BCa Intervals And Jackknife Acceleration

Why Adjust Bootstrap Percentiles?

A percentile bootstrap interval takes lower and upper quantiles of the bootstrap estimates. This is direct, but it treats the observed estimate as if it were centered appropriately and does not adjust for changes in sampling variability across the parameter range. For a skewed loss statistic or a biased estimator, those features can make the percentile cutoffs poorly placed.

The bias-corrected and accelerated (BCa) interval adjusts the percentile probabilities before selecting bootstrap quantiles. Its bias-correction term measures where the observed estimate lies within the bootstrap distribution. Its acceleration term estimates how the estimator's variability changes with the underlying parameter. The method is useful when those features matter, but it does not remove the need to check the data, the resampling design, or the adequacy of the number of replicates.

Definition: Let $\hat{\theta}$ be the estimate from the observed sample, and let $\hat{\theta}^{*}_1,\ldots,\hat{\theta}^{*}_B$ be $B$ bootstrap estimates produced by resampling the observational units and recalculating the same statistic. The BCa bias-correction term is $$ z_0=\Phi^{-1}\left(\frac{\#\{\hat{\theta}^{*}_b<\hat{\theta}\}}{B}\right), $$ where $\Phi$ is the standard normal cumulative distribution function. The jackknife acceleration estimate, based on $n$ leave-one-out estimates $\hat{\theta}_{(i)}$, is $$ \hat{a}= \frac{\displaystyle\sum_{i=1}^{n}\left(\bar{\theta}_{(.)}-\hat{\theta}_{(i)}\right)^3} {\displaystyle 6\left[\sum_{i=1}^{n}\left(\bar{\theta}_{(.)}-\hat{\theta}_{(i)}\right)^2\right]^{3/2}}, \qquad \bar{\theta}_{(.)}=\frac{1}{n}\sum_{i=1}^{n}\hat{\theta}_{(i)}. $$

These formulas presume that the bootstrap resampling units match the sampling units relevant to the estimator. For example, if policies are independent observational units, resampling individual claims separately may not preserve policy-level dependence. If bootstrap estimates tie the observed estimate, a convention is needed; a common adjustment counts each tie as one-half in the proportion defining $z_0$. State and use a tie convention consistently. The examples below have no ties.

From Bias and Acceleration to Adjusted Probabilities

For a two-sided interval with confidence level $1-\alpha$, the nominal tail probabilities are $\alpha/2$ and $1-\alpha/2$. Write $z_p=\Phi^{-1}(p)$ for a standard normal quantile. BCa transforms each nominal probability $p$ to

p_{\mathrm{BCa}}= \Phi\left[ z_0+\frac{z_0+z_p}{1-\hat{a}(z_0+z_p)} \right].

The adjusted probabilities replace the nominal ones: the interval endpoints are the corresponding quantiles of the bootstrap estimates. A positive $z_0$ indicates that more than half the bootstrap estimates lie below $\hat{\theta}$; it shifts the correction in the direction of the observed estimate's relative position. Acceleration changes the two tails differently when jackknife estimates are asymmetric.

Calculate $z_0$.
Count bootstrap estimates below $\hat{\theta}$, divide by $B$, and take the standard normal quantile.

Calculate $\hat{a}$.
Find each leave-one-out estimate, their average, and the centered jackknife deviations; substitute those deviations into the acceleration formula.

Transform both tail probabilities.
Apply the BCa formula separately to $p=\alpha/2$ and $p=1-\alpha/2$.

Read the bootstrap quantiles.
Use the adjusted probabilities to select endpoints from the ordered bootstrap estimates, using a stated quantile convention.

The acceleration formula is not an arbitrary skewness summary. Its numerator records the signed cubic asymmetry of the jackknife estimates around their average; its denominator scales that quantity so the result is dimensionless. In the BCa approximation, this standardized third-order contribution estimates how the standard error changes as the parameter changes. The factor $6$ is part of the BCa expansion and must be retained.

A short way to see why the jackknife enters is to view $\bar{\theta}_{(.)}-\hat{\theta}_{(i)}$ as an estimate of the effect of omitting unit $i$, with sign reversed. The sum of their cubes captures whether a few omissions move the statistic much more in one direction than the other. The sum of squares supplies the corresponding scale. Thus a balanced set of jackknife effects gives $\hat{a}=0$, while asymmetric effects produce a nonzero adjustment. This is an approximation for smooth statistics; with a highly discrete or nonsmooth statistic, the jackknife acceleration may be unstable or uninformative.

Worked Example: A 95% BCa Interval With Positive Acceleration

A hypothetical portfolio statistic has observed estimate $\hat{\theta}=100$, in thousands of dollars. There are $B=100$ bootstrap estimates, of which 60 are less than 100. The four leave-one-out estimates are 101, 101, 101, and 97. In the ordered bootstrap estimates, the 12th value is 98 and the 100th value is 109. Calculate a 95% BCa interval. Use the empirical inverse-CDF convention: at probability $p$, select ordered value $\lceil Bp\rceil$. Use normal quantiles rounded to four decimals and report endpoints in thousands of dollars to the nearest cent.

Method and bias correction. The bootstrap proportion below the observed estimate is $60/100=0.60$. Therefore $z_0=\Phi^{-1}(0.60)=0.2533$.

Jackknife acceleration. The average leave-one-out estimate is $\bar{\theta}_{(.)}=(101+101+101+97)/4=100$. The centered deviations $\bar{\theta}_{(.)}-\hat{\theta}_{(i)}$ are $-1,-1,-1,3$. Their squared deviations sum to $1+1+1+9=12$, and their cubed deviations sum to $-1-1-1+27=24$. Hence

\hat{a}=\frac{24}{6(12)^{3/2}}=0.0962.

Adjusted lower probability. For a 95% interval, the nominal lower probability is $p=0.025$, with $z_p=-1.9600$. Then $z_0+z_p=-1.7067$, and

p_{\mathrm{BCa,L}} =\Phi\left(0.2533+\frac{-1.7067}{1-0.0962(-1.7067)}\right) \approx\Phi(-1.2126) \approx 0.1127.

The selected rank is $\lceil100(0.1127)\rceil=12$, whose bootstrap estimate is 98.

Adjusted upper probability. For $p=0.975$, $z_p=1.9600$, so $z_0+z_p=2.2133$. Thus

p_{\mathrm{BCa,U}} =\Phi\left(0.2533+\frac{2.2133}{1-0.0962(2.2133)}\right) \approx\Phi(3.0656) \approx 0.9989.

The selected rank is $\lceil100(0.9989)\rceil=100$, whose estimate is 109. The 95% BCa interval is therefore $(98.00,109.00)$ thousand dollars. The method has moved the nominal cutoffs substantially, especially in the upper tail. Because that endpoint is the largest of only 100 replicates, it is particularly sensitive to Monte Carlo variation. The common mistake to avoid is selecting the 97.5th percentile after doing the BCa calculation: the adjusted probability, not the nominal probability, determines the endpoint.

Worked Example: Zero Acceleration but Nonzero Bias Correction

For a hypothetical estimate $\hat{\theta}=25.0$, 55 of $B=100$ bootstrap estimates are below the observed estimate. The jackknife estimates are 25.1, 25.1, 24.9, and 24.9. The 16th and 94th ordered bootstrap estimates are 18.4 and 27.1. Find an 80% BCa interval, using the empirical inverse-CDF convention $\lceil Bp\rceil$.

Bias correction. The proportion below the estimate is $0.55$, giving $z_0=\Phi^{-1}(0.55)=0.1257$.

Acceleration. The jackknife average is $25.0$. The centered deviations are $-0.1,-0.1,0.1,0.1$. Their cubes sum to zero, so the acceleration numerator is zero and $\hat{a}=0$.

For an 80% interval, the nominal tail probabilities are 0.10 and 0.90, with normal quantiles $-1.2816$ and $1.2816$. With $\hat{a}=0$, the adjusted normal scores are $z_0+(z_0+z_p)=2z_0+z_p$. Therefore,

p_{\mathrm{BCa,L}}=\Phi(2(0.1257)-1.2816) =\Phi(-1.0302)\approx0.1514,

and

p_{\mathrm{BCa,U}}=\Phi(2(0.1257)+1.2816) =\Phi(1.5330)\approx0.9373.

The ranks are $\lceil100(0.1514)\rceil=16$ and $\lceil100(0.9373)\rceil=94$. The BCa interval is $(18.4,27.1)$. Here the interval differs from a percentile interval because the bias correction is nonzero, even though the symmetric jackknife deviations give no acceleration adjustment. The trap is to conclude that $z_0=0$ merely because $\hat{a}=0$; these terms measure different features.

Worked Example: Acceleration for a Sample Mean

A hypothetical sample of four claim amounts, in thousands of dollars, is $1,2,3,6$. The estimator is the sample mean, so $\hat{\theta}=3$. A bootstrap run has $B=200$ estimates, 130 of which are below 3. Its 43rd and 200th ordered estimates are 2.4 and 5.8. Calculate a 90% BCa interval, using $\lceil Bp\rceil$ for empirical quantiles.

Jackknife estimates. Omitting each observation in turn gives means $(2+3+6)/3=11/3$, $(1+3+6)/3=10/3$, $(1+2+6)/3=3$, and $(1+2+3)/3=2$. Their average is $(11/3+10/3+3+2)/4=3$. The centered deviations are $-2/3,-1/3,0,1$. Their squared deviations sum to $4/9+1/9+0+1=14/9$, and their cubed deviations sum to $-8/27-1/27+0+1=2/3$. Thus

\hat{a}= \frac{2/3}{6(14/9)^{3/2}} \approx0.0573.

Bias correction. The proportion of bootstrap estimates below 3 is $130/200=0.65$, so $z_0=\Phi^{-1}(0.65)=0.3853$.

For a 90% interval, the nominal tail normal scores are approximately $-1.6449$ and $1.6449$. For the lower endpoint, $z_0+z_p=-1.2596$; the transformed score is $0.3853+(-1.2596)/(1-0.0573(-1.2596))\approx-0.7896$. Thus $p_{\mathrm{BCa,L}}\approx\Phi(-0.7896)=0.2150$. For the upper endpoint, $z_0+z_p=2.0302$; the transformed score is $0.3853+2.0302/(1-0.0573(2.0302))\approx2.6834$. Thus $p_{\mathrm{BCa,U}}\approx\Phi(2.6834)=0.9964$.

The ranks are $\lceil200(0.2150)\rceil=43$ and $\lceil200(0.9964)\rceil=200$. The interval is $(2.4,5.8)$ thousand dollars. The result illustrates why the acceleration calculation must use the leave-one-out estimates rather than the original sample values directly. It also illustrates a practical limitation: the upper endpoint comes from the largest bootstrap replicate, so more replicates may be needed for a stable estimate of that tail.

Interpretation and Exam Traps

The BCa procedure changes probabilities, not bootstrap estimates. Once $z_0$, $\hat{a}$, and the adjusted probabilities are calculated, endpoints still come from the empirical bootstrap distribution of the same statistic. The approach assumes the bootstrap replicates represent the estimator's sampling behavior and that the jackknife provides a useful approximation to its local influence.

Using the wrong sign for $z_0$. The stated convention counts replicates below $\hat{\theta}$. Reversing the count changes the sign and can shift both endpoints in the wrong direction.
Using the nominal quantiles at the end. After transforming the tail probabilities, select quantiles at $p_{\mathrm{BCa,L}}$ and $p_{\mathrm{BCa,U}}$, not at $\alpha/2$ and $1-\alpha/2$.
Dropping the factor 6. The acceleration denominator is $6$ times the sum of squared deviations raised to the $3/2$ power.
Confusing bias correction with acceleration. $z_0$ comes from the bootstrap distribution; $\hat{a}$ comes from leave-one-out jackknife estimates. One can be zero while the other is not.
Ignoring numerical and Monte Carlo limits. If the proportion defining $z_0$ is 0 or 1, the normal quantile is infinite. If an adjusted probability is extremely close to 0 or 1, a modest $B$ may not estimate the required tail reliably. Increase $B$, inspect the replicates, and do not present an endpoint beyond the information the simulation supports.

The acceleration formula also fails if every jackknife estimate is identical, because its denominator is zero. That situation does not yield a usable jackknife acceleration; it should be investigated rather than handled by dividing by zero. BCa intervals are most informative when the statistic is sufficiently regular for resampling and jackknife approximations to be meaningful.

Key takeaway: BCa first estimates bootstrap bias with $z_0$, estimates local skewness-related change with jackknife acceleration $\hat{a}$, and then maps nominal tail probabilities to adjusted bootstrap quantiles. Keep the two estimates distinct, apply the transformation to both tails, and check whether the resulting quantiles are supported by enough bootstrap replicates.

Check Your Understanding

For each problem, use the stated conventions and retain enough precision in intermediate calculations.

A bootstrap run has 200 replicates, 140 below the observed estimate, and no ties. Find $z_0$ to four decimals. (Answer: $z_0=\Phi^{-1}(0.70)=0.5244$.)
The centered jackknife deviations are $-2,-1,1,2$. Calculate the acceleration. (Answer: $\hat{a}=0$, since the cubed deviations sum to zero.)
The centered jackknife deviations are $-1,-1,2$. Calculate $\hat{a}$. (Answer: $\hat{a}=1/(6\sqrt{6})\approx0.0680$.)
There are 500 ordered bootstrap estimates. The adjusted probabilities are 0.18 and 0.91. Under the convention $\lceil Bp\rceil$, which order statistics are the endpoints? (Answer: the 90th and 455th order statistics.)
An analyst reports $z_0=0$, $\hat{a}=0.05$, and nominal tail normal scores $-1.6449$ and $1.6449$. Calculate the adjusted probabilities to four decimals. (Answer: approximately 0.0590 and 0.9428.)

What You'll Learn