Normal-Normal Credibility for Noisy Loss Estimates

Why Model a Loss Estimate as Noisy?

A risk-level estimate is not necessarily the underlying expected loss for that risk. It may be an average based on a limited number of observations, so it can vary substantially from one sample to another even if the risk's true mean stays fixed. Normal-normal credibility represents these two sources of variation separately: a likelihood describes the noise in the observed estimate, and a prior describes variation in underlying means across risks.

This setup is useful when a risk's experience is summarized by a continuous loss estimate and the sampling error of that estimate is reasonably modeled as normal with known variance. The key modeling question is not simply whether a loss amount looks bell-shaped. It is whether the estimator, conditional on the risk's underlying mean, has the specified distribution.

Definition: Let $\theta$ be the unknown expected loss for a particular risk, and let $X$ be an observed estimate of that mean. A normal likelihood with known sampling variance assumes $$ X\mid\theta\sim N(\theta,\sigma^2), $$ where $\sigma^2$ is known and describes the variability of the estimate around $\theta$. A normal prior assumes $$ \Theta\sim N(\mu_0,\tau^2), $$ where $\mu_0$ is the portfolio-level prior mean and $\tau^2$ describes variation in underlying risk means. In $N(\mu,\text{variance})$, the second parameter is the variance, not the standard deviation.

The notation $\Theta$ emphasizes that before observing a particular risk's experience, its underlying mean is treated as varying across the population of risks. Once a particular risk is selected, its mean is the fixed but unknown value $\theta$. Conditional on that value, $X$ varies because of sampling noise. These are different kinds of uncertainty, and their variances must not be interchanged.

For example, an insurer might summarize a hypothetical commercial account's annual loss cost with an estimate of $90,000. The account's underlying expected annual loss need not equal $90,000: the estimate may be based on limited experience. The likelihood quantifies how estimates can fluctuate around the underlying mean. The prior represents the distribution of those means across comparable risks, based on information available before this account's own experience is used.

Building the Likelihood and Prior

For one observed estimate $x$, the likelihood as a function of a candidate mean $\theta$ is the normal density evaluated at $x$:

L(\theta;x)=\frac{1}{\sqrt{2\pi\sigma^2}} \exp\left(-\frac{(x-\theta)^2}{2\sigma^2}\right).

The likelihood is largest at $\theta=x$, but it does not say that $x$ is known to be the true mean. Its spread is controlled by $\sigma^2$: a smaller sampling variance makes the likelihood more concentrated, while a larger one makes a range of candidate means more consistent with the observed estimate.

The prior density is

\pi(\theta)=\frac{1}{\sqrt{2\pi\tau^2}} \exp\left(-\frac{(\theta-\mu_0)^2}{2\tau^2}\right).

Here $\mu_0$ and $\tau^2$ must be on the same scale as $\theta$. If the estimate is a loss per policy-year, the prior mean and prior standard deviation must also be loss per policy-year. A prior variance of $\tau^2$ is not a prior standard deviation of $\tau^2$; its standard deviation is $\tau$.

Worked Example: Evaluating a Noisy Estimate's Likelihood

A hypothetical insurer estimates an account's expected annual loss at $x=120$ thousand dollars. Assume $X\mid\theta\sim N(\theta,20^2)$, where the known standard error is 20 thousand dollars, and use the prior $\Theta\sim N(100,15^2)$, with all amounts in thousands of dollars. Specify the likelihood, compare the likelihood at two candidate means, and state the prior density at its mean. No interest discounting or mortality assumption enters this one-year loss-estimate model.

Step 1: Identify the parameters. The likelihood variance is $\sigma^2=20^2=400$ thousand-dollars-squared. The prior mean is $\mu_0=100$ thousand dollars and prior variance is $\tau^2=15^2=225$ thousand-dollars-squared. Both distributions are normal, with the parameters explicitly stated.

Step 2: Write the likelihood.

L(\theta;120)=\frac{1}{20\sqrt{2\pi}} \exp\left(-\frac{(120-\theta)^2}{800}\right).

At $\theta=120$, the exponent is zero, so the likelihood density is $1/(20\sqrt{2\pi})$, approximately $0.019947$ per thousand dollars. At $\theta=110$, the exponent is $-100/800=-0.125$. Thus $L(110;120)=0.019947e^{-0.125}$, approximately $0.01760$ per thousand dollars.

Step 3: Compare the likelihoods. The ratio is $L(110;120)/L(120;120)=e^{-0.125}$, approximately $0.8825$. The observed estimate is more likely at $\theta=120$ than at $\theta=110$, but the likelihood at 110 is still substantial relative to its maximum.

Step 4: Evaluate the prior at its mean. The prior density is largest at 100. Its value there is $1/(15\sqrt{2\pi})$, approximately $0.02660$ per thousand dollars. This is a density, not the probability that the mean equals exactly 100; a continuous distribution assigns probability zero to any single point.

Check. The likelihood is centered at the observed estimate, 120, and has standard deviation 20. The prior is centered at 100 and has standard deviation 15. The method avoids treating either the estimate or the prior mean as known without uncertainty.

Variance of an Estimate and the Prior Predictive Distribution

The known variance in the likelihood is often obtained from the construction of the estimate. If $n$ independent observations have conditional mean $\theta$ and known variance $v$, their sample average has conditional variance $v/n$, not $v$. This relationship applies when the observations share the same conditional mean and are independent with the stated common variance. It does not automatically apply when observations are dependent, have different means, or have unequal variances.

The prior and likelihood also imply a distribution for an estimate before the risk's own data are observed. This is the prior predictive distribution. Under the stated normal model, $X=\Theta+\varepsilon$, where $\varepsilon\sim N(0,\sigma^2)$ is independent of $\Theta$. Therefore the estimate has mean $\mu_0$ and variance $\tau^2+\sigma^2$:

X\sim N(\mu_0,\tau^2+\sigma^2).

Justification: Conditional on $\Theta$, write $X=\Theta+\varepsilon$, with $E[\varepsilon]=0$, $\operatorname{Var}(\varepsilon)=\sigma^2$, and $\varepsilon$ independent of $\Theta$. Then $E[X]=E[\Theta]+E[\varepsilon]=\mu_0$. Independence gives $\operatorname{Var}(X)=\operatorname{Var}(\Theta)+\operatorname{Var}(\varepsilon)=\tau^2+\sigma^2$. The sum of independent normal variables is normal, giving the stated distribution. This predictive variance includes both variation among risk means and sampling noise.

Worked Example: Recovering the Standard Error From Sample Size

A hypothetical insurer uses the mean loss from 25 independent policy-years to estimate one risk's annual expected loss. Conditional on that risk's mean $\theta$, each policy-year loss has known variance 2,500 thousand-dollars-squared. The portfolio prior is $\Theta\sim N(80,12^2)$, and the observed sample mean is 92 thousand dollars. Find the correct likelihood variance and the prior predictive standard deviation. No interest discounting or mortality assumption applies; all values are annual losses in thousands of dollars.

Step 1: Calculate the sampling variance of the mean. Independence and a common conditional mean give $ \operatorname{Var}(X\mid\theta)=2{,}500/25=100 $. Its standard error is $\sqrt{100}=10$ thousand dollars. The likelihood is therefore $X\mid\theta\sim N(\theta,100)$, not $N(\theta,2{,}500)$.

Step 2: Write the likelihood at the observed value.

L(\theta;92)=\frac{1}{10\sqrt{2\pi}} \exp\left(-\frac{(92-\theta)^2}{200}\right).

For example, at the prior mean $\theta=80$, the standardized difference is $(92-80)/10=1.2$, and the exponential factor is $e^{-1.2^2/2}=e^{-0.72}$, approximately $0.4868$. This evaluates how the estimate compares with that candidate mean; it is not a posterior estimate.

Step 3: Calculate the prior predictive distribution. The prior variance is $12^2=144$. Adding the likelihood variance gives $144+100=244$, so before observing the sample mean, $X\sim N(80,244)$. Its standard deviation is $\sqrt{244}\approx15.6205$, or 15.62 thousand dollars to two decimals.

Check. The individual policy-year variance is 2,500, but the sample-mean variance is 100. The predictive variance adds 100 to the prior variance 144. The method avoids both using an individual variance as an estimator variance and adding standard deviations instead of variances.

Several Estimates for the Same Risk

If multiple estimates concern the same $\theta$, their likelihoods can be multiplied when they are conditionally independent given $\theta$. For observations $x_j$ with known conditional variances $\sigma_j^2$, the likelihood is proportional to

L(\theta;x_1,\ldots,x_k)\propto \exp\left[-\frac{1}{2}\sum_{j=1}^{k}\frac{(x_j-\theta)^2}{\sigma_j^2}\right].

The estimate with smaller variance contributes more precision, where precision means reciprocal variance. The likelihood alone can be summarized by a precision-weighted estimate with variance equal to the reciprocal of total precision. This is a way to combine observations in the likelihood; it is not yet the combination of the likelihood with a prior.

The result follows by expanding the exponent and completing the square in $\theta$. The coefficient of $\theta^2$ is $\sum_j1/\sigma_j^2$, and the coefficient of $-2\theta$ is $\sum_jx_j/\sigma_j^2$. Consequently, the likelihood is centered at $(\sum_jx_j/\sigma_j^2)/(\sum_j1/\sigma_j^2)$, with variance $1/(\sum_j1/\sigma_j^2)$. Conditional independence is essential: correlated estimates cannot generally be combined by this formula without accounting for their covariance.

Worked Example: Combining Two Conditional Estimates

For a hypothetical risk, two independent studies estimate the same annual expected loss. They report $x_1=92$ with known standard error 4 and $x_2=108$ with known standard error 6, in thousands of dollars. Assume $X_1\mid\theta\sim N(\theta,16)$, $X_2\mid\theta\sim N(\theta,36)$, conditionally independently, and the prior is $\Theta\sim N(100,10^2)$. Find the equivalent likelihood estimate and variance, then compare the likelihood at $\theta=100$ with its maximum. No interest or mortality assumption enters.

Step 1: Convert standard errors to variances. The two likelihood variances are $4^2=16$ and $6^2=36$. Their precisions are $1/16=0.0625$ and $1/36=0.0277778$. Total precision is $0.0625+0.0277778=0.0902778=13/144$.

Step 2: Calculate the equivalent likelihood estimate. The precision-weighted numerator is $92/16+108/36=5.75+3=8.75$. Dividing by $13/144$ gives $8.75(144/13)=1260/13\approx96.9231$ thousand dollars. The equivalent likelihood variance is $1/(13/144)=144/13\approx11.0769$ thousand-dollars-squared.

Step 3: Compare the likelihood at the prior mean to its maximum. The maximum occurs at $\theta=1260/13$. The difference from 100 is $100-1260/13=40/13$. The likelihood ratio is $ \exp[-(40/13)^2/(2(144/13))]=\exp(-0.42735)\approx0.6523 $. The exact exponent magnitude is $25/58\approx0.4310$? Checking the arithmetic is important: $(40/13)^2/(2(144/13))=1600/169\cdot13/288=1600/3744=25/58\approx0.4310$. Therefore the corrected ratio is $e^{-25/58}\approx0.6499$.

Check. The equivalent likelihood is centered near 96.92, between the two estimates and closer to 92 because its variance is smaller. Recomputing the exponent gives the likelihood ratio 0.6499; the method avoids a precision-weighting error and illustrates why checking intermediate arithmetic matters.

Assumptions, Limits, and Exam Traps

A normal likelihood is a modeling choice about the estimate conditional on the underlying mean. It is most defensible when the estimator is approximately symmetric and its standard error is known or treated as known. It can be a poor model for strongly skewed, bounded, or highly discrete estimates, especially with sparse data. In those settings, a likelihood suited to the underlying observations may be preferable. A normal prior likewise permits any real-valued mean, including negative values; if negative loss means are impossible and the prior places material probability below zero, that feature may be inappropriate.

A frequent trap is to use the prior standard deviation where the prior variance is required, or to use an observation-level variance for an average. Another is to mistake a likelihood value or predictive distribution for a posterior result. The likelihood describes data compatibility across candidate $\theta$; the prior predictive distribution describes possible observations before seeing the risk's data. Updating the prior with the likelihood is a distinct step.

Define the target.
State what $\theta$ measures and the units of the observed estimate.

Specify the conditional model.
Write $X\mid\theta\sim N(\theta,\sigma^2)$, and identify how the known variance of the estimate was obtained.

Specify the prior separately.
Write $\Theta\sim N(\mu_0,\tau^2)$, checking that the mean, variance, and units are consistent.

Check assumptions before combining.
Verify the normal approximation and, for multiple estimates, conditional independence or an appropriate dependence model.

Key takeaway: In the normal-normal setup, $\sigma^2$ measures noise in the observed estimate conditional on a risk mean, while $\tau^2$ measures heterogeneity of risk means before observing that risk's experience. Keeping these roles, units, and assumptions distinct is the foundation for correct credibility updating.

Check Your Understanding

For each problem, identify the relevant model quantities and check the variance arithmetic before giving the requested result.

A hypothetical account estimate is 64 thousand dollars, with known standard error 8 thousand dollars, and a candidate mean is 68 thousand dollars. Under $X\mid\theta\sim N(\theta,8^2)$, calculate the likelihood ratio $L(68;64)/L(64;64)$. (Answer: $e^{-0.125}\approx0.8825$.)
A mean of 25 independent policy-year losses estimates $\theta$. Each loss has conditional variance 2,500 thousand-dollars-squared, and the prior is $N(80,12^2)$. State the likelihood variance and prior predictive variance. (Answer: Likelihood variance $100$; prior predictive variance $244$, in thousand-dollars-squared.)
Two conditionally independent estimates of the same mean are 40 with standard error 5 and 52 with standard error 10, all in thousands of dollars. Find the equivalent likelihood center and variance. (Answer: Center $42.4$ thousand dollars; variance $20$ thousand-dollars-squared.)
A prior is $\Theta\sim N(30,4^2)$, and the sampling model has standard error 3. For an observation of 39, calculate its prior predictive standard deviation and standardized value relative to the predictive mean. (Answer: Standard deviation $5$; standardized value $1.8$.)
A study reports $x_1=70$ with standard error 3 and $x_2=76$ with standard error 4, conditionally independently for the same risk mean. Find the equivalent likelihood variance and center. (Answer: Variance $144/25=5.76$ thousand-dollars-squared; center $71.44$ thousand dollars.)

What You'll Learn