Statistics: describing a best critical region of size alpha.

[Artusartos]

Homework Statement

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 $H_1 : \theta =1$.

Homework Equations

The Attempt at a Solution

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...

Thanks in advance

 

[Ray Vickson]

Homework Statement

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 $H_1 : \theta =1$.

Homework Equations

The Attempt at a Solution

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...

Thanks in advance

You should not have $x_1, x_2, \ldots, x_{10}$. All samples are drawn from the same distribution, with some fixed but unknown $x$.

 

[Artusartos]

You should not have $x_1, x_2, \ldots, x_{10}$. All samples are drawn from the same distribution, with some fixed but unknown $x$.

Thanks a lot.

We have,

$\frac{L(0)}{L(1)} = \frac{(1-x)^{10}}{x^{10}} = (\frac{1-x}{x})^{10}$

So,

$(\frac{1-x}{x})^{10} \leq k$

$\implies (\frac{1-x}{x}) \leq k^{1/10}$

$\implies \ln (\frac{1-x}{x}) \leq \frac{1}{10} \ln(k)$

Let $k' = \frac{1}{10} \ln(k)$, and let $g(x) = \ln (\frac{1-x}{x})$. We can see that the second derivative of $g(x)$ is $\frac{1}{1-x^2} + \frac{1}{x^2}$, which is positive for $0 < x < 1$. So the function is concave upward.

But then $g(x) \leq k'$ is equivalent to $c < x < c'$ for some $c$ and $c'$ such that, for all $x \leq c$ and $x \geq c'$, we have $g(x) > k'$.

But we already have the distribution for $X$, right?

 

[Ray Vickson]

Thanks a lot.

We have,

$\frac{L(0)}{L(1)} = \frac{(1-x)^{10}}{x^{10}} = (\frac{1-x}{x})^{10}$

So,

$(\frac{1-x}{x})^{10} \leq k$

$\implies (\frac{1-x}{x}) \leq k^{1/10}$

$\implies \ln (\frac{1-x}{x}) \leq \frac{1}{10} \ln(k)$

Let $k' = \frac{1}{10} \ln(k)$, and let $g(x) = \ln (\frac{1-x}{x})$. We can see that the second derivative of $g(x)$ is $\frac{1}{1-x^2} + \frac{1}{x^2}$, which is positive for $0 < x < 1$. So the function is concave upward.

But then $g(x) \leq k'$ is equivalent to $c < x < c'$ for some $c$ and $c'$ such that, for all $x \leq c$ and $x \geq c'$, we have $g(x) > k'$.

But we already have the distribution for $X$, right?

You claim that $g(x) = \ln (\frac{1-x}{x})$ is a convex function (I refuse to use the old-fashioned terms concave up or concave down), but that is false: you have computed $g''(x)$ incorrectly. The function g(x) switches from convex to concave as x increases from 0 to 1.

 

[Artusartos]

You claim that $g(x) = \ln (\frac{1-x}{x})$ is a convex function (I refuse to use the old-fashioned terms concave up or concave down), but that is false: you have computed $g''(x)$ incorrectly. The function g(x) switches from convex to concave as x increases from 0 to 1.

Thanks. But we do know that $\frac{1-x}{x}$ is convex...so we can come up with a similar result, right? (without taking the ln of both sides). Is that right?

 

[LCKurtz]

(I refuse to use the old-fashioned terms concave up or concave down)

Damn! I must of missed when they went out of fashion.

 

[statdad]

You should not have $x_1, x_2, \ldots, x_{10}$. All samples are drawn from the same distribution, with some fixed but unknown $x$.

The samples DO NOT have the same fixed but unknown x.
He should have the subscripts. It is true that the x values are from the same distribution, but calling each one x is not correct. In particular, the ratio of likelihoods at 0 and 1 IS NOT

<br /> \frac{(1-x)^{10}}{x^{10}}<br />

Look, for example, at the relevant sections of an introductory text such as Hogg/Craig or intermediate texts like Bickel and Doksum, or any more recent mathematical stat text.

 

[Ray Vickson]

Damn! I must of missed when they went out of fashion.

Most modern optimization textbooks (say, written after about the mid 1960s) seem to have abandoned the concave up/down nomenclature. Calculus texts often still use it, though. I guess it has not been standardized.

 

[Ray Vickson]

Thanks a lot.

We have,

$\frac{L(0)}{L(1)} = \frac{(1-x)^{10}}{x^{10}} = (\frac{1-x}{x})^{10}$

So,

$(\frac{1-x}{x})^{10} \leq k$

$\implies (\frac{1-x}{x}) \leq k^{1/10}$

$\implies \ln (\frac{1-x}{x}) \leq \frac{1}{10} \ln(k)$

Let $k' = \frac{1}{10} \ln(k)$, and let $g(x) = \ln (\frac{1-x}{x})$. We can see that the second derivative of $g(x)$ is $\frac{1}{1-x^2} + \frac{1}{x^2}$, which is positive for $0 < x < 1$. So the function is concave upward.

But then $g(x) \leq k'$ is equivalent to $c < x < c'$ for some $c$ and $c'$ such that, for all $x \leq c$ and $x \geq c'$, we have $g(x) > k'$.

But we already have the distribution for $X$, right?

Sorry: I take back what I said at first: I guess $X$ is the random variable and $\theta$ the parameter, so you should have $x_1, x_2, \ldots, x_{10}$! I am going to stop posting very late at night as an insomnia cure.

I think I see your problem, but can only tell you what I would do if I were a Bayesian. As a Bayesian, I would essentially view $\theta$ as a random quantity with prior probabilities
P\{\theta = 0 \} = p_0, \: P\{ \theta = 1 \} = q_0 \equiv 1 - p_0.
The posterior probabilities would be
P\{ \theta = 0 | x_1, x_2 , \ldots, x_n \} = <br /> \frac{p_0 f(x_1,x_2,\ldots,x_n|\theta = 0)}{f(x_1,x_2, \ldots, x_n)}, \text{ where}\\<br /> f(x_1,x_2, \ldots, x_n) = p_0 f(x_1,x_2,\ldots,x_n|\theta = 0) <br /> + q_0 f(x_1,x_2,\ldots,x_n|\theta = 1) \text{ and}\\<br /> f(x_1,x_2,\ldots,x_n|\theta = 0) = 2^n (1-x_1)(1-x_2) \cdots (1-x_n), \;<br /> f(x_1,x_2,\ldots,x_n|\theta = 1) = 2^n x_1 x_2 \cdots x_n
with $P\{ \theta = 1 | x_1, x_2 , \ldots, x_n \}= 1-P\{ \theta = 0 | x_1, x_2 , \ldots, x_n \}.$
Thus, if we set $P = x_1 x_2 \cdots x_n$, $Q = (1-x_1)(1-x_2) \cdots (1-x_n),$ and if we had a uniform prior ($p_0 = q_0 = 1/2$) we would have
P\{\theta = 1|X\} = \frac{Q}{Q+P}.

I'm not sure what you would do if you were not a Bayesian.