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.


1 Answer 1


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:

x = rexp(25, 1/6)
[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 $$\left(\frac{\bar X}{U},\,\frac{\bar X}{L}\right) = (4.634,10.229).$$

[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.

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
  • $\begingroup$ 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 $\endgroup$ Commented Dec 13, 2020 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
    Commented Dec 13, 2020 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
    Commented Dec 13, 2020 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. $\endgroup$ Commented Dec 13, 2020 at 4:05
  • $\begingroup$ The endpoint of the curve corresponds to $Z_{n}$. I have edited my post to reflect this. $\endgroup$ Commented Dec 13, 2020 at 5:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.