I'm struggling with a problem from Lehmann & Romano's book *Testing Statistical Hypothesis."

Suppose $X_i$ is a random sample from $$f(x) = \frac{1}{b}e^{-(x-a)/b}\mathbf{1}_{x>a}$$

The problem concerns finding the UMP of the two sided hypothesis:

$$H_0: a=a_0\quad; \quad H_1: a\neq a_0\,,$$

where $b$ is assumed known.

I'm not sure how I am supposed to derive a two-sided test since Lehmann/Romano only seem to introduce Neyman-Pearson lemma (simple vs simple) and Monotone Likelihood ratio tests.

I know I am supposed to present working or an attempt, but honestly I have no idea where to begin.

What is the strategy I'm supposed to use here? Because I don't think the preceding chapter has made it clear.

  • $\begingroup$ You might have a look at en.wikipedia.org/wiki/…. $\endgroup$ – StubbornAtom May 11 '19 at 15:55
  • $\begingroup$ Much appreciated! For your tip + answer. Thanks greatly $\endgroup$ – Xiaomi May 11 '19 at 16:13
  • $\begingroup$ You can also look at the two-sided UMP test for $U(0,\theta)$ distribution (see math.stackexchange.com/questions/1736322/…). The test in your question can be derived as a special case of this test by a transformation from uniform to shifted exponential. $\endgroup$ – StubbornAtom May 16 '19 at 11:48

Pdf of the sample $(X_1,\ldots,X_n)$ is $$f_a(x_1,\ldots,x_n)=\frac{1}{b^n}\exp\left(-\frac{1}{b}\sum_{i=1}^n (x_i-a)\right)\mathbf1_{x_{(1)}>a}\quad,\,a\in\mathbb R\,,b>0$$

Rewrite the alternative as $H_1:a=a_1\,(\ne a_0)$.

It is better to derive UMP tests separately for the alternatives $a_1>a_0$ and $a_1<a_0$. If you can show that both the tests are same (they will be), then that test is also UMP for the alternative $a_1\ne a_0$.

For $a_1>a_0$, we have the likelihood ratio

\begin{align} \Lambda(x_1,\ldots,x_n)&=\frac{f_{H_1}(x_1,\ldots,x_n)}{f_{H_0}(x_1,\ldots,x_n)} \\\\&=\frac{\exp\left(-\frac{1}{b}\sum\limits_{i=1}^n(x_i-a_1)\right)\mathbf1_{x_{(1)}>a_1}}{\exp\left(-\frac{1}{b}\sum\limits_{i=1}^n(x_i-a_0)\right)\mathbf1_{x_{(1)}>a_0}} \\\\&=e^{n(a_1-a_0)/b}\frac{\mathbf1_{x_{(1)}>a_1}}{\mathbf1_{x_{(1)}>a_0}} \\\\&=\begin{cases}e^{n(a_1-a_0)/b}&,\text{ if }x_{(1)}>a_1\\0&,\text{ if }a_0<x_{(1)}< a_1 \end{cases} \end{align}

Find $\Lambda$ similarly for $a_1<a_0$.

We have to carefully study the nature of $\Lambda$ as a function of $x_{(1)}$ to apply the Neyman-Pearson lemma to get an MP test, and eventually extend that test to a UMP test by making it free of $a_1$.

Alternatively one can solve the equivalent exercise for $U(0,\theta)$ distribution since the shifted exponential distribution in this question can be transformed to $U(0,\theta)$.

The UMP test of size $\alpha$ for testing $\theta=\theta_0$ against $\theta\ne \theta_0$ for a sample $Y_1,\ldots,Y_n$ from $U(0,\theta)$ distribution has the form

$$\psi(y_1,\ldots,y_n)=\begin{cases}1&,\text{ if }y_{(n)}< \theta_0\alpha^{1/n}\,\text{ or }\,y_{(n)}>\theta_0 \\ 0 &,\text{ otherwise }\end{cases}$$

Now if $X$ has the pdf $f_a(x)=\frac{1}{b}e^{-(x-a)/b}\mathbf1_{x>a}$, it can be verified that $Y=e^{-X/b}$ has the $U(0,\theta)$ distribution with $\theta=e^{-a/b}$.

So the required UMP test here must be $$\phi(x_1,\ldots,x_n)=\begin{cases}1&,\text{ if }x_{(1)}<\frac{a_0}{b}\,\text{ or }\,x_{(1)}>\frac{a_0}{b}-\frac{1}{n}\ln\alpha \\ 0&,\text{ otherwise }\end{cases}$$


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