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Statistics: describing a best critical region of size alpha.

  1. Jan 4, 2014 #1
    1. The problem statement, all variables and given/known data

    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##.

    2. Relevant equations



    3. 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
     
    Last edited: Jan 4, 2014
  2. jcsd
  3. Jan 4, 2014 #2

    Ray Vickson

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    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##.
     
  4. Jan 4, 2014 #3
    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?
     
  5. Jan 4, 2014 #4

    Ray Vickson

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    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.
     
  6. Jan 5, 2014 #5
    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?
     
  7. Jan 5, 2014 #6

    LCKurtz

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    Damn! I must of missed when they went out of fashion.
     
  8. Jan 5, 2014 #7

    statdad

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    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

    [tex]
    \frac{(1-x)^{10}}{x^{10}}
    [/tex]

    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.
     
  9. Jan 5, 2014 #8

    Ray Vickson

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    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.
     
  10. Jan 5, 2014 #9

    Ray Vickson

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    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
    [tex] P\{\theta = 0 \} = p_0, \: P\{ \theta = 1 \} = q_0 \equiv 1 - p_0.[/tex]
    The posterior probabilities would be
    [tex] P\{ \theta = 0 | x_1, x_2 , \ldots, x_n \} =
    \frac{p_0 f(x_1,x_2,\ldots,x_n|\theta = 0)}{f(x_1,x_2, \ldots, x_n)}, \text{ where}\\
    f(x_1,x_2, \ldots, x_n) = p_0 f(x_1,x_2,\ldots,x_n|\theta = 0)
    + q_0 f(x_1,x_2,\ldots,x_n|\theta = 1) \text{ and}\\
    f(x_1,x_2,\ldots,x_n|\theta = 0) = 2^n (1-x_1)(1-x_2) \cdots (1-x_n), \;
    f(x_1,x_2,\ldots,x_n|\theta = 1) = 2^n x_1 x_2 \cdots x_n[/tex]
    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
    [tex] P\{\theta = 1|X\} = \frac{Q}{Q+P}.[/tex]

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