# Two-sided UMP test for exponential densities?

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.

• You might have a look at en.wikipedia.org/wiki/…. – StubbornAtom May 11 '19 at 15:55
• Much appreciated! For your tip + answer. Thanks greatly – Xiaomi May 11 '19 at 16:13
• 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. – 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 $$a_1. If you can show that both the tests are same (they will be), then that test is also UMP for the alternative $$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

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

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