Normal-Approximation Bootstrap Intervals

Why Use a Normal-Approximation Bootstrap Interval?

A bootstrap distribution can show how an estimator varies when the observed data are resampled. Its spread provides a bootstrap estimate of the estimator’s standard error. If the estimator’s sampling distribution is reasonably close to normal, that standard error can be combined with a normal critical value to form a confidence interval.

This approach is direct, but it relies on an approximation: a symmetric interval may not represent uncertainty well when the estimator’s distribution is strongly skewed, has a long tail, or is constrained to a range that the interval ignores. The calculation and the diagnostic judgment are both important.

Definition: Let $\hat{\theta}$ be an estimate of a parameter $\theta$, and let $\hat{\theta}^{*(1)},\ldots,\hat{\theta}^{*(B)}$ be the values of the same estimator from $B$ bootstrap resamples. The bootstrap standard error is the sample standard deviation of these bootstrap estimates:

\widehat{\operatorname{SE}}_{\text{boot}}(\hat{\theta}) = \sqrt{ \frac{1}{B-1} \sum_{b=1}^{B} \left(\hat{\theta}^{*(b)}-\bar{\theta}^*\right)^2 }, \qquad \bar{\theta}^*=\frac{1}{B}\sum_{b=1}^{B}\hat{\theta}^{*(b)}.

For a two-sided confidence level $1-\alpha$, the normal-approximation bootstrap interval is

\left[ \hat{\theta}-z_{1-\alpha/2}\widehat{\operatorname{SE}}_{\text{boot}}(\hat{\theta}), \; \hat{\theta}+z_{1-\alpha/2}\widehat{\operatorname{SE}}_{\text{boot}}(\hat{\theta}) \right],

where $z_{1-\alpha/2}$ is the corresponding standard normal quantile. The interval is centered on the original estimate $\hat{\theta}$, not automatically on the bootstrap mean $\bar{\theta}^*$.

From Bootstrap Resamples to an Interval

For a nonparametric bootstrap, resamples are drawn with replacement from the observed data, usually keeping the resample size equal to the original sample size. The estimator is recalculated on every resample. The standard deviation of those recalculated estimates estimates the sampling standard error of the original estimator.

A useful way to justify the interval is to begin with the approximate standardized-error statement

\frac{\hat{\theta}-\theta}{\operatorname{SE}(\hat{\theta})} \ \approx\ N(0,1).

Under this approximation, the probability that the standardized error lies between $-z_{1-\alpha/2}$ and $z_{1-\alpha/2}$ is approximately $1-\alpha$. Replacing the unknown standard error by its bootstrap estimate and rearranging gives

P\!\left( -z_{1-\alpha/2} \leq \frac{\hat{\theta}-\theta}{\widehat{\operatorname{SE}}_{\text{boot}}(\hat{\theta})} \leq z_{1-\alpha/2} \right) \approx 1-\alpha

\Longrightarrow \quad P\!\left( \hat{\theta}-z_{1-\alpha/2}\widehat{\operatorname{SE}}_{\text{boot}}(\hat{\theta}) \leq \theta \leq \hat{\theta}+z_{1-\alpha/2}\widehat{\operatorname{SE}}_{\text{boot}}(\hat{\theta}) \right) \approx 1-\alpha.

The justification therefore depends on the normal approximation being suitable and the bootstrap resampling scheme adequately representing the data-generating process. The formula does not make a skewed or unrepresentative bootstrap distribution normal.

Identify the target and point estimate.
Write down the parameter $\theta$, the original estimate $\hat{\theta}$, and the units. Confirm that each bootstrap replicate recalculates the same estimator.

Calculate the bootstrap standard error.
Use the sample standard deviation of the $B$ bootstrap estimates, with denominator $B-1$.

Select the normal critical value.
For a two-sided 95% interval, use $z_{0.975}=1.96$; for a 90% interval, use $z_{0.95}\approx1.645$.

Form and assess the interval.
Use $\hat{\theta}\pm z\widehat{\operatorname{SE}}_{\text{boot}}$. Check whether the interval’s symmetry, endpoints, and scale make sense for the parameter.

Worked Calculations

Worked Example: A Bootstrap Interval for an Average Claim Cost

A hypothetical insurer estimates the mean claim cost from its observed claim-level data as $\hat{\theta}=\$250$. A nonparametric bootstrap with $B=1001$ resamples gives a bootstrap mean of $\bar{\theta}^*=\$252$ and a sum of squared deviations from that bootstrap mean of $\$100{,}000$. Construct a 95% normal-approximation bootstrap interval. Give the endpoints to the cent.

Method. Calculate the sample standard deviation of the bootstrap estimates, then use $z_{0.975}=1.96$. The supplied squared deviations are centered on the bootstrap mean, as required by the standard deviation formula.

There are $B-1=1000$ degrees of freedom in the denominator. Thus the bootstrap variance estimate is

\widehat{\operatorname{SE}}_{\text{boot}}^2 = \frac{100{,}000}{1000} = 100\text{ dollars}^2.

Taking the square root gives $\widehat{\operatorname{SE}}_{\text{boot}}=\$10$. The 95% margin of error is

1.96(\$10)=\$19.60.

Center the interval on the original estimate, $\$250$:

[\$250-\$19.60,\ \$250+\$19.60] = [\$230.40,\ \$269.60].

Check. The interval is symmetric around $\$250$, and its half-width is $\$19.60$, matching the calculated margin. The bootstrap mean is $\$2$ above the original estimate, but the normal-approximation formula does not replace the center with that bootstrap mean. The common mistake of centering at $\$252$ would produce $[\$232.40,\$271.60]$, a different interval.

Worked Example: A Confidence Interval for Claim Frequency

For a hypothetical portfolio, the original estimate of annual claim frequency is $\hat{\theta}=1.70$ claims per policy-year. Bootstrap resampling of policy-level records produces $B=401$ replicate frequency estimates, with mean $1.72$ and sum of squared deviations from that mean equal to $16$. Construct a 90% normal-approximation interval and give the endpoints to four decimal places.

Method. Use the sample standard deviation of the bootstrap replicates and the two-sided 90% critical value $z_{0.95}\approx1.645$.

The denominator is $B-1=400$. The bootstrap variance estimate and standard error are

\widehat{\operatorname{SE}}_{\text{boot}}^2 = \frac{16}{400} = 0.04, \qquad \widehat{\operatorname{SE}}_{\text{boot}} = \sqrt{0.04} = 0.20 \text{ claims per policy-year}.

The margin of error is

1.645(0.20)=0.3290 \text{ claims per policy-year}.

Using the original frequency estimate as the center gives

[1.70-0.3290,\ 1.70+0.3290] = [1.3710,\ 2.0290] \text{ claims per policy-year}.

Check. The midpoint is $(1.3710+2.0290)/2=1.7000$, and both endpoints are $0.3290$ from the midpoint. This interval is positive at both endpoints; that is reassuring about its scale, but it does not by itself establish that the bootstrap distribution is sufficiently symmetric.

Worked Example: A Symmetric Interval for a Skewed Estimate

A hypothetical analyst estimates a high quantile of claim severity as $\hat{\theta}=\$8.00$ thousand. In $B=1001$ bootstrap replicates, the bootstrap mean is $\$12.00$ thousand, the sum of squared deviations from that mean is $28{,}000$ thousand-dollars squared, and the 2.5th and 97.5th percentiles of the replicate estimates are $\$7.10$ thousand and $\$29.40$ thousand. Calculate the 95% normal-approximation interval, then assess whether the normal approximation is credible.

Method. First calculate the bootstrap standard error using $B-1=1000$. Then use $z_{0.975}=1.96$ and center the symmetric interval on the original estimate of $\$8.00$ thousand.

\widehat{\operatorname{SE}}_{\text{boot}}^2 = \frac{28{,}000}{1000} = 28 \text{ thousand-dollars squared}, \qquad \widehat{\operatorname{SE}}_{\text{boot}} = \sqrt{28} \approx 5.2915 \text{ thousand dollars}.

The margin of error is

1.96(5.2915)\approx10.3713 \text{ thousand dollars}.

Therefore, the normal-approximation interval is

[8.00-10.3713,\ 8.00+10.3713] \approx [-2.37,\ 18.37] \text{ thousand dollars}.

Assessment and check. The calculation is internally consistent: the endpoints are approximately $10.37$ thousand dollars below and above the original estimate. But the lower endpoint is negative, which is not possible for a claim-severity quantile. The replicate distribution’s upper percentile, $\$29.40$ thousand, is also far from its lower percentile, $\$7.10$ thousand, and its bootstrap mean is well above the original estimate. These features warn that a symmetric normal approximation may be poor. The interval is the requested calculation, not evidence that negative severity is possible or that the approximation is trustworthy.

Interpretation and Limitations

A 95% confidence interval is a procedure intended to cover the fixed parameter in about 95% of repeated samples when its assumptions and approximation are adequate. It is not a statement that, after observing the data, the fixed parameter has a 95% probability of lying in this particular interval.

The bootstrap standard error estimates the variability of the estimator; it is not the standard error of the bootstrap calculation itself. Increasing $B$ can make the estimated bootstrap standard error more stable, but it cannot fix an unrepresentative original sample, an unsuitable resampling unit, or a strongly nonnormal estimator distribution. If observations are dependent or clustered, resampling individual records independently may fail to preserve that structure.

The normal-approximation interval is most plausible when the estimator’s sampling distribution is approximately symmetric and not dominated by a few extreme outcomes. Be cautious when the bootstrap estimates show a long tail, marked asymmetry, strong outliers, or an estimate near a boundary. An interval extending beyond a known parameter range is a particularly visible warning. A large bootstrap bias relative to the standard error can also indicate that simply centering a symmetric interval on $\hat{\theta}$ needs scrutiny; calculating this interval does not itself correct bias.

Common Exam Traps

Centering on the wrong value. Use the original estimate $\hat{\theta}$ in the normal-approximation interval. The bootstrap mean is used to calculate deviations in the standard error; it is not automatically the interval center.
Using the wrong critical value. A two-sided 95% interval uses $1.96$, not $1.645$. The latter is the usual critical value for a two-sided 90% interval.
Dividing the bootstrap standard error by $\sqrt{B}$. That would describe the Monte Carlo precision of a bootstrap average in a different calculation, not the estimated sampling standard error of $\hat{\theta}$.
Ignoring the units or parameter range. The margin of error has the same units as the estimate. A negative lower endpoint for a quantity that cannot be negative is a warning about the approximation, not a valid negative value of that quantity.
Treating more replicates as a cure for skewness. More resamples can reduce simulation noise in the bootstrap calculation; they do not remove skewness present in the estimated bootstrap distribution.

Key takeaway: Calculate the bootstrap sample standard deviation, multiply it by the normal critical value, and add and subtract that margin from the original estimate. Then assess whether a symmetric normal interval is credible for the estimator and parameter in context.

Check Your Understanding

For each problem, use the stated normal critical value and assess any warning explicitly requested.

A hypothetical estimate of average repair cost is $\$42$. From $B=201$ bootstrap estimates, the sum of squared deviations from their bootstrap mean is $2000$ dollars squared. Construct a 90% normal-approximation interval using $z=1.645$. Give endpoints to the nearest cent. (Answer: Bootstrap SE $=\sqrt{10}\approx\$3.16$; interval $[\$36.80,\$47.20]$.)
A hypothetical probability estimate is $\hat{\theta}=0.18$. With $B=501$, the bootstrap sum of squared deviations from the bootstrap mean is $125$. Construct a 95% interval using $z=1.96$, and state the range warning. (Answer: Bootstrap SE $=0.50$; interval $[-0.80,1.16]$; it extends outside the possible probability range $[0,1]$, so the normal approximation is questionable.)
A hypothetical claim-frequency estimate is $2.40$ claims per policy-year. For $B=1001$ bootstrap estimates, the sum of squared deviations from their mean is $3600$. Construct a 90% interval using $z=1.645$, to two decimals. (Answer: Bootstrap SE $=\sqrt{3.6}\approx1.8974$; interval $[-0.72,5.52]$ claims per policy-year.)
A hypothetical mean-severity estimate is $\$1{,}800$. For $B=801$, the bootstrap sum of squared deviations from the bootstrap mean is $320{,}000$ dollars squared. Construct a 99% interval using $z=2.576$, to the cent. (Answer: Bootstrap SE $=\$20$; interval $[\$1{,}748.48,\$1{,}851.52]$.)
A hypothetical estimator has original estimate $75$, bootstrap mean $77$, and bootstrap sum of squared deviations from the bootstrap mean $2500$, based on $B=101$ replicates. Construct its 95% normal-approximation interval using $z=1.96$, and identify the correct center. (Answer: Bootstrap SE $=5$; center on the original estimate $75$, giving $[65.20,84.80]$, not an interval centered on $77$.)

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