# Describing a best critical region of size $\alpha$ for testing $H_0 : \theta = 0$ against the alternative hypothesis $H_1 : \theta =1$

Suppose a sample of size 10 is drawn from a distribution with probability density function $f(x, \theta) = 2x^{\theta}(1-x)^{1-\theta}$ if $0<x<1$ and $0$ otherwise, where $\theta \in \{0,1\}$. Describe a best critical region of size $\alpha$ for testing $H_0 : \theta = 0$ against the alternative hypothesis $H_1 : \theta =1$.

We need $\frac{L(0)}{L(1)} \leq k$ for some $k < 1$.

We find that $$\frac{L(0)}{L(1)} = \frac{2(1-x_1)2(1-x_2)...2(1-x_{10})}{2x_12x_2...2x_{10}} = \frac{(1-x_1)(1-x_2)...(1-x_{10})}{x_1x_2...x_{10}}.$$ Now I want to simplify $\frac{(1-x_1)(1-x_2)...(1-x_{10})}{x_1x_2...x_{10}} \leq k$ to see if I can turn this into a distribution that looks more familiar. But for some reason I wasn't able to do anything, so I was wondering if somebody would give me a hint so I can continue...

• You are not seeking a "distribution": the answer to this question is a set of samples of size $10$. If you would like a convenient way to describe such a set, why not re-express the data? Instead of writing a value as $x$, use the quantity $\log(1-x)-\log(x)$. – whuber Jan 6 '14 at 15:21
• @whuber I was just trying to manipulate $\frac{(1-x_1)...(1-x_{10})}{x_1...x_{10}} \leq k$ so that I would have a familiar distribution on the left side of the inequality (or something related to a familiar distribution so it would be easier to find the critical region)...but I guess I don't need that, right? Even if I write it the way you said $\log(1-x) - \log(x)$, I can't really see how that would help to simplify it. – Artus Jan 6 '14 at 15:47
• Let $y_i = \log(1-x_i) - \log(x_i)$. Then the region you found is the set $\{(y_i)\ |\ \sum_{i=1}^n y_i \le k\}$. In vector notation, writing $y=(y_1, y_2, \ldots, y_n)$ and $1=(1, 1, \ldots, 1)$, it can be abbreviated $1\cdot y\le k$. It's a half-space: the multidimensional equivalent of an unbounded interval. That's about as simple as anything ever gets. – whuber Jan 6 '14 at 15:55