# Construct hypothesis test

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.

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

• 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? – Xarrus Jul 20 '17 at 19:01
• 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? – Glen_b -Reinstate Monica Jul 20 '17 at 19:38
• In a first step, yes. I just want to construct a consistent test. I don't care about $\alpha$. – Xarrus Jul 20 '17 at 19:42
• Then simply reject every time; you'll always reject the null when its false. – Glen_b -Reinstate Monica Jul 20 '17 at 19:50
• 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? – Xarrus Jul 20 '17 at 19:53