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Math Help - hypothesis testing: parameter 0 against parameter positive.

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    hypothesis testing: parameter 0 against parameter positive.

    I failed at the test in statistics recently and I'm preparing to another term. This a problem from the test. I'm translating the problem from Polish; I'm sorry if I'm getting some words wrong.

    The problem:

    We observe a single variable X with the following pdf:

    f_{\theta}(x)=  \begin{cases} {\theta}e^{-x}+2(1-\theta)e^{-2x} & \mbox{dla }  x \ge 0;  \\ 0  & \mbox{dla } x < 0, \end{cases}

    where \theta\in[0,1] is an unknown parameter.

    (a) Find the uniformly most powerful test with the significance level of \alpha = 0.05 to verify the null hypothesis H_0\,:\,\theta=0 against the alternative hypothesis H_1 \,:\,\theta > 0.

    (b) Find the power function of the test.

    My attempt:

    First, I want to find the most powerful test with the level of significance of 0.05 to verify the hypothesis H_0 against the alternative hypothesis H_2\,:\,\theta=\theta_1, where \theta_1 is a fixed number larger than 0. This I want to achieve by using the Neyman-Pearson lemma. Then I want to say that this test is also most powerful against H_1, which I would like to be implied by the arbitrarity of \theta_1>0 and a simple reasoning. I'm not entirely convinced about this simple reasoning. It comes from my book and althought I thought I understood it, I'm not so sure now. I will say what I mean on request, but now I'd like to give my other trouble.

    The trouble:

    The calculations seem more complicated than usually on such tests, which makes me think I'm doing something wrong.

    The calculations:

    We have:
    f_0(x)=\begin{cases} 2e^{-2x} & \mbox{dla }  x \ge 0;  \\ 0  & \mbox{dla } x < 0, \end{cases}

    f_2(x)=\begin{cases} {\theta}_{1}e^{-x}+2(1-\theta_1)e^{-2x} & \mbox{dla }  x \ge 0;  \\ 0  & \mbox{dla } x < 0, \end{cases}

    where \theta_1\in(0,1].

    The likelihood ratio is given by the following formula, for x\ge0:

    \frac{f_2(x)}{f_0(x)}=\frac{2e^{-2x}}{{\theta}_{1}e^{-x}+2(1-\theta_1)e^{-2x}}

    N-P tells me I want to look for the rejection region of the following form:

    K^*=\left\{x\ge0\,:\,\frac{2e^{-2x}}{{\theta}_{1}e^{-x}+2(1-\theta_1)e^{-2x}}>c\right\},

    where c>0 is a constant we will have to choose correctly.

    We want

    0.05=\mathbb{P}_0\left(\left\{x\ge0\,:\,\frac{2e^{-2x}}{{\theta}_{1}e^{-x}+2(1-\theta_1)e^{-2x}}>c\right\}\right)=\mathbb{P}_0\left(\left\{x\g  e0\,: 2e^{-2x}\left[1-c(1-\theta_1)\right]>\theta_{1}ce^{-x}\right\}\right)=\mathbb{P}_0\left(\left\{x\ge0\,  : e^{-x}\left[1-c(1-\theta_1)\right]>\frac{1}{2}\theta_{1}c\right\}\right).

    Here's my trouble. To solve the inequality with respect to x, I need to divide by \left[1-c(1-\theta_1)\right], whose sign I don't know, so I have to consider two cases and some unpleasant calculations seem to follow. As far as I know my professor, it's impossible, so I think I'm doing it the wrong way.

    It's unfortunately my first attempt at solving a problem with a composite hypothesis.

    I'll be very grateful for any help.

    PS: I'm sorry about the size of the formulas, but I have no idea what to do about it. They show correctly in preview.
    Last edited by ymar; February 10th 2011 at 02:59 PM. Reason: PS
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