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 $\lambda$ .

  • 1
    $\begingroup$ How is it possible that the great Kolmogorov doesn't know the asnwer to this question? $\endgroup$
    – Zen
    Nov 15, 2020 at 23:52
  • 1
    $\begingroup$ I know the answer. Just testing others' skills. ;) $\endgroup$
    – Kolmogorov
    Nov 16, 2020 at 3:19

1 Answer 1


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 $\tilde x/\bar x<c$ 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.


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