I'm dealing with a simple inference problem which involve a PDF I've never dealt with before.

We have $X_1, ..., X_n$ iid variables where

$X_i$ has a PDF $$f(x) = (\frac{x}{\theta} + \frac{1}{2}) \space \mathbb{1}_{(-1,1)} (x)$$

The $\theta$ parameter is constrained to $|\theta|>2$.

The problem asks to construct a test for: $H_0: \theta = \theta_0 \space vs \space H_1: \theta = -\theta_0$ for a given $\theta_0 > 2$

I've tried applying LRT (Likelihood ratio test) and Neymann-Pearson theorem, but I end up dealing with expression such as

$$\prod_i (\frac{x_i}{\theta} + \frac{1}{2})$$

which are kind of hard to manipulate.

The exercise also asks to find an estimator $\bar{\theta}$ for $\theta$ using the moment method. Using first order moment of $X$ I could find:

$$\bar{\theta} = \frac{3}{2\bar{X}}$$

However I don't know how to test for consistency nor asymptotic normality.

The reason lies in the fact I don't know how to manipulate the given PDF, especially the conjunct PDF for the vector $\vec{X}$ the way I would do with other PDFs.

Could you please give me a hint?

  • 1
    $\begingroup$ You should note that in general the Likelihood ratio doesn't simplify very far, but the Neyman-Pearson lemma still applies. [I make no claims either way about it simplifying a lot in your case, but I have noticed many students - exposed only to examples with neat simplifications - come to think that something is wrong when it doesn't simplify easily in some instance -- or even conclude that there isn't a likelihood ratio test because they don't see how to write the rejection rule as a simple inequality involving some statistic (like "reject if T(X)>c"). The Lemma may still be ... ctd $\endgroup$ – Glen_b Feb 10 at 6:34
  • $\begingroup$ ctd... successfully applied on real data even when there's no simplification in the ratio at all; the only potential difficulty being the need to figure out a null distribution for the ratio (or some transformation of it like minus twice its log, say) if one isn't applying an asymptotic approximation. In practice this difficulty can be navigated (e.g. by simulation at the specific values of the parameter under the two hypotheses). $\endgroup$ – Glen_b Feb 10 at 6:36
  • $\begingroup$ Of course with independence we can write the log of the likelihood ratio as a sum (over observations) of a difference of terms in the log-density (one term for the null and one for the alternative), but whether you can go much further is not automatically guaranteed. $\endgroup$ – Glen_b Feb 10 at 6:43

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