Suppose I have a probability density function for x

$$ f_n( {\bf x}; \theta) = \exp(n\theta - \sum_{i = 1}^n x_i) * I(\min{{\bf x}} \ge \theta) $$

Where I is an indicator function ( = 1 if the condition inside the brackets is satisfied, 0 otherwise).

I wish to find the maximum likelihood estimator for this density function. Would I be correct in saying the MLE is the minimum observation in x? $$\hat{\theta}_n = \min({\bf x})$$

Next, I consider a hypothesis test $$ H_0: \theta = \theta_0 \text{ vs. } H_1: \theta \text{ > } \theta_0$$

And want to find the Neyman Pearson test statistic for the altered test $$ H_0': \theta = \theta_0 \text{ vs. } H_1': \theta = \theta_1$$

I think I am correct in thinking such a test is $$ T_{NP}({\bf x}) = \exp(n(\theta_1 - \theta_0))$$

Apparently I should be able to show that the Neyman Pearson test is equivalent to the test $$T({\bf x}) = \hat{\theta}_n$$

If someone could show me how to do this, that would be great. Then I have to determine a critical value to obtain a test at significance level alpha. But I don't know the distribution of theta hat.

Many thanks!


Your $T_{NP} (\bf x)$ looks strictly wrong as it stands since you don't seem to have accounted for the conditions under which it, in fact, is the expression you give (the conditions are embodied by the indicator functions that should still appear in the Neyman-Pearson ratio - let this be the lesson: the indicator functions are very relevant to any solution). Have a look again carefully at what happens with the indicator function in your likelihood function ($\mathcal{L}(\theta;\bf x))=f_n( {\bf x}; \theta)$): it doesn't simply cancel out when you find the Neyman-Pearson ratio, which is what you seem to have done, because $I(\min{{\bf x}} > \theta_0)$ is not the same as $I(\min{{\bf x}} > \theta_1)$, necessarily (some manipulations become easier if you assume $\theta_1>\theta_0$, although it isn't strictly necessary).

You should be able to obtain a piecewise function which is either $0$ or the expression you offer depending on the relative positions of the MLE $\hat{\theta}_n=\min(\bf x)$, $\theta_0$ and $\theta_1$ (there is also an indeterminate case when the sample minimum is below both $\theta_0$ and $\theta_1$, but presumably there would be no point in testing the given $H_0$ in such a case as it is immediately invalidated by the sample obtained). You can now impose the N-P criterion on the ratio.

You can obtain the distribution of the MLE $\hat{\theta}_n=\min(\bf X)$ from expanding the distribution function $$P(\hat{\theta}_n<t)=1-P(\min(\bf X)\ge t)=1-P(X_1\ge t\cap X_2\ge t\cap...\cap X_n\ge t)$$ You might like to see how far you get from there using the independence of the $X_i$s. However, due to the nature of this problem (draw, for example, the N-P ratio as a function of the sample minimum), I do not see that you are going to be able to find a critical value in the usual way. You could, however, calculate the exact size and power for this test.

Have fun.


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