Let $x_1,\dots,x_n$ be iid samples from a $N(\mu,\sigma^2)$ and consider the hypotesis

$$H_0:\theta\in\Theta_0,\,\,\,\,\,\,vs\,\,\,\,\,\,H_1:\theta\in\Theta _{0}^c$$

where $\Theta _{0} = \{\mu>0,\sigma^2>1\}$, $\theta=(\mu,\sigma)$. Let $\Lambda(x_1,\dots,x_n)$ be the likelihood ratio test

$${\displaystyle \Lambda (x_1,\dots,x_n)={\frac {\sup\{\,{\mathcal {L}}(\theta \mid x):\theta \in \Theta _{0}\,\}}{\sup\{\,{\mathcal {L}}(\theta \mid x):\theta \in \Theta \,\}}},}$$

What is the number of degrees of freedom of the $\chi^2$ distribution here? Usually, it is the dimension of $\Theta$ minus the dimension of $\Theta_0$, but in this case, both spaces have dimension 2.


  • $\begingroup$ The likelihood ratio approaches a $\chi^2$ distribution for nested models. $\endgroup$ – Sextus Empiricus Dec 7 '18 at 15:48
  • $\begingroup$ @MartijnWeterings Isn't $\Theta_0\subset\Theta$? $\endgroup$ – Chim Dec 7 '18 at 15:49
  • 2
    $\begingroup$ The problem isn't "nesting" per se, but the fact that one model is differentiated from the other in terms of inequality constraints. That is, there is no difference at all between the two models unless a constraint applies, in which case your estimate is at the boundary of $\Theta_0,$ implying the $\chi^2$ approximation will be (probably grossly) wrong. In effect, the concept of "degrees of freedom" becomes both irrelevant and misleading. $\endgroup$ – whuber Dec 7 '18 at 16:25
  • 1
    $\begingroup$ (Continued) Ordinarily situations like this are handled by reducing the submodel to a lower-dimensional space, but this is such an unnatural pair of null and alternative hypotheses (since they are not complementary) it is not obvious how to proceed. Are you sure you have correctly stated both hypotheses? $\endgroup$ – whuber Dec 7 '18 at 16:31
  • 1
    $\begingroup$ @whuber The fact that they are not complementary was a typo, sorry. So, is the LRT applicable in this case? $\endgroup$ – Chim Dec 7 '18 at 16:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.