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I have an estimator $\hat{\theta}$=$\frac{XY}{Z}$ where $X$ and $Y$ are constants and $Z$ is a random variable. $Z$ ranges from [1, $Y$]. Further $Z$ $\rightarrow$ $Y$ in the limit (asymptotically), and therefore $X$ $\rightarrow$ $\hat{\theta}$.

I can use the Delta Method to produce a Wald-type (symmetric) confidence interval (CI) for $\theta$ through relying on the asymptotic normality of $Z$. The interval is

$\hat{\theta} \pm z_{1-\alpha/2}\frac{\hat{\sigma}}{Z_{n}}\sqrt{\hat{\theta}}$

where $Z_{n}$ is the last realization corresponding to $Z$ and $\hat{\sigma}$ is the estimated (sample) standard deviation. Unfortunately, at small sample sizes, the resulting confidence interval will likely have very poor coverage due to the nonlinear nature of $\theta$ as a function of $Z$. Thus, I suspect that an asymmetric interval would be better.

Is there a simple method (easily coded up in R) to produce such an asymmetric interval?

I found the below link to a post by @Cliff AB which suggests that direct sampling from $N(\hat{\theta}, \hat{\sigma^2}$) would work. A user points out that this can be accomplished via resampling, since techniques like bootstrapping are equivalent to the delta method.

Sampling instead of delta method

Once resampling is done, a simple 95% CI could be formed from the 2.5th and 97.5th percentiles of the resulting estimated sampling distribution of $\hat{\theta}$.

Any suggestions would be greatly appreciated.

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The particular distribution of the population and the parameter being estimated will determine the exact answer to this question. This information is not provided. I will illustrate what I hope is the key issue in terms of a more transparent situation.

Suppose you have $n = 25$ observations $X_i$ from $\mathsf{Exp}(\mathrm{rate} = 1/6),$ which has $E(X_i) = \mu = 6.$ The point estimate of $\mu$ is $\hat\mu = \bar X.$ Because $SD(X_i) = E(X_i),$ a symmetrical Wald style 95% (approximate) CI for $\mu$ would be of the form $\bar X \pm 1.96\bar X/\sqrt{n}.$

For a particular sample, we might have the following:

set.seed(1212)
x = rexp(25, 1/6)
mean(x)
[1] 6.619402

Thus a 95% CI with limits symmetrically located about $\bar X$ is $(4.02, 9.21)$.

mean(x) + qnorm(c(.025,.975))*mean(x)/sqrt(25)
[1] 4.024644 9.214160

However, $\frac{\bar X}{\mu} \sim \mathsf{Gamma}(\mathsf{shape}=25;\mathsf{rate}=25),$ so $P\left(L \le \frac{\bar X}{\mu} \le U\right) = 0.95,$ where $L$ and $U$ cut probability $0.025,$ respectively from the lower and upper tails of this gamma distribution. Upon pivoting, we have an exact 95% CI of the form $$\left(\frac{\bar X}{U},\,\frac{\bar X}{L}\right) = (4.634,10.229).$$

mean(x)/qgamma(c(.975,.025),25,25)
[1]  4.634125 10.228587

Note: The following simulation illustrates that the approximate symmetrical "95%" CI actually covers $\mu = 6$ only 93% of the time, even for $n$ as large as $n = 25.$ By contrast, the exact interval has 95% coverage.

set.seed(2020)
n = 25; lam = 1/6
a = replicate(10^6, mean(rexp(n,lam)))
#aprx CI
mean( (6 > (1-1.96/sqrt(n))*a)  & (6 < (1+1.96/sqrt(n))*a) )
[1] 0.930849
#exact CI
mean((6 > a/qgamma(.975,25,25)) & (6 < a/qgamma(.025,25,25)))
[1] 0.950385
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  • $\begingroup$ Thanks. This is helpful. For my particular case, I know nothing of the population's underlying distribution, so likely the best I can do is use the estimated quantiles of the sampling distribution of the statistic, as you nicely illustrate here, to come up with a plausible nonparametric interval. Also, you write Exp(rate = 1/5), but in the R code, you use Exp(rate = 1/6). Perhaps this might be confusing for future readers $\endgroup$ Dec 13 '20 at 2:48
  • $\begingroup$ Thanks. Changed from mean 5 to 6 in midstream. Fixed now. // Working on a simulation to show Wald interval doesn't really have 95% 'coverage' even for $n$ as large as 25. Hope to post that soon. $\endgroup$
    – BruceET
    Dec 13 '20 at 3:22
  • $\begingroup$ The information that $Z$ has support $(1,\infty)$ makes me think of delayed exponential or Pareto family of distributions. If this is a textbook problem, what distributions have been mentioned in the pages before the problem appears? $\endgroup$
    – BruceET
    Dec 13 '20 at 3:31
  • $\begingroup$ Actually, it's not from a textbook of any kind. I ask for my PhD. research (i hope this is OK, but I may need the self-study tag). Basically, I plot monotonically-increasing curves for a bunch of species and wish to find the $x$ value ($X$ in my estimator) past which there is no change in the $y$ value (hence reaching a horizontal asymptote equal to $Y$). This suggests the need for an asymmetric interval. $\endgroup$ Dec 13 '20 at 4:05
  • $\begingroup$ The endpoint of the curve corresponds to $Z_{n}$. I have edited my post to reflect this. $\endgroup$ Dec 13 '20 at 5:08

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