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Take the original observations $(1,2,4)$, with sample mean $\overline{x}=7/3$. All $3^3=27$ ordered bootstrap samples have probability $1/27$. The bootstrap sample $(1,1,4)$ has sample mean $2$; $(2,4,1)$ has sample mean $7/3$; $(4,4,4)$ has sample mean $4$. These are illustrative possibilities, not three simulated measurements used to estimate the distribution.

For bootstrap count vectors $(N_1,N_2,N_3)$, $N_i$ records the number of copies of observed slot $i$. Their multinomial distribution isand ${\overline{X}}^{\ast }=(N_1+2N_2+4N_3)/3$. Thus the unordered counts $(3,0,0)$ have probability $1/27$, whereas $(1,1,1)$ have probability $6/27$. Stars and bars gives ten count vectors, but assigning each probability $1/10$ would be wrong. Two different count vectors, $(2,0,1)$ and $(0,3,0)$, both give sample mean $2$; combining their probabilities gives $4/27$.

The conditional bootstrap variance of a sample mean gives a compact check on the enumeration. Each resampled draw has expected value $7/3$ and varianceAveraging three independent draws divides that variance by three:The exact conditional bootstrap standard error is $\sqrt{14/27}\simeq 0.720$. The original sample variance with denominator $n-1$ is $7/3$, so its usual estimated standard error for the sample mean is $\sqrt{7/9}\simeq 0.882$. Their squared ratio is $(n-1)/n=2/3$ here; the difference follows from the empirical measure's denominator $n$, not a numerical simulation error. Neither number is the known true sampling standard error of the unknown source probability distribution.

The empirical distribution gives each observed value mass $1/n$, so its empirical distribution function isFor each of $B$ independent replications, draw $n$ values with replacement from this empirical distribution, recompute the same estimator ${\hat{\theta }}^{\ast (b)}$, and use their sample standard deviation:This estimates the conditional bootstrap standard error; dividing it by $\sqrt{B}$ would instead estimate simulation uncertainty in the bootstrap mean and is not the requested sampling standard error.

When the original observations are distinct, the first resampled index may be arbitrary. After $j$ distinct indices have been drawn, the next index avoids them with probability $1-j/n$. Multiplying conditional probabilities givesHere $P_{\ast }$ denotes probability conditional on the data.

The nonparametric bootstrap replaces the unknown sampling law by the empirical distributionGenerate $B$ independent bootstrap samples, each containing $n$ independent draws with replacement from that law. Recompute the original estimator in each sample, obtaining ${\hat{\theta }}_1^{\ast },\dots ,{\hat{\theta }}_B^{\ast }$. With ${\overline{\theta }}^{\ast }=B^{-1}{\sum }_b{\hat{\theta }}_b^{\ast }$, the bootstrap standard error estimate isThis estimates the conditional bootstrap standard deviation, which approximates the estimator's sampling standard deviation when the bootstrap is consistent. Dividing this expression by $\sqrt{B}$ would instead estimate the Monte Carlo error in the bootstrap mean, a different quantity.

For distinct original observations, each resample is an ordered index sequence with one of $n^n$ equally likely values. It contains no repeats precisely when it is a permutation of all $n$ original indices, giving $n!$ possibilities. HenceThe distinctness assumption is needed to equate repeated values with repeated indices.