Given some Data $X_{1},X_{2},\ldots ,X_{n}$ we are interested in constructing a $\textbf{consistent}$ hypothesis test for $H_{0}:\theta =\theta _{0}$ vs. $H_{1}:\theta \neq \theta_{0}$. Suppose that there is a weak convergence result such as under $H_{0}$ $\alpha_{n}(T_{n}-\theta)\rightarrow X$ in distribution holds. Furthermore, the distribution of $X$ may be known. So is the following testing procedure appropriate?

reject $H_{0}$ if $\left|T_{n}\right|\geq c\alpha_{n}^{-1}+\theta_{0}$ for some $c$?

If yes, why? if not, why? Again, I am only interested in constructing a consistent test.

  • $\begingroup$ Is this for some class? Can you be more explicit about where your difficulties are? $\endgroup$ – Glen_b -Reinstate Monica Jul 20 '17 at 10:24
  • $\begingroup$ @Glen_b I added some further information. $\endgroup$ – Xarrus Jul 20 '17 at 14:59

No that doesn't work, because your manipulation of terms didn't maintain the relationship between the components correctly.

You have something that's asymptotically a pivotal quantity $Q_n=α_n(T_n−θ)$ (asymptotically it's distributed as $X$, where $F_X(x)$ doesn't depend on $\theta$). Work out your limits on $Q_n(\theta)$ using $F$ and back out the asymptotic limits on $T_n$.

Consider the situation under the null ($\theta=\theta_0$). You can find two quantiles $x_l$ and $x_u$ where $F_X(x_l) + 1-F_X(x_u)$ is the desired significance level (I'm there assuming this is continuous, its a teeny bit more fiddly in the general case, because you would include $p(x_u)$ in there, or write it as $1-F_X(x_u^-)\,$), and then reject when $\alpha_n(T_n-\theta_0)$ lies outside $(x_l,x_u)$.

You can then convert that to a rejection rule directly in terms of $T_n$.

Beware - there's nothing in your question that established that the distribution of $X$ is symmetric.

  • $\begingroup$ For a level $\alpha$ test it should work as follows: The Type I Error should equal $\alpha$. That is, $1-(F_{X}(\frac{x_{i}}{\alpha_{n}}+\theta_{0})-F_{X}(\frac{x_{u}}{\alpha_{n}}+\theta_{0}))=\alpha$ yields the choice of $(x_{l},x_{u})$. But is it possible to simplify the whole stuff, if I am only interested in consistency of the test? $\endgroup$ – Xarrus Jul 20 '17 at 19:01
  • $\begingroup$ Your rejection rule seems to rely on an assumption that $|T_n|$ will converge to $\theta_0$ if the null hypothesis is true. ... but even leaving that aside are you really saying you don't care what the level is? $\endgroup$ – Glen_b -Reinstate Monica Jul 20 '17 at 19:38
  • $\begingroup$ In a first step, yes. I just want to construct a consistent test. I don't care about $\alpha$. $\endgroup$ – Xarrus Jul 20 '17 at 19:42
  • $\begingroup$ Then simply reject every time; you'll always reject the null when its false. $\endgroup$ – Glen_b -Reinstate Monica Jul 20 '17 at 19:50
  • $\begingroup$ Ok, then it doesn't make sense what I have intended. How then I have to work out the test, if I want a consistent one? $\endgroup$ – Xarrus Jul 20 '17 at 19:53

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.