Deriving the Normal-Normal Posterior Mean

Why Precision Determines the Weight

When an insurer estimates the underlying risk level of a class or account, it may have both a collective estimate and a noisy observation of that risk’s own experience. The Bayesian normal-normal model gives a precise way to combine them. Its central result is that the observation receives more weight when it is more precise, while the prior receives more weight when the observation is noisy.

The derivation matters because the credibility weight is not an arbitrary blend. It follows directly from the information in the prior and the observation, measured by their precisions (the reciprocals of their variances). The result applies to a continuous quantity modeled as normal; it is not a general formula for every kind of claim data.

Definition: Let $\theta$ be an unknown risk-level quantity. Assume the prior distribution is $\theta\sim N(\mu_0,\tau^2)$, and, conditional on $\theta$, a noisy estimate $Y$ satisfies $Y\mid\theta\sim N(\theta,s^2)$. The prior mean is $\mu_0$, the prior variance is $\tau^2$, and the observation variance is $s^2$. These variances are assumed known, positive, and measured in squared units of $\theta$.

This model assumes that the observation is unbiased for $\theta$, its error is normal with constant variance $s^2$, and its error is conditionally independent of the prior information given $\theta$. The prior and observation variances describe different uncertainties: $\tau^2$ represents uncertainty about the risk level before seeing $Y$, while $s^2$ represents noise in the observation even if $\theta$ were known.

A frequent special case is a sample mean. If $X_1,\ldots,X_n$, conditional on $\theta$, are independent with $X_j\mid\theta\sim N(\theta,\sigma^2)$, then their mean $\bar X$ is normal with mean $\theta$ and variance $\sigma^2/n$. Thus $Y=\bar X$ and $s^2=\sigma^2/n$. The relevant uncertainty for the mean is not the variance of an individual observation.

Deriving the Posterior Distribution

Bayes’ rule says that the posterior density is proportional to the prior density times the likelihood. For a realized observation $Y=y$, discard factors that do not depend on $\theta$. The prior contributes an exponential term involving $(\theta-\mu_0)^2/\tau^2$, and the likelihood contributes one involving $(y-\theta)^2/s^2$. Therefore the part of the posterior density that depends on $\theta$ is

\pi(\theta\mid y)\ \propto\ \exp\left\{-\frac{1}{2}\left[ \frac{(\theta-\mu_0)^2}{\tau^2} +\frac{(y-\theta)^2}{s^2} \right]\right\}.

Expanding the two squares, the coefficient of $\theta^2$ inside the brackets is $1/\tau^2+1/s^2$, and the coefficient of $-2\theta$ is $\mu_0/\tau^2+y/s^2$. Define $A=1/\tau^2+1/s^2$ and $B=\mu_0/\tau^2+y/s^2$. Completing the square gives $A(\theta-B/A)^2$ plus a term that does not depend on $\theta$. A normal density with this quadratic term has mean $B/A$ and variance $1/A$. Hence the posterior is normal, with the following parameters:

\theta\mid y\sim N\left( \frac{\mu_0/\tau^2+y/s^2}{1/\tau^2+1/s^2}, \frac{1}{1/\tau^2+1/s^2} \right).

This derivation explains why precision, rather than standard deviation, is added: the quadratic terms in the log densities have coefficients equal to inverse variances. Rewriting the mean in a more recognizable form gives the posterior mean and variance:

E[\theta\mid y] =Zy+(1-Z)\mu_0,\qquad Z=\frac{\tau^2}{\tau^2+s^2} =\frac{1/s^2}{1/\tau^2+1/s^2}, \qquad \operatorname{Var}(\theta\mid y) =\frac{\tau^2s^2}{\tau^2+s^2}.

The observation’s credibility weight $Z$ is its precision divided by total precision. The prior’s weight is $1-Z=s^2/(\tau^2+s^2)$. These weights are between zero and one and sum to one. If the observation variance is small relative to prior variance, the observation is more precise and $Z$ is larger. If the observation is very noisy, $Z$ approaches zero and the posterior mean stays near the prior mean.

Theorem: Under the stated normal prior and normal observation model, the posterior mean is a weighted average of the prior mean and the observed estimate, with the observation weight equal to its share of total precision. The posterior variance is smaller than either input variance: $\tau^2s^2/(\tau^2+s^2)<\tau^2$ and $\tau^2s^2/(\tau^2+s^2)<s^2$.

For a sample mean from $n$ conditionally independent normal observations, substitute $s^2=\sigma^2/n$. The credibility weight on $\bar X$ then becomes $n\tau^2/(\sigma^2+n\tau^2)$. Increasing the sample size increases the precision of the sample mean, so its weight rises. This is an application of the same result, not a separate weighting rule.

Worked Applications

In each calculation, keep the units consistent and square standard deviations before using the formulas. The posterior mean estimates the unknown risk level; the posterior variance measures remaining uncertainty about that level under the model.

Worked Example: Combining a Prior With a Noisy Loss Estimate

A hypothetical account has an underlying annual loss-cost level $\theta$, measured in dollars. Before observing the account’s latest estimate, the actuary uses $\theta\sim N(1{,}000,200^2)$. The latest estimate is $y=1{,}300$, with observation standard deviation $300$ dollars, so $Y\mid\theta\sim N(\theta,300^2)$. Find the posterior mean and variance.

Method. Use the normal-normal update. The prior variance is $\tau^2=200^2=40{,}000$ dollars squared, and the observation variance is $s^2=300^2=90{,}000$ dollars squared. The observation weight is

Z=\frac{40{,}000}{40{,}000+90{,}000} =\frac{4}{13} \approx 0.307692.

The posterior mean is the prior mean plus the weight times the difference between the observed estimate and prior mean:

E[\theta\mid y] =1{,}000+\frac{4}{13}(1{,}300-1{,}000) =1{,}000+\frac{1{,}200}{13} \approx 1{,}092.31.

The posterior variance is

\operatorname{Var}(\theta\mid y) =\frac{40{,}000(90{,}000)}{130{,}000} \approx 27{,}692.31\text{ dollars squared}.

Check. The posterior mean, about $\$1{,}092.31$, lies between $\$1{,}000$ and $\$1{,}300$, and is closer to the prior because the observation has lower precision. The posterior variance is below both $40{,}000$ and $90{,}000$. The posterior standard deviation is about $\$166.41$, the square root of $27{,}692.31$. Common mistake avoided: weighting the observation by $s^2/(\tau^2+s^2)$ would assign more credibility to the noisier input; the observation weight is $\tau^2/(\tau^2+s^2)$.

Worked Example: Updating From a Sample Mean

For a hypothetical group of risks, let the unknown mean annual cost be $\theta$. The prior is $\theta\sim N(500,100^2)$. Conditional on $\theta$, 16 independent annual-cost observations are normally distributed with mean $\theta$ and standard deviation $200$. The observed sample mean is $\bar x=560$. Calculate the posterior mean and variance.

Method. The observation is the sample mean, whose variance is the individual-observation variance divided by sample size:

s^2=\frac{200^2}{16}=\frac{40{,}000}{16}=2{,}500.

The prior variance is $100^2=10{,}000$. Thus the credibility weight on the sample mean is

Z=\frac{10{,}000}{10{,}000+2{,}500}=0.8.

The posterior mean and variance are

E[\theta\mid\bar x]=500+0.8(560-500)=548, \qquad \operatorname{Var}(\theta\mid\bar x) =\frac{10{,}000(2{,}500)}{12{,}500}=2{,}000.

Check. The sample mean receives 80% weight because its variance is much smaller than the prior variance. The posterior mean $548$ lies between $500$ and $560$, and the posterior standard deviation is $\sqrt{2{,}000}\approx44.72$. Common mistake avoided: using the individual variance $40{,}000$ as the variance of $\bar X$ would ignore the 16 observations and underweight the data. The sample-mean variance is $40{,}000/16$.

Worked Example: Avoiding the Standard-Deviation Trap

A hypothetical portfolio’s loss-cost level is expressed in thousands of dollars. The prior is $\theta\sim N(70,12^2)$, and an experience-based estimate is $y=94$ with standard deviation $9$. Find the posterior mean, variance, and standard deviation.

Method. First convert both standard deviations to variances: $\tau^2=12^2=144$ and $s^2=9^2=81$, in squared thousands of dollars. The observation weight is

Z=\frac{144}{144+81}=\frac{144}{225}=0.64.

The posterior mean is

E[\theta\mid y]=70+0.64(94-70)=70+15.36=85.36.

The posterior variance and standard deviation are

\operatorname{Var}(\theta\mid y)=\frac{144(81)}{225} =\frac{11{,}664}{225}=51.84, \qquad \operatorname{SD}(\theta\mid y)=\sqrt{51.84}=7.2.

Check. The posterior estimate is $\$85{,}360$, with posterior standard deviation $\$7{,}200$; the variance is $51.84$ squared thousands of dollars. The mean lies between the prior mean of $\$70{,}000$ and observed estimate of $\$94{,}000$. Common mistake avoided: using 12 and 9 directly as variances gives an incorrect weight. The formula uses squared standard deviations, not standard deviations.

Assumptions, Interpretation, and Exam Traps

The model is useful when a continuous risk-level estimate and its sampling error are reasonably represented by normal distributions. The conditional observation must be centered at $\theta$, and the variances used in the formulas must be known or treated as fixed for the calculation. A materially biased estimate, dependent observation errors, or substantial uncertainty in the variance requires a different or expanded model.

A normal prior also assigns some probability to every real number. For a strictly positive quantity such as a loss cost, the normal approximation may be acceptable when its negative tail probability is negligible. If that probability is material, an untruncated normal model can produce implausible negative risk levels; a positive-support model may be more appropriate. Do not apply the normal-normal formulas mechanically when their assumptions are not credible.

Do not confuse variance and standard deviation. The update uses $\tau^2$ and $s^2$. Square each standard deviation first.
Do not reverse the credibility weight. The observation weight is its precision divided by total precision; equivalently, it is $\tau^2/(\tau^2+s^2)$.
Use the variance of the actual observation. For a sample mean, this is $\sigma^2/n$, not $\sigma^2$.
Keep the meaning of the posterior variance clear. It is uncertainty about $\theta$ after observing the data, not the variance of a future individual loss.
Check direction and bounds. The posterior mean must lie between the two means, and the posterior variance must be less than both input variances.

These calculations are aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus. In an exam calculation, identifying the observation and its variance before substituting is often the decisive step: it determines both the credibility weight and the uncertainty left after updating.

Check Your Understanding

For each problem, identify the observation variance before calculating the posterior quantities.

A hypothetical risk level has prior $N(40,10^2)$. A noisy estimate is 52 with standard deviation 15. Find the observation weight, posterior mean, and posterior variance. (Answer: $Z=4/13\approx0.3077$; mean $\approx43.6923$; variance $\approx69.2308$.)
A prior for a hypothetical annual cost is $N(120,30^2)$. An estimate is 105 with standard deviation 20. Calculate the posterior mean and variance. (Answer: mean $\approx109.6154$; variance $\approx276.9231$.)
Assume $X_1,\ldots,X_{25}\mid\theta$ are independent $N(\theta,50^2)$, the prior is $N(60,20^2)$, and the sample mean is 68. Find the sample-mean variance, posterior weight, posterior mean, and posterior variance. (Answer: sample-mean variance $100$; $Z=0.8$; mean $66.4$; variance $80$.)
A rate estimate is 3.0 with standard deviation 0.4, and its prior is $N(2.4,0.3^2)$. Find the posterior mean, variance, and standard deviation. (Answer: mean $2.616$; variance $0.0576$; standard deviation $0.24$.)
A prior is $N(110,20^2)$. Two conditionally independent estimates of the same risk level are 100 with standard deviation 10 and 130 with standard deviation 20. Combine the prior and both observations to find the posterior mean and variance. (Answer: posterior mean $\approx106.6667$; variance $\approx66.6667$.)

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