I am considering a random sample $X_1, \ldots, X_m; \ m \geq 2$, from a 2-parameter exponential distribution with pdf $$f_X(x; \mu, \sigma) = \frac{1}{\sigma} \exp \left( -\frac{x-\mu}{\sigma} \right) I(x \geq \mu),$$

where $I$ is an indicator function, $\mu \in \mathbb{R}$ and $\sigma > 0$. I got that the unconstrained MLE of the parameters is $$\hat{\mu} = X_{(1)} = \min\{X_1, \ldots, X_m\}; \ \hat{\sigma} = \sum_{i=1}^m \left(x_i - X_{(1)}\right).$$

I am interested in testing the hypothesis $H_0: \sigma = \sigma_0$ vs. $H_1: \sigma \neq \sigma_0$ for some constant value $\sigma_0$ using a likelihood ratio test. In this case the constrained MLE over $H_0$ is $$\hat{\mu}_0 = X_{(1)}; \ \hat{\sigma}_0 = \sigma_0.$$

Then the likelihood ratio is $$\Lambda = \frac{(\sigma_0)^{-m}\exp\left( -\frac{1}{\sigma_0}\sum_{i=1}^m\left(X_i - X_{(1)}\right) \right)}{(\hat{\sigma})^{-m}\exp\left(-\frac{1}{\hat{\sigma}}\sum_{i=1}^m\left(X_i - X_{(1)}\right)\right)}$$ which I got to simplify to $$\Lambda = \left( \frac{\hat{\sigma}}{\sigma_0} \right)^m \exp \left( -\frac{1}{\sigma_0}\sum_{i=1}^m\left(X_i - X_{(1)}\right)-m \right). $$

However I am not sure that this is right because even if I set $$\Lambda \leq c$$

I have no idea what to do next or how to get a quantity with a known distribution. What do I do for the LRT? I cannot appeal to the asymptotic result as I have to use this test to get a confidence interval for $\sigma$.


1 Answer 1


Consider $Z = \frac{1}{\sigma_0} \sum (X_i - X_{(1)})$, then: $$ \Lambda = e^m Z^{m} e^{-Z} $$

Note that $g(z) = z^me^{-z} = e^{m \log z - z}$ has the following shape: enter image description here

Therefore, $\Lambda < c$ for some $c$ if and only if $Z < c_1$ or $Z > c_2$ for some $c_1$ and $c_2$. Therefore, the problem is reduced to finding $c_1$ and $c_2$ such that: $$ P(Z < c_1) + P(Z > c_2) = \alpha, \qquad \text{under }\sigma = \sigma_0 $$ Now, under the null hypothesis: $$ Z = \frac{1}{\sigma_0} \sum (X_i - X_{(1)}) = \frac{1}{\sigma_0} \sum[ (X_i - a) - (X_{(1)} - a)] $$ Therefore $Z \sim \Gamma(n-1,1)$, i.e. $2Z \sim \chi^2_{2(n-1)}$. Now all we need to do is to find appropriate quanitles $a,b$ of the chi squared distribution such that we reject the null hypothese if $2Z < a$ or $2Z > b$.

  • 1
    $\begingroup$ Btw The details of how Z is a gamma distribution can be found on stack overflow $\endgroup$
    – blooraven
    Commented Apr 23, 2021 at 14:40
  • $\begingroup$ thanks, I understand that part! I was stuck on getting the RR but from this I see I can use the critical point to show the RR has this form. $\endgroup$
    – wzbillings
    Commented Apr 23, 2021 at 18:19

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