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Given $\mathcal{X}=\{0,1\}$, $\Theta= \{ 1/2,1/4 \}$ and $\mathcal{P}= \{ B(\theta):\theta \in \Theta \}$, I am trying to construct an UMP test of size $\alpha \in (0,1)$ for $$H_0: \theta=1/4 \,H_1: \theta=1/2$$

I figured to evaluate $$L_{1/2,\ 1/4}(x)=\frac{f_{1/2}(x)}{f_{1/4}(x)}=\begin{cases} 2 & x = 1 \\\ \frac{2}{3} & \, x=0\end{cases} $$

Using this I tried to solve for $c \geq 0$ such that:

$$\varphi(x)= \mathbb{1}(L_{1/2,\ 1/4}(x)>c)$$

and $E_{1/4}(\varphi)=\alpha$, so I have a Neyman-Pearson Test which by NP-Lemma would be UMP.

But evaluating the expected value got me:

$\begin{align*} E_{1/4}(\varphi) &= \mathbb{P}_{1/4}(L_{1/2,\ 1/4}(x)>c) \\ &= \mathbb{P}_{1/4}(L_{1/2,\ 1/4}(X)>c, X=0) + \mathbb{P}_{1/4}(L_{1/2,\ 1/4}(X)>c, X=1) \\ &= \mathbb{P}(\frac{2}{3}>c)\frac{3}{4}+ \mathbb{P}(2>c)\frac{1}{4} \\ &= \mathbb{1}(\frac{2}{3}>c)\frac{3}{4}+\mathbb{1}(\frac{2}{3}>c)\frac{1}{4} \stackrel{!}{=} \alpha \end{align*} $

Here I don't know how to proceed, to solve for c. I would be glad about a tip and/or pointer!

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  • $\begingroup$ One flip of a coin is not really enough to distinguish between the two hypotheses. You essentially have three possible strategies: $2 \le c$ in which case you never reject $H_0$ and $\alpha=0$, $\frac23 \le c \lt 2$ in which case you reject $H_0$ if $x=1$ but not if $x=0$ so $\alpha = \frac14$, or $c \lt \frac23$ in which case you always reject $H_0$ and $\alpha=1$. $\endgroup$
    – Henry
    Commented Jul 3 at 23:03
  • $\begingroup$ @Henry, yes this what I essentially concluded, but I thought I missed something. Thank you for your insight! I will try to construct an UMP test, for any $\alpha$, with more than one sample. $\endgroup$
    – tychonovs-scholar
    Commented Jul 3 at 23:21

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