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Given the hyphotesis

$$ H_0 : \theta_0 \space vs\space H_1 : \theta_1 $$

I want find the Uniformely Most Powerful test (UMP).

Well, we know by the Neyman-Pearson lemma that: $$ \frac{f(\widetilde{x}/\theta_1)}{f(\widetilde{x}/\theta_0)} > k $$ to some $$ k > 0 $$ we will have a UMP with size $$ \alpha $$ such that $$ P(X \space \epsilon \space R | \theta = \theta_0) = \alpha $$

Doing this to the Geometric distribution I will find something like this:

$$ \sum{X_i} \geq \frac{k + n ln(\frac{\theta_1}{\theta_0})}{ln(\frac{1 - \theta_1}{1 - \theta_0})} $$

My doubt is how to proced by here. I know that $$ \frac{k + n ln(\frac{\theta_1}{\theta_0})}{ln(\frac{1 - \theta_1}{1 - \theta_0})} = c $$ such that

$$ P(\sum{X_i} \geq c) = \alpha $$ but how can I sum all the geometric distribution? Is this new distribution a normal or another one?

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