Confidence interval on ratio of estimates for exponential random variables

Question

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 $\frac{\hat{\beta_x}}{\hat{\beta_y}}$?

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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 $2n_x \hat{\beta_x} \sim \chi^2_{2n_x}$.

Answer 1

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 calculation of 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. You can use relationships between F distributions to put an F quantile on the denominator but I think it's best left as is, I think it keeps what goes on in constructing the interval more clear.