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Uniformly Most Powerful Tests.

  1. Mar 26, 2013 #1
    1. The problem statement, all variables and given/known data

    Let X have the pdf [itex]f(x, \theta) = \theta^x(1-\theta)^{1-x}[/itex], x = 0, 1, zero elsewhere. We test [itex]H_0 = \theta = 1/2[/itex] against [itex]H_1 : \theta < 1/2[/itex] by taking a random sample [itex]X_1, X_2, ... , X_5[/itex] of size n=5 and rejecting [itex]H_0[/itex] if [itex]Y = \sum^n_1 X_i[/itex] is observed to be less than or equal to a constant c. Show that this is a uniformly most powerful test.

    2. Relevant equations



    3. The attempt at a solution


    [itex]L(\theta; x_1, ... ,x_n) = \theta^{x_1}(1 - \theta)^{1-x_1}...\theta^{x_n}(1-\theta)^{1-x_n}[/itex]

    [itex]\frac{L(\theta'; x_1, ... ,x_n)}{L(\theta''; x_1, ... ,x_n)} \leq k[/itex]

    [itex]\frac{\theta'^{x_1}(1 - \theta')^{1-x_1}...\theta'^{x_n}(1-\theta')^{1-x_n}}{\theta''^{x_1}(1 - \theta'')^{1-x_1}...\theta''^{x_n}(1-\theta'')^{1-x_n}}[/itex]

    [itex](\frac{\theta'}{\theta''})^{x_1+...+x_n}(\frac{1-\theta'}{1-\theta''})^{n-\sum^n_1 x_n} \leq k[/itex]

    Now I was trying to transform the left side of the equality into a binomial distribution, but I was kind of stuck...
     
  2. jcsd
  3. Mar 26, 2013 #2
    Why do you need to obtain a binomial distribution? The question does not require this.
     
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