# Method for obtaining an asymmetric confidence interval

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.

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.

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)
 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)
 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)
  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) )
 0.930849
#exact CI
mean((6 > a/qgamma(.975,25,25)) & (6 < a/qgamma(.025,25,25)))
 0.950385

• 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 Dec 13 '20 at 2:48
• 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. Dec 13 '20 at 3:22
• 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? Dec 13 '20 at 3:31
• 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. Dec 13 '20 at 4:05
• The endpoint of the curve corresponds to $Z_{n}$. I have edited my post to reflect this. Dec 13 '20 at 5:08