cross terms vanish by the between-year independence assumption.
For fixed , the Cauchy-Schwarz inequality gives
with equality when . Minimize the remaining quadratic . Its minimizing value is the Bühlmann–Straub credibility factor
The experience term is exposure-weighted:
Since the future conditional mean count is , multiplying the optimal estimate by the known future exposure gives
It is the best affine linear least-squares projection, not a claim that the exact conditional expectation given all data is always affine. If , the conditional mean is a known constant almost surely and one takes ; if , the observations reveal it without process noise and . The ordinary case has .