UMP unbiased test I was wondering if any of you folks could help me with this statistics problem. Here is the problem :
set T an exhaustive statistic which $T \sim  \Gamma(n,1)$. We need to find a UMP unbiased test for $H_0 : \tau = \tau_0 \ vs \  H_1 : \tau \neq \tau_0$ .
I know that the form of reject zone must be $\mathcal{R} = \{T \notin [q_1, q_2]\}$ and I found two equations relating $q_1$ and $q_2$.
I have :
$$
F(q_2) - F(q_1) = 1 -\alpha 
$$
and $$
q_1^ne^{-q_1} = q_2^ne^{-q_2} 
$$
\tau is the parameter of the density $$f(x)=\frac{\tau}{x^{\tau +1}}$$ and I know that $$T= \tau \sum_{i=1}^{n} \log(X_i) \sim \Gamma(n,1)$$
I need to find a single equation and that solve it numerically.
Thank you for your help
 A: Assuming $X_1,X_2,\ldots,X_n$ are i.i.d with pdf
$$f(x)=\frac{\tau}{x^{\tau +1}}\mathbf1_{x>1} \,,\quad \tau>0$$
I suggest an alternative way to go about this.
A UMPU test as you said is of the form
$$\phi(t)=\begin{cases}1&,\text{ if }t<q_1 \text{ or }t>q_2 \\ 0 &,\text{ otherwise }\end{cases}$$
For a size $\alpha$ test, $(q_1,q_2)$ are such that $E_{H_0} \phi(T)=\alpha$ and $\operatorname{Cov}_{H_0}(T,\phi(T))=0$.
Note that $2T\sim \chi^2_{2n}$ under $H_0$. The two restrictions above then reduce to
$$P(q_1'<\chi^2_{2n}<q_2')=1-\alpha $$
and
\begin{align}
\operatorname{Cov}_{H_0}(T,\phi(T))=0 &\iff  E_{H_0}(T(1-\phi(T))=E_{H_0}(T)E_{H_0}(1-\phi(T))
\\& \iff E_{H_0}\left[2T(1-\phi(T))\right]=(1-\alpha)E_{H_0}(2T)
\\& \iff \int_{q_1'}^{q_2'} \frac{e^{-x/2}x^n}{2^{n+1}\Gamma(n+1)}\,dx=1-\alpha
\\& \iff P(q_1'<\chi^2_{2n+2}<q_2')=1-\alpha 
\end{align}
with $$(q_1',q_2')=(2q_1,2q_2)$$
So for $\alpha=0.05$ (say), if $G_{2n}$ is the cdf of $\chi^2_{2n}$ distribution, you have
$$G_{2n}(q_2')-G_{2n}(q_1')=0.95 \tag{1}$$
and $$G_{2n+2}(q_2')-G_{2n+2}(q_1')=0.95 \tag{2}$$
Now for example, we can write $0.95=0.975-0.025$, so one choice of $(q_1',q_2')$ from $(1)$ would be that pair which satisfies $G_{2n}(q_2')=0.975$ and $G_{2n}(q_1')=0.025$. Similarly, with the decomposition $0.95=0.995-0.045$, you have another choice of $(q_1',q_2')$. Take four or five of these choices and choose that pair which satisfies $(2)$ the closest. If this works out, you have an approximate solution to $(q_1',q_2')$.
You can also try the same procedure with the two equations you derived involving $q_1,q_2$.
