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Given exponential random variable X, the MLE for the scale parameter is $\hat{\beta_x} = \bar{x}$, and the confidence interval for that estimate is: $$\frac{2n\bar{x}}{\chi^2_{\frac{\alpha}{2},2n}} < \hat{\beta_x} < \frac{2n\bar{x}}{\chi^2_{1-\frac{\alpha}{2},2n}}$$

Given samples from two independent exponential variables X, Y, is there a closed form expression for the confidence interval on the ratio of their parameter estimates $\frac{\hat{\beta_x}}{\hat{\beta_y}}$?


I'm optimistic that there is because of the neat characteristics of the exponential distribution, including the fact that $\frac{\hat{\beta_x}}{\hat{\beta_y}} \sim F(2n_x, 2n_y)$. So for example we can go to log space and get $\log(\hat{\beta_x})-\log(\hat{\beta_y}) \sim 2 \operatorname{FisherZ}(2n_x, 2n_y)$. (But getting a Fisher z-distribution doesn't open any doors that I can find.)

And I think but can't confirm that $2n_x \hat{\beta_x} \sim \chi^2_{2n_x}$.

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    $\begingroup$ Since you're dealing with a scale parameter, not a location parameter, I (again) strongly advise focusing on the ratio, not the difference. Percentage changes - closely related to the ratio - make much more sense for scale parameters than absolute change. (As a bonus, it's also more convenient to deal with a ratio, but that's not why I am saying to do it.) ... among other things, note that a change of say 0.25 in $\beta$ means a very different thing when the mean of the Rayleigh is say 0.5 than when it's 8, and different again when it's 100 $\endgroup$
    – Glen_b
    Commented Dec 5, 2023 at 23:02
  • $\begingroup$ @Glen_b Yes, thank you: that makes much more sense here! I just revised to ask about the ratio instead of the difference. $\endgroup$
    – feetwet
    Commented Dec 5, 2023 at 23:33

1 Answer 1

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Note:

  1. The CI is an interval for a population parameter not a sample estimate. In a nice case like this, the estimates will crop up in the endpoints of your interval.

  2. The (ordinary) F distribution for the ratio of estimates that you mention only holds when the parameters are equal. Directly useful for a test but with a CI - while it's still relevant - you need the population parameter in there since that's the thing you need an interval for.

For CIs it helps if you can find a pivotal quantity. A pivot is a function of the variables and the parameter whose distribution doesn't depend on the parameter. A simple one exists in this case:

Let $X_i$ be iid $\sim\text{Exp}(\beta_X)$, $i=1,2,...,n_X$, similarly for $Y_j$, where here the $\beta$ parameter represents the scale, not the rate; $E(X_i)=\beta_X$. We have $X_i / \beta_X\sim \text{Exp}(1)$, so $X_i / \beta_X$ is pivotal for $\beta_X$.

Further, $2X_i/\beta_X \sim \text{Exp}(2)$ which is $\chi^2_2$.

Consequently $q_{X} = 2\sum_i X_i/\beta_{X}$ $\sim$ $\chi^2_{2n_{X}}$, or $2n_X \bar{X}/\beta_X \sim \chi^2_{2n_X}$

Now consider:

$$Q=\frac{q_X}{q_Y}=\frac{2n_X \bar{X}/\beta_X}{2n_Y \bar{Y}/\beta_Y} = \frac{\frac{n_X}{n_Y} \frac{\bar{X}}{\bar{Y}}}{\frac{\beta_X}{\beta_Y}}$$

Let $\theta=\frac{\beta_X}{\beta_Y}$ and $R =\frac{\bar{X}}{\bar{Y}}$, and let $m=n_X/n_Y$, so we have $Q = mR/\theta$, and then $Q\sim F_{2n_x,2n_y}$ is a pivotal quantity for $\theta$.

Immediately, $P(a<Q<b)=G(b)-G(a)$ where $G$ is the cdf of the $F$ distribution with $2n_x,2n_y$ d.f., which probability we will choose to be $1-\alpha$ in order to attain that coverage (with a continuous pivot we can attain the coverage exactly).

We can choose $a$ and $b$ in a variety of ways, but for now let's just take the "symmetric" (i.e. equal-tailed) interval, with $\alpha/2$ in each tail, with $a$ in the left tail of $Q$ and $b$ in its right tail. I will assume this part presents no difficulty.

Then we have $P(a\,<\,mR/\theta \,<\,b) = 1-\alpha$ as the long run (i.e. frequentist) probability under repeated sampling. I am somewhat abusing notation (if I'm repeatedly sampling, I should have my notation talk about a distinct ratio random variable like $R$ for each sample, etc) but I don't want to go far into the weeds here; I'll hand-wave that detail for now.

Now if $a\,<\,mR/\theta \,<\,b$ then $\frac{1}{a}\,>\, \frac{\theta}{mR} \,>\frac{1}{b}$, or $\frac{mR}{b} < \theta < \frac{mR}{a}$. Note that here the endpoints are now the random quantity, not the term in the middle.

So given a sample, $\frac{mr}{b} < \theta < \frac{mr}{a}$ is a $(1-\alpha)$ CI for $\theta$, where $r = \frac{\bar{x}}{\bar{y}}$ is the ratio of sample means, $m$ is the corresponding ratio of sample sizes, $a$ is the $\frac{\alpha}{2}$ quantile of an $F_{2n_x,2n_y}$ distribution and $b$ is the $1-\frac{\alpha}{2}$ quantile of the same distribution.

Note that the lower limit of the interval for $Q$ (the $\alpha/2$ quantile of the $F$ distribution) appears on the denominator in the upper limit of the CI for $\theta$, the ratio of the exponential means.

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  • $\begingroup$ I confirmed via simulation that $P(\frac{\bar{x}}{\bar{y}} F^{-1}(\frac{\alpha}{2}, 2n_x, 2n_y) < \frac{\hat{\beta_x}}{\hat{\beta_y}} < \frac{\bar{x}}{\bar{y}} F^{-1}(1-\frac{\alpha}{2}, 2n_x, 2n_y)) = \alpha$ if sample sizes are approximately equal. As the sample sizes become increasingly unequal that probability diverges. I don't see why that causes a complication, given that the F-dist models the difference in sample size. Can you explain why and show how to handle unequal sample sizes? $\endgroup$
    – feetwet
    Commented Dec 6, 2023 at 19:10
  • $\begingroup$ Correction to the previous comment: The CI probability is on the true parameters, not the estimates; and it is = $(1-\alpha)$, not $\alpha$. $\endgroup$
    – feetwet
    Commented Dec 6, 2023 at 19:37
  • $\begingroup$ It's important to first write the inequality for the pivot and then back out the CI from it, not just jump straight to what you think the result should be. The F quantiles are not how you have them there. The reciprocal of the parameter of interest is in the pivot, so when you flip that to get an interval for the parameter you get $1/F^{-1}(...)$ and the inequality directions reverse. Unscrambling that puts the upper tail quantile of F in the lower limit and vice versa $\endgroup$
    – Glen_b
    Commented Dec 6, 2023 at 22:09
  • $\begingroup$ Of course the reciprocal of an F r.v. is another F which goes half way to explaining why your incorrect interval worked when sample sizes are equal (the rest has to do with the log-symmetry of F with equal dfs) $\endgroup$
    – Glen_b
    Commented Dec 6, 2023 at 22:12
  • $\begingroup$ Ah ha! Yep, that appears to solve it. As for the derivation: I keep re-reading everything and it's not obvious to me – especially in light of your point #2 in the answer – how I could get from the facts stated in the question to this final answer. (I don't doubt that it's straightforward. Rather, at this point I'm not proficient enough to do it even with all of the guidance you've given. :( $\endgroup$
    – feetwet
    Commented Dec 6, 2023 at 22:32

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