Two sample test for exponential distribution with only two observations

Suppose we have two independent random variables $$X_1 \sim \exp(\lambda_1)$$ and $$X_2 \sim \exp(\lambda_2)$$ .

Now, we are given just one observation each from the two distributions above, say $$S_1$$ and $$S_2$$ respectively. On the basis of these two observed numbers $$S_1$$ and $$S_2$$ , I am interested in testing the following two different hypotheses, viz.

• Null hypothesis $$~~H_0 : ~~\lambda_1 = \lambda_2~~$$ against the alternative hypothesis $$~~H_A : ~~\lambda_1 \neq \lambda_2$$
• Null hypothesis $$~~H_0 : ~~\lambda_1 = \lambda_2~~$$ against the alternative hypothesis $$~~H_A : ~~\lambda_1 < \lambda_2$$

Can someone please suggest appropriate ways to test these two hypotheses at some given level, say $$\alpha\in(0,1)$$ ?

Note : I guess, under the assumption of the null hypothesis (in both the problems), we may play with the random variable $$Z = \frac{X_1}{X_1+X_2}$$ because it'll probably be uniformly distributed. Anyway, I'm unable to think any further. Also, here $$\exp(\lambda)$$ refers to an exponential distribution with parameter $$\lambda$$ .

• How is it possible that the great Kolmogorov doesn't know the asnwer to this question?
– Zen
Nov 15 '20 at 23:52
• I know the answer. Just testing others' skills. ;) Nov 16 '20 at 3:19

Consider a likelihood ratio test. Since the likelihood maximizing $$\hat\lambda_0$$, $$\hat\lambda_1$$, $$\hat\lambda_2$$, are respectively $$1/\bar{x}=2/(x_1+x_2)$$, $$1/x_1$$, and $$1/x_2$$, the likelihood ratio $$\Lambda$$ is
\begin{align} \Lambda &= \frac{f_{\hat\lambda_0}(x_1,x_2)}{f_{\hat \lambda_1}(x_1)f_{\hat \lambda_2}(x_2)} \\ &= \frac{\hat\lambda_0^2 e^{-\hat\lambda_0(x_1+x_2)}}{\hat\lambda_1 e^{-\hat\lambda_1x_1} \hat\lambda_2 e^{-\hat\lambda_2x_2}} \\ &= \frac{\hat\lambda_0^2}{\hat\lambda_1 \hat\lambda_2}e^{-\hat\lambda_0(x_1+x_2)+\hat\lambda_1x_1+\hat\lambda_2x_2} \\ &= \frac{x_1x_2}{\bar x^2}e^{-2+1+1} \\ &= \left(\frac{\tilde x}{\bar x}\right)^2 \end{align}
where $$\tilde x$$ is the geometric mean of $$x_1$$ and $$x_2$$.
That gives the rule for the first hypothesis test to reject $$H_0$$ when $$\tilde x/\bar x an appropriate $$c$$. The same rule works for the second test but needs the accompanying condition that $$x_1>x_2$$.
The ratio of geometric to algebraic mean makes intuitive sense since the geometric is always less and only equals the algebraic mean with $$x_1=x_2$$. If the two observations are close, then the null hypothesis looks reasonable and the likelihood ratio will be high. If $$x_1$$ and $$x_2$$ are very different then the geometric mean will be a good deal less than the algebraic mean and the likelihood ratio will be low, indicating that the null hypothesis should be rejected.