How to find one-sided confidence interval for exponential law Question:
Let $X_1,\ldots,X_n \sim \operatorname{Exp}(\lambda)$ iid for some $\lambda>0$.
Find an expression for the "left" one-sided confidence uniformly most powerful interval for a threshold $1-\alpha$ for the parameter $\lambda>0$
My attempt:
We're looking for an interval of the form $[\delta_\alpha,\infty)$ such that $\Bbb P[\lambda\in [\delta_\alpha,\infty)]\geq1-\alpha\ (\iff\Bbb P[\lambda<\delta_\alpha]\leq1-\alpha)$ 
first we find the $1-\alpha$-quantile: $$\lambda e^{-\lambda x}=1-\alpha\iff -\lambda x=\log \left( \frac{1-\alpha}{\lambda} \right)\iff x=q_{1-\alpha}=-\frac{\log(\frac{1-\alpha}{\lambda})}{\lambda}$$
Then the statistic test we're looking for is $\delta=\Bbb 1\{\tau>q_{1-\alpha}\}$ where $\tau=\sum\limits_{i=1}^n T(X_i)$ where $T$ is from the exponential family form of $\lambda e^{-\lambda x}=\exp\{\eta(\lambda)T(x)-d(\lambda)+S(x)\}=\exp\{\log(\lambda)-\lambda x\}$ 
But in order to apply the theorem that tells us what the wirght test function is $\eta$ must be strictly increasing and in $C^1$ so we have to choose $\eta(\lambda)=\lambda$ and $T(x)=-x\implies \tau=-\sum\limits_{i=1}^nX_i$ and at this point I feel like I am doing something wrong... Any help is welcome!
 A: The joint density is
$$
\prod_{i=1}^n \left(\lambda e^{-\lambda x_i} \right) = \lambda^n e^{-\lambda (x_1 + \cdots + x_n)} \text{ for } x_1,\ldots,x_n\ge0.
$$
This depends on $x_1,\ldots,x_n$ only through their sum, which is therefore a sufficient statistic for $\lambda.$ Even if it were not only through their sum, the sum would be sufficient if the density could be factored in such a way that one factor depends on $x_1,\ldots,x_n$ only through their sum and the other does not depend on $\lambda.$ That is Fisher's factorization theorem.
Recall (?) that
$$
\Pr(X_1+\cdots +X_n \ge x) = \frac 1 {(n-1)!} \int_x^\infty (\lambda u)^{n-1} e^{-\lambda u} (\lambda \, du)
$$
i.e. the sum has a gamma distribution. Then we have
\begin{align}
& \Pr(\lambda (X_1+\cdots+X_n) \ge \lambda x)=\Pr(X_1+\cdots+X_n \ge x ) = \frac 1 {(n-1)!} \int_{\lambda x}^\infty w^{n-1} e^{-w} \, dw \\[10pt]
& \Pr(\lambda(X_1+\cdots+X_n) \ge y) = \frac 1 {(n-1)!} \int_y^\infty w^{n-1} e^{-w} \, dw \\[10pt]
& \Pr\left( \lambda \ge \frac y {X_1+\cdots+X_n} \right) = \frac 1 {(n-1)!} \int_y^\infty w^{n-1} e^{-w} \, dw. 
\end{align}
So $\left( \dfrac y {X_1+\cdots +X_n},\infty \right)$ is a confidence interval for $\lambda,$ whose confidence level is the integral on the right.
