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Let $X \sim \mathrm{Bernoulli}(\vartheta)$ for some unknown $\vartheta \in (0,1)$, and let $(X_1, …, X_n)$ be a moderately large IID sample for $X$.

Let $\vartheta_0 \in (0,1)$. I want to test $H_0 \colon \vartheta \leq \vartheta_0$ versus $H_1 \colon \vartheta > \vartheta_0$.

An UMP size $\alpha$ test for this problem is the one that $$ \begin{cases} \text{rejects $H_0$ if $\overline X > c$} \\ \text{rejects $H_0$ with probability $\gamma$ if $\overline X = c$} \\ \text{does not reject $H_0$ if $\overline X < c$} \end{cases}$$ for some constants $c \in (0,1)$ and $\gamma \in [0,1]$ such that $\mathbb P_{\vartheta_0}(\overline X > c) + \gamma \, \mathbb P_{\vartheta_0}(\overline X = c) = \alpha$, where $\mathbb P_{\vartheta_0}$ denotes the probability calculated under the assumption that $\vartheta = \vartheta_0$.

Now, both $\dfrac{\overline X - \vartheta_0}{\sqrt{\frac{\vartheta_0(1-\vartheta_0)}{n}}}$ and $\dfrac{\overline X - \vartheta_0}{\sqrt{\frac{\overline X(1-\overline X)}{n}}}$ converge in distribution to a $N(0,1)$ as $n \to +\infty$.

Therefore we can approximate $\mathbb P_{\vartheta_0}(\overline X > c)$ with

  1. $\mathbb P \left( N(0,1) > \dfrac{c - \vartheta_0}{\sqrt{\frac{\vartheta_0(1-\vartheta_0)}{n}}} \right)$
  2. $\mathbb P \left( N(0,1) > \dfrac{c - \vartheta_0}{\sqrt{\frac{\overline x(1-\overline x)}{n}}} \right)$ where $\overline x$ is the realization of $\overline X$

to be able to determine $c$, while $\mathbb P_{\vartheta_0}(\overline X = c)$ is either exactly $0$ or very close to $0$.

My question is the following: intuitively I would say that the first one is the best approximation for $\mathbb P_{\vartheta_0}(\overline X > c)$, because it contains fewer estimates; but how can I prove that this must be the case?

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