**Fictitious data.** Suppose you have a sample `x` of size $n = 50$ from a population
with an unknown mean and distribution. Then in R we have:

    x
     [1]   7.1  26.9  41.1  22.8  18.2  19.5  37.7  39.1  17.5   3.3
    [11]   6.1   2.3  12.5  11.7  29.1   9.5   6.5  26.1  33.0   9.5
    [21]   6.5   0.5   8.0  24.1  79.4   4.3  39.8   0.3  36.8   2.2
    [31]   2.1   3.0   9.9   5.0   9.4 181.3   0.7   4.3  14.8   0.4
    [41]   3.1   7.3   4.7   1.6  26.5   6.9   2.7   3.6  10.1   0.4

    summary(x)
       Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
      0.300   3.375   8.700  17.584  23.775 181.300 

    stripchart(x, pch="|")

[![enter image description here][1]][1]

There are many styles of nonparametric and parametric bootstrap confidence intervals. I will
compare three of them with two "traditional" CIs.

**Questionable t CI.** Obviously, the observations are strongly right-skewed. But suppose
we believe, somewhat too naively and strongly, in the legendary robustness of
t methods against departure from normality. So we try a 95% t confidence interval, which is $(9.57, 25.59).$ In R, this is part of the `t.test` procedure.

    t.test(x)$conf.int
    [1]  9.574129 25.593871
    attr(,"conf.level")
    [1] 0.95

**Nonparametric bootstrap CI.** Not knowing the family of distributions from which this sample was randomly chosen, we might try a 95% nonparametric
confidence interval for the population mean $\mu$ (which we
assume exists). To get an idea how variable the sample mean
$\bar X$ is as an estimate of $\mu,$ we re-sample many samples of size $50$ from `x` with replacement. For each re-sample,
we find the the distance between the observed mean $\bar X = 17.584$ and the mean of the re-sample. The distribution of
these many differences `d.re` can be used to find the
95% nonparametric bootstrap CI $(9.12, 23.82).$ 

    set.seed(2021) 
    # non-parametric bootstrap, re-sample from sample
    a.obs = mean(x); a.obs
    [1] 17.584

    d.re = replicate(3000, mean(sample(x, 50, rep=T))-a.obs)
    UL = quantile(d.re,c(.975,.025))
    a.obs - UL
       97.5%     2.5% 
     9.12105 23.81885 

**Parametric bootstrap CI.** Now suppose that we know that the population is exponentially distributed, so that $X_i \stackrel{iid}{\sim}\mathsf{EXP = \mathrm{rate}}).$  

Then we can make
a 95% parametric CI for $\mu$ by taking re-samples from
a population with mean $1/\bar X = 1/17.584.$ [Instead of
re-sampling from the sample `x`, we re-sample from
an exponential distribution 'suggested by' the sample `x`.] Of course, it would be better to know 
the exact $\mu,$ but knowing $\hat\mu = 1/17.584$ is better than nothing.

For my fictitious data `x` the resulting 95% parametric
bootstrap CI is $(12.44, 22.13).$ This interval is narrower
than the nonparametric bootstrap CI because it is based
on the additional information that the population is exponential. [I did more re-samples here because
parametric bootstrap CIs with larger numbers of resamples
may be noticeably more accurate.]

    set.seed(2021)
    # parametric bootstrap, sample 50 from EXP(rate=1/a.obs)
    a.obs = mean(x); a.obs
    [1] 17.584
    d.re = replicate(10000, mean(rexp(50,1/a.obs))-a.obs)
    UL = quantile(d.re,c(.975,.025))
    a.obs - UL
       97.5%     2.5% 
    12.44381 22.13479 


**Parametric CI, treating the mean as a scale parameter.** For some right-skewed distributions,
the mean $\mu$ is more accurately viewed as a scale
parameter than a location parameter. If we take this
point of view, it makes more sense to look at ratios of re-sampled means to observed means $\bar X^*/\bar X_{obs}$ rather than differences $\bar X^* - \bar X_{obs},$ for each re-sample. This style of parametric
bootstrap gives the reault $(13.66, 23.77).$

    set.seed(2021)
    # parametric bootstrap of ratios, sample 50 from EXP(rate=1/a.obs)
    r.re = replicate(3000, mean(rexp(50,1/obs.a))/a.obs)
    UL = quantile(r.re,c(.975,.025))
    a.obs / UL
       97.5%     2.5% 
    13.66134 23.76732 


**If you know it: Exact CI.** However, if the population is known to be exponential, then
we know that $\frac{\bar X}{\mu} \sim\mathsf{Gamma}(\mathrm{shape}=1/n, \mathrm{rate}=1/n).$ We can 'pivot' this
relationship to make an _exact_ 95% CI for $\mu$ of
the form $\left(\frac{\bar X}{U}, \frac{\bar X}{L}\right),$
where $L$ and $U$ cut probability $0.025$ from the lower
and upper tails of $\mathsf{Gamma}(1/50, 1/50).$
This exact 95% CI for $\mu$ is $(13.57, 23.69).$

    mean(x)/qgamma(c(.975,.025), 50, 50)
    [1] 13.57196 23.69111

Of course, this is the best 95% CI of the four on this
page because is strictly based on statistical theory.
Sometimes one may not know (or remember) that an
exact CI is available.

_Note:_ The following R code was used to 
sample the fictitious data used in this illustration:

    set.seed(1203)
    x = round(rexp(50,1/20),1)



  [1]: https://i.sstatic.net/1BZJV.png