I have the following question about the Neyman Pearson Lemma. It is an example in my lecture notes of testing $H_0:\theta=\theta_0$ against $H_1:\theta=\theta_1$ for the exponential distribution.

The likelihoods are given by \begin{equation*} L(\theta_i) = \theta_i^{-n}\exp(-\sum x_i/\theta_i); \quad i=0,1 \end{equation*}

Therefore, from the NP lemma, reject $H_0$ if and only if \begin{equation*} \left(\frac{\theta_1}{\theta_0}\right)^n \exp \left[-\left(\frac{1}{\theta_0}-\frac{1}{\theta_1}\right)\sum x_i \right]\leq k \end{equation*} or equivalently when $\theta_1<\theta_0$ \begin{equation*} \sum x_i \leq \left(\frac{1}{\theta_1}-\frac{1}{\theta_0}\right)^{-1} \ln\left(\frac{\theta_0}{\theta_1}\right)^n k = k^* \end{equation*} where $k^*$ is a positive constant. Since $\sum x_i$ has a gamma$(\theta,n)$ distribution, $K$ can be set to give a particular size $\alpha$, using quantiles of the gamma distribution (this test can be based on $\bar{x}$).

The question is: what happens when $\theta_0<\theta_1$? I don't really see how this changes things ...


1 Answer 1


You reject $H_0$ iff $$ \left( \frac 1 {\theta_1} - \frac 1 {\theta_0} \right) \sum_i x_i \le \text{something}. $$ Multiplying both sides by the reciprocal of $\displaystyle\left( \frac 1 {\theta_1} - \frac 1 {\theta_0} \right)$, you get $$ \sum_i x_i \quad \left.\begin{cases} \le \\[10pt] \ge \end{cases}\right\} \quad \left( \frac 1 {\theta_1} - \frac 1 {\theta_0} \right)^{-1} \cdot\text{something}. $$

You get "$\le$" if the thing you multiplied by is positive, and "$\ge$" if it's negative.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .