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The Bühlmann credibility premium is the best affine mean squared error estimate of a risk’s conditional claim expected value. In the Bühlmann model it is $Z\overline{X}+(1-Z)m$, where $Z=na/(na+v)$. The linear least-squares projection equations use $\mathrm{Var}(X_j)=a+v$ and $\mathrm{Cov}(X_i,X_j)=a$ for distinct years.

At first order, the transverse drag depends on $g_t+V_1+{\Omega }_{1}x$. Its zero force and torque conditions make this residual orthogonal to $1$ and $x$. Thus $-V_1-{\Omega }_{1}x$ is the linear least-squares projection of the prescribed transverse deformation velocity onto the span of $1$ and $x$. This recasts a force-balance calculation as a projection.

Write $C_j={\sum }_{t=1}^T{y}_t\cos ({\omega }_{j}t)$ and $S_j={\sum }_{t=1}^T{y}_t\sin ({\omega }_{j}t)$. The discrete Fourier transform coefficient is $C_j-i{S}_j$. Alternatively, by the given orthogonality relations, the harmonic obtained by linear least-squares projection has coefficients $a_j=2C_j/T$ and $b_j=2S_j/T$. Thus the source's periodogram is the squared amplitude of a fitted harmonic, scaled as power per unit angular frequency:It measures how much of the centered record varies at frequency ${\omega }_j$. For odd $T$, Fourier Parseval identity givesThe positive-frequency spacing is $2\pi /T$, so summing the power ordinates times this spacing estimates the total variance. The factor $1/(\pi T)$ corresponds to the one-sided spectral density on $(0,\pi )$; the usual two-sided periodogram is half as large.

If the observations are Gaussian white noise with common variance ${\sigma }^2$, the $2m$ harmonic sums are jointly normally distributed. Their means are zero even with a nonzero constant series mean, because each sine and cosine sums to zero. Orthogonality givesUncorrelated coordinates of a multivariate normal distribution are independent. Therefore all $C_j/\sqrt{T{\sigma }^2/2}$ and $S_j/\sqrt{T{\sigma }^2/2}$ are independent standard normals, andSince the chi-squared distribution with two degrees of freedom has mean two and variance four,Here $f_{+}$ is the one-sided white-noise spectral density. Its exact unbiasedness is accompanied by a nonconcentrating distribution: along Fourier frequencies tending to any interior frequency, $I/f_{+}$ has the same nondegenerate exponential law at every $T$. It therefore does not converge in probability to one, proving failure of statistical consistency, not just nonvanishing variance.

For a general stationary process, exact finite-sample unbiasedness is not implied. Expanding the Fourier square instead giveswhere $\gamma$ is the autocovariance function. Under suitable short-memory regularity this tends to $f_{+}(\omega )$ at frequencies ${\omega }_j\to \omega$. The periodogram is thus generally asymptotically unbiased, while the exact assertion above is for white noise.

Use local periodogram smoothing to reduce variance: average $L_T$ neighboring ordinates with weights summing to one, where $L_T\to \infty$ but $L_T/T\to 0$. At an interior frequency the shrinking frequency window makes the bias tend to zero for smooth $f_{+}$, while averaging many weakly dependent ordinates makes the variance of order $1/L_T$. For white noise with equal weights, the ordinates are exactly independent and the variance is exactly $f_{+}^2/L_T$. For a general process, impose suitable short-memory and fourth-order dependence assumptions, such as a stationary Gaussian model with sufficiently summable autocovariances. These conditions, together with smoothness, give a consistent smoothed estimator. Boundary frequencies require an appropriately one-sided or reflected window.

Smooth spectral density alone is not enough. For example let $y_t=A{ϵ}_t$, where $A>0$ is a nondegenerate random variable constant throughout the record, independent of independent unit-variance Gaussian ${ϵ}_t$, with $E[A^2]<\infty$. The unconditional autocovariance is zero off lag zero and its one-sided spectral density is the smooth constant $E[A^2]/\pi$. Nevertheless a long local average estimates the realized $A^2/\pi$, not that deterministic unconditional value. This nonergodic example explains why a dependence/ergodicity condition accompanies the smoothing argument.

For the turning point test of randomness, the null is independent identically distributed continuous observations. Let $J_t$ be one if $y_t$ is a strict local maximum or minimum among $y_{t-1},y_t,y_{t+1}$, for $2\le t\le T-1$, and set $P_T={\sum }_t{J}_t$. Three ranks are equally likely; the middle observation is an extremum in four of the six orders, giving $E[J_t]=2/3$ and $\mathrm{Var}(J_t)=2/9$. ThereforeIndicators more than two positions apart depend on disjoint observations and are independent. For neighboring indicators, both are one precisely when four observations alternate up-down-up or down-up-down. By the probability-integral transform, use four independent uniforms. The first pattern has probabilityhere $b=y_t$, $c=y_{t+1}<b$, with the two outer inequalities having probabilities $b$ and $1-c$. The reverse pattern has the same probability, so $E[J_t{J}_{t+1}]=5/12$ and $\mathrm{Cov}(J_t,J_{t+1})=5/12-4/9=-1/36$.

At separation two, the triples share only one endpoint value $c$. Conditional on it, their other observations are independent. The probability that either triple's central observation is a turning point isIt follows that $E[J_t{J}_{t+2}]={\int }_0^{1}r(c{)}^{2}dc=9/20$, so this covariance is $9/20-4/9=1/180$. Combining all terms for $T\ge 4$ givesThe standardized count is approximately standard normal for large $T$, by a central limit theorem for bounded 2-dependent indicators. A two-sided randomness test rejects for large absolute values ofToo few turns suggests persistent movement, while too many suggests alternation. Exact rank-permutation calibration is possible at small $T$. Ties invalidate the continuous-rank probabilities unless their handling is specified. The test probes this feature of randomness; nonrejection does not establish independence.

Write $m(\lambda )=\mathbb{E}(X_i\mid \lambda )$ and $v(\lambda )=\mathrm{Var}(X_i\mid \lambda )$. The Bühlmann model uses the finite structural parametersHere $v$ is the expected process variance and $a$ is the variance of hypothetical means. The target is the latent conditional mean $m(\lambda )$, rather than the realized next count. The Bühlmann credibility estimate is the best affine predictor of that target under mean squared error.

The law of total expectation and law of total variance give $\mathbb{E}{X}_i=m_0$ and $\mathrm{Var}(X_i)=v+a$. Conditional independence gives $\mathrm{Cov}(X_i,X_j\mid \lambda )=0$ for $i\ne j$, and the law of total covariance therefore yieldsTo derive the optimal predictor, consider $d=A+{\sum }_{i=1}^n{b}_i{X}_i$. Minimizing with respect to the intercept gives $A=m_0(1-{\sum }_i{b}_i)$. Thus $d=m_0+{\sum }_i{b}_i(X_i-m_0)$. The linear least-squares projection normal equations areFor $v>0$, subtraction of any two equations forces all $b_i$ equal. Substituting a common coefficient then gives $b_i=a/(v+na)$. Equivalently, the mean squared error of the centered predictor is $a-2a{\sum }_i{b}_i+v{\sum }_i{b}_i^2+a({\sum }_i{b}_i{)}^2$, a convex quadratic with precisely these normal equations. HenceThe ratio notation assumes $a>0$; the credibility factor formula $na/(na+v)$ also handles $a=0,v>0$. If $a=0$, the target is the constant $m_0$ almost surely. If $v=0,a>0$, one observation already equals $m(\lambda )$ almost surely, and the average gives it exactly. If both vanish, the target and observations are constant.

The result optimizes over affine functions of the observations. It need not equal the unrestricted posterior mean; exact Bayesian inference generally depends on the whole prior and likelihood, whereas the Bühlmann credibility estimate uses these second-moment structural parameters.

Write the year-$j$ count as a sum of its $m_j$ individual counts, with $m_j>0$. The specified conditional independence within that year givesDividing by $m_j$ and scaling the conditional variance by $m_j^{-2}$ therefore givesThis is the Bühlmann–Straub model: larger exposures reduce the process noise in a year's average.

Define the population parameters and total observed exposure byHere $v$ is the expected process variance and $a$ the variance of hypothetical means. Assume finite second moments. The law of total variance gives $\mathrm{Var}(X_j)=a+v/m_j$, and conditional independence between years gives $\mathrm{Cov}(X_j,X_k)=a$ for $j\ne k$: their only shared variation is the latent conditional mean.

To derive the Bühlmann–Straub credibility estimate, minimize mean squared error among affine estimates of $\mu (\Theta )$. For coefficients $b_j$ and $B={\sum }_j{b}_j$, the optimal intercept is $m_0(1-B)$. Centering at $m_0$ and conditioning on $\Theta$ shows that the resulting error isThe cross term vanishes because $X_j-\mu (\Theta )$ has conditional mean zero; the conditional noise cross terms vanish by the between-year independence assumption.

For fixed $B$, the Cauchy-Schwarz inequality giveswith equality when $b_j=B{m}_j/W$. Minimize the remaining quadratic $a(1-B{)}^2+(v/W)B^2$. Its minimizing value is the Bühlmann–Straub credibility factorThe experience term is exposure-weighted:Since the future conditional mean count is $m_{n+1}\mu (\Theta )$, multiplying the optimal estimate by the known future exposure givesIt is the best affine linear least-squares projection, not a claim that the exact conditional expectation given all data is always affine. If $a=0$, the conditional mean is a known constant almost surely and one takes $Z=0$; if $v=0<a$, the observations reveal it without process noise and $Z=1$. The ordinary case has $a,v>0$.

In the Bühlmann model, a latent risk parameter $\Theta$ is drawn from a population distribution. Conditional on $\Theta$, the yearly observations are independent and identically distributed random variables, with conditional expectation $m(\Theta )$ and conditional variance $v(\Theta )$. Define the structural parametersHere $v$ is the expected process variance, while $a$ is the variance of hypothetical means. The Bühlmann credibility premium is the best affine estimate of $m(\Theta )$ from the observed claims, under mean squared error. Predicting the next claim gives the same affine estimate: the extra conditional observation noise contributes the constant $v$ to the prediction error.

The law of total variance and conditional independence giveAn affine estimate can be written as $\hat{m}=m+{\sum }_{j=1}^n{b}_j(X_j-m)$: for any chosen $b_j$, optimizing the constant makes its expected value equal to $m$. The normal equations for the linear least-squares projection areFor $v>0$ they force all $b_i$ to agree, with $b_i=a/(v+na)$. Thus the credibility factor and premium areThe credibility factor increases with the observation count and between-risk variance, and decreases with within-risk variance. If $a=0$, the risk mean is known and $Z=0$; if $v=0$ and $a>0$, one observation reveals it and $Z=1$. If both vanish, the premium is the fixed value $m$ and the factor is immaterial.

In the specified model, the conditional law is a gamma distribution with shape $\alpha$ and scale $\theta$. ThereforeThe prior is an inverse-gamma distribution with shape $k$ and scale $\lambda$. To obtain its moments directly, substitute $y=\lambda /\theta$ in the defining integral, obtainingThe Gamma function recurrence yieldsThe assumption $k>2$ makes both structural variances finite. HenceIt follows that the model-specific credibility estimate is

For the Bayes estimator under squared error loss, the quantity to estimate is $\alpha \Theta$, so the optimum is its posterior mean. The likelihood function, viewed as a function of $\theta$, is proportional toMultiplication by the prior shows gamma scale inverse-gamma conjugacy:Although the printed hint only mentions integer shapes, the same substitution and gamma integral normalize this posterior for every positive real shape, so no integrality of $\alpha$ is needed. Its posterior mean gives the Bayesian estimate and comparisonThis exact Bühlmann credibility for gamma claims holds for every observed sample, not merely on average. Here the posterior mean is affine in the sample mean, so the best affine Bühlmann credibility premium is also the unrestricted Bayes estimator under squared error loss.

Imagine an instrument measuring the same fixed quantity twice on each unit. Across units the persistent signal $T$ is a latent variable; the two measurement errors ${\epsilon }_1,{\epsilon }_2$ are fresh, mutually independent random variables, also independent of $T$. Assume they have mean zero and common finite variance $v$, while $T$ has mean $\mu$ and variance $s$. ThenThus the Pearson correlation coefficient is $\rho =s/(s+v)$ when $s+v>0$. Both measurements retain the same marginal mean and variance. Repetition does not make the entire population collapse toward $\mu$; it changes what we expect after selecting on $X$.

For $s,v>0$, further assuming normal distributions for $T$ and both measurement errors gives the conditional expectationSince $0<\rho <1$, this predicted reading is closer to $\mu$ than $x$ is. Independence and zero mean errors alone do not justify this linear conditional expectation: without the normal distribution assumption the same expression is the linear least-squares projection, and the actual conditional expectation can be nonlinear. Correlated errors, a changing signal, or different error variances require their own model; a noisy reading by itself supplies no universal shrinkage factor.