Hypothesis testing involving two parameters in a 2-parameter exponential family

Question

Let $(X, Y)$ be distributed according to the exponential family density $f_{\theta_{1}, \theta_{2}}(x,y)=c( \theta_1 , \theta_2 )h(x,y)exp( \theta_1x+ \theta_2y)$. Show that the only unbiased test for testing $H_{0} : \theta_{1} \leq a, \theta_{2} \leq b$ against $H_{1}:\theta_{1} > a$ or $\theta_{2} > b_1$ or both, is $\phi(x, y) = \alpha$ where $\alpha$ is the level of the test.

I am unable to combine the given hypothesis into a single parametric hypothesis meaningfully, since, for instance, if I consider $H_0: \theta_1+\theta_2 \leq a+b$, it might be accepted but $H_1: \theta_1>a$ might still be true. Is it correct to test the hypotheses separately and see which region is common or something similar? Please help.

Edit:

In particular, I'm stuck at finding points in H1 such that the power becomes < alpha, contradicting unbiasedness for a non-constant test function.

Answer 1

One point missed out a bit here (IMHO) is the joint density belonging to the exponential family. As $f_{\theta_{1}, \theta_{2}}(x,y)=c(\theta_1 ,\theta_2) h(x,y) \exp(\theta_1x+\theta_2y)$ and considering the curved exponential parametrization $L(X;c(\theta))=h(X)\exp\left(c(\theta)^TT(X)-A(c(\theta))\right)$ we get

$$h(X)=h(x,y),\quad c(\theta)=\begin{pmatrix} \theta_1 \\ \theta_2 \end{pmatrix},\quad T(X)=\begin{pmatrix} x \\ y \end{pmatrix},\quad A(c(\theta))=-\log c( \theta_1 , \theta_2 )$$

keep the sufficient statistic $T(X)$ in mind, we'll get back there in a second.

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Consider a univariate hypothesis testing problem for $f_\theta(X)$ that belongs to the exponential family, for example $H_0:\theta\le 0,\quad H_1:\theta>0$. The Karlin–Rubin theorem tells us that the threshold test $$\varphi(x) = \begin{cases} 1 & \text{if } x > x_0 \\ 0 & \text{if } x < x_0 \end{cases}$$ with $x_0$ chosen so ${E}_{\theta_0}\varphi(X)=\alpha$ is UMP of size $\alpha$ for testing the composite hypotheses $H_0: \theta \leq \theta_0 \text{ vs. } H_1: \theta > \theta_0$. Moreover, if $c(\theta)$ is a strictly increasing function then $T(X)$ is the optimal test statistic for testing $H_0~\text{vs.}~H_1$ noted above. Optimal, in the sense that the $\alpha$-level test $\{T(X)\ge c_\alpha\}$ is UMP.

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One of the major setbacks of the Karlin-Rubin theorem is that it is not generalizable for the multivariate case. However, given the form of $T(X)$, we can do a naughty trick here: We write the functions

$$L(x,c(\theta))=\sqrt{h(x,y)}\exp\left(\theta_1x-\frac{1}{2}\log c(\theta_1,\theta_2)\right) \\ L(y,c(\theta))=\sqrt{h(x,y)}\exp\left(\theta_2y-\frac{1}{2}\log c(\theta_1,\theta_2)\right)$$

so $f_{\theta_{1}, \theta_{2}}(x,y)=L(x,c(\theta))\cdot L(y,c(\theta))$. For each of the "marginal" distributions we can formulate the relevant univariate hypotheses (e.g. $H_0:\theta_1\le a,\quad H_1: \theta_1>a$) and construct a corresponding threshold test, $\varphi(x),\varphi(y)$ of level $\frac{\alpha}{2}$ (Bonferroni) and eventually we can write the "double-threshold" test:

$$\varphi(x,y)=\varphi(x)\cdot\varphi(y)=\begin{cases} 1 & \text{if } x > x_0 \text{ AND } y>y_0 \\ 0 & otherwise \end{cases}$$

As we used the Bonferroni correction, the test is UMP and is at least $\alpha$-level.