Why Refit in Every Parametric Bootstrap Replicate?
A parametric bootstrap starts with a model fitted to observed data. It uses that fitted model to generate new data sets, each with the same structure and sample size as the original, and then applies the estimator of interest to each generated data set. The key point in this tutorial is the last step: the model parameters are estimated again in every replicate.
The parameter values used to generate a replicate are not the replicate's estimates. For example, a fitted exponential model might be used to generate a claim sample with rate \(\hat{\lambda}\), but the rate calculated from that new sample is \(\hat{\lambda}^{*}\). The values can differ because each generated sample differs. If the original estimate is simply copied into every replicate, the procedure does not measure the sampling variation of that estimator.
The procedure is conditional on the model family and on the observed estimate used as the generating value. It therefore relies on the model being an adequate description of the data and on the specified sampling structure being appropriate. A parametric bootstrap does not remove model risk, and a small number of replicates can make its numerical summaries unstable.
The Refitting Procedure
For each replicate, retain the model form and the sample structure, but replace the original observations by newly simulated observations. Refit every parameter that was estimated from the original data and is part of the estimator under study. If there are fixed, known quantities, such as a known exposure or a known model constraint, keep those fixed as appropriate.
Use the observed data \(D\) to calculate \(\hat{\theta}\). Specify the model, the estimator, and any known design quantities.
Simulate \(D^{*(b)}\) from the fitted model \(P_{\hat{\theta}}\), with the same sample size and observation structure as \(D\).
Apply the original estimation method to \(D^{*(b)}\) and record \(\hat{\theta}^{*(b)}\). Do not substitute the generating value \(\hat{\theta}\) for this estimate.
After \(B\) replicates, use the replicate estimates to describe the bootstrap distribution and, when appropriate, estimate the standard error.
For a scalar parameter estimate, the bootstrap average and the usual sample standard deviation across replicates are
The denominator \(B-1\) is used because the replicate standard deviation estimates the spread of the bootstrap distribution from a finite set of simulated estimates. The bootstrap average can also help describe the center of that distribution; it need not equal the original estimate exactly, especially when \(B\) is small.
Why the Replicate Must Be Re-estimated
The reason for refitting follows directly from what sampling variation means. Imagine repeatedly collecting data under a model with parameter \(\theta\), and applying an estimator \(T\) each time. The resulting values \(T(D)\) vary because the data vary. Since \(\theta\) is unknown, the parametric bootstrap substitutes \(\hat{\theta}\) for \(\theta\), simulates data under that fitted model, and applies \(T\) again to each simulated data set.
Thus the bootstrap variation comes from the variation in the generated data passing through the estimation procedure. If instead each replicate is assigned \(\hat{\theta}^{*(b)}=\hat{\theta}\), every estimate is identical and the estimated standard error is zero. That result describes no sampling variation in the reported parameter estimate; it merely repeats the input value.
This argument explains the plug-in approximation, not a guarantee that the approximation is accurate. Its quality depends on the fitted model, the estimator, and the sample size. In practice, use enough replicates for stable summaries and inspect whether estimation failures or boundary estimates occur. Do not silently discard problematic replicates: their frequency may indicate that the estimator or model behaves poorly for samples of the observed size.
Worked Example: Refitting an Exponential Severity Rate
A hypothetical portfolio has four independent, uncensored claim severities, measured in thousands of dollars: \(2,4,6,8\). Assume these are independent observations from an exponential distribution with rate \(\lambda\) per thousand dollars. The maximum likelihood estimate is \(n/\sum x_i\). Three parametric bootstrap samples generated from the fitted model are \((1,3,5,7)\), \((2,4,6,8)\), and \((3,5,7,9)\), also in thousands of dollars. Find the refitted rates and the bootstrap standard error. This is a small illustrative set of replicates, not a recommendation for the number to use in practice.
Fit the original sample. The sample size is \(n=4\), and the total severity is \(2+4+6+8=20\) thousand dollars. Therefore
Refit each replicate. For the first replicate, the total is \(1+3+5+7=16\), so \(\hat{\lambda}^{*(1)}=4/16=0.25\). For the second, the total is \(2+4+6+8=20\), so \(\hat{\lambda}^{*(2)}=4/20=0.20\). For the third, the total is \(3+5+7+9=24\), so \(\hat{\lambda}^{*(3)}=4/24=1/6\), or approximately \(0.1667\), per thousand dollars.
Calculate the bootstrap standard error. Using the unrounded third estimate, the replicate average is
The sum of squared deviations is
Dividing by \(B-1=2\) and taking the square root gives a bootstrap standard error of \(\sqrt{114/32400}\approx 0.0593\) per thousand dollars. The calculation is internally consistent: the estimates lie on both sides of their average, and the estimated spread is positive. The method is parametric simulation followed by maximum likelihood refitting. The common mistake it avoids is reporting \(0.20\) for all three replicates just because \(0.20\) was used to generate them.
Worked Example: Refitting a Poisson Claim Frequency
A hypothetical insurer observes claim counts over five equal, one-year exposure periods: \(0,1,2,3,4\). Assume independent Poisson counts with a common annual mean \(\lambda\). Two replicate count samples generated under the fitted model are \((0,1,1,2,2)\) and \((1,2,2,2,3)\). Calculate the original estimate, the two refitted estimates, and the bootstrap standard error based on these two replicates.
Fit the original sample. For independent Poisson observations with equal exposure, the maximum likelihood estimate is the sample mean. The original total is \(0+1+2+3+4=10\), so \(\hat{\lambda}=10/5=2\) claims per year.
Refit the replicates. The first replicate has total \(0+1+1+2+2=6\), giving \(\hat{\lambda}^{*(1)}=6/5=1.2\). The second has total \(1+2+2+2+3=10\), giving \(\hat{\lambda}^{*(2)}=10/5=2\). The bootstrap average is \((1.2+2)/2=1.6\).
Calculate the standard error. The deviations from \(1.6\) are \(-0.4\) and \(0.4\). Their squared sum is \(0.16+0.16=0.32\). With \(B-1=1\), the sample variance is \(0.32\), and the sample standard deviation is \(\sqrt{0.32}\approx 0.5657\) claims per year.
This standard deviation is only a rough illustration because two replicates provide little information about the bootstrap spread. The estimation method is the Poisson maximum likelihood estimate, refitted as a sample mean for each generated data set. The common trap is to confuse the generating mean, \(2\), with a required answer for every refitted mean; here one replicate's estimate is \(1.2\), not \(2\).
Worked Example: Refitting Both Lognormal Parameters
Suppose four hypothetical positive severities are modeled as independent lognormal observations. Their logarithms are \(0,1,2,3\). Under the lognormal model, assume the logarithms are normal with unknown mean \(\mu\) and variance \(\sigma^2\). Use the maximum likelihood estimates, for which \(\hat{\mu}\) is the mean of the log observations and \(\hat{\sigma}^2\) is their mean squared deviation using divisor \(n\). Three simulated bootstrap samples are specified by their log observations: \((0,1,2,2)\), \((0,1,3,4)\), and \((1,1,2,3)\). Refit both parameters in every replicate.
Fit the original model. The log observations sum to \(0+1+2+3=6\), so \(\hat{\mu}=6/4=1.5\). The deviations from \(1.5\) are \(-1.5,-0.5,0.5,1.5\). The squared deviations sum to \(2.25+0.25+0.25+2.25=5\), so \(\hat{\sigma}^2=5/4=1.25\).
Refit the first replicate. Its log-observation mean is \((0+1+2+2)/4=1.25\). Its deviations are \(-1.25,-0.25,0.75,0.75\), whose squared sum is \(1.5625+0.0625+0.5625+0.5625=2.75\). Thus \(\hat{\mu}^{*(1)}=1.25\) and \(\hat{\sigma}^{2*(1)}=2.75/4=0.6875\).
Refit the second replicate. Its mean is \((0+1+3+4)/4=2\). The deviations are \(-2,-1,1,2\), and the squared sum is \(4+1+1+4=10\). Thus \(\hat{\mu}^{*(2)}=2\) and \(\hat{\sigma}^{2*(2)}=10/4=2.5\).
Refit the third replicate. Its mean is \((1+1+2+3)/4=1.75\). The deviations are \(-0.75,-0.75,0.25,1.25\); the squared sum is \(0.5625+0.5625+0.0625+1.5625=2.75\). Thus \(\hat{\mu}^{*(3)}=1.75\) and \(\hat{\sigma}^{2*(3)}=2.75/4=0.6875\).
The bootstrap averages are \((1.25+2+1.75)/3=1.6667\) for \(\mu\), and \((0.6875+2.5+0.6875)/3=1.2917\) for \(\sigma^2\), rounded to four decimals. Both parameters were refitted because both were estimated from the original data. The method uses the normal maximum likelihood estimates on the log scale; the common mistake is to keep \(\hat{\mu}=1.5\) and \(\hat{\sigma}^2=1.25\) fixed in all replicates, thereby eliminating the parameter-estimation variation this procedure is intended to show.
Common Traps and Practical Checks
The distinction between generating and refitting values is easy to lose when a simulation program stores both in one place. Label them separately: \(\hat{\theta}\) is the original fitted value used by the generator, while \(\hat{\theta}^{*(b)}\) is the estimate calculated from replicate \(b\). Then verify that the fitting function is actually called on the new replicate data.
- Copying the fitted value into every replicate. This gives zero spread for that reported parameter. Re-estimate it from each simulated data set.
- Refitting only some estimated parameters. If the estimator is a vector, refit each component that would be estimated from a real sample. Keep genuinely known quantities fixed.
- Changing the data structure. The simulated replicate should preserve the original sample size and relevant design, such as equal exposure periods. A different structure may produce a different estimator and sampling distribution.
- Using \(B\) rather than \(B-1\) for the replicate sample variance. The usual estimated bootstrap standard error uses the sample standard deviation across the \(B\) replicate estimates, with denominator \(B-1\).
- Rounding too early. Keep full precision in replicate estimates and deviations, and round only the final reported summary.
- Ignoring failed fits. Check for nonconvergence, estimates on boundaries, and undefined estimates. Report and investigate such cases rather than quietly removing them.
Refitting matters whenever the purpose is to study how an estimated model parameter varies across samples under a specified fitted model. By contrast, simulating new observations while keeping the fitted parameters fixed can be useful for questions about variation conditional on those parameter values, but it is not a bootstrap distribution of the parameter estimator. Clear identification of the target quantity prevents these two simulation tasks from being confused. This distinction is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.
Check Your Understanding
For each problem, identify what is generated and what must be refitted. Give the requested final answer.
- A hypothetical exponential severity sample has observations \(3,5,7\), in thousands of dollars. Use the rate MLE \(n/\sum x_i\). Two bootstrap samples are \((2,4,6)\) and \((4,6,8)\). What are the two refitted rates and their average? (Answer: \(0.2500\) and \(0.1667\) per thousand dollars; average \(0.2083\) per thousand dollars.)
- Four equal-exposure Poisson count observations are \(1,1,2,4\). Two simulated replicates have totals 6 and 10 across the four periods. What are the refitted annual means and their bootstrap average? (Answer: \(1.5\) and \(2.5\) claims per period; average \(2.0\).)
- For a lognormal model, replicate log observations are \(0,0,2,2\). Using maximum likelihood on the log scale, refit \(\mu\) and \(\sigma^2\), with divisor \(n\) for the variance. (Answer: \(\hat{\mu}^{*}=1\), \(\hat{\sigma}^{2*}=1\).)
- A fitted Poisson mean is \(3\). A program generates 500 replicate count samples, but assigns the value \(3\) as the estimated mean in every replicate without calculating the sample mean of each sample. What bootstrap standard error for the estimated mean does this produce, and why is it not the refitted bootstrap standard error? (Answer: \(0\); it repeats the generating value instead of estimating the mean from each replicate.)
- Three refitted parameter estimates are \(1.2, 1.6, 2.0\). Calculate their bootstrap average and their sample standard deviation using \(B-1\). (Answer: average \(1.6\); standard deviation \(0.4\).)