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Math Help - question on hypothesis

  1. #1
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    question on hypothesis

    Under H0, a random variable has the cumulative distribution function F_0(x)=x^2, 0\leq x\leq 1; and under H1, it has the cumulative distribution function F_1(x)=x^3, 0\leq x\leq 1

    Question: If the two hypotheses have equal prior probability, for what values of x is the posterior probability of H0 greater than that of H1?

    This is what I did:

    Since I know that \frac{P(H_0|x)}{P(H_1|x)}=\frac{P(H_0)}{P(H_1)}\fr  ac{P(x|H_0)}{P(x|H_1)}, Given that the prior probability is equal will yield \frac{P(H_0|x)}{P(H_1|x)}=\frac{P(x|H_0)}{P(x|H_1)  } However, x^2 \geq x^3 given that 0<x<1, so it seems that for all values the statement is true, but the answer is apparent not as above.

    Hope someone can help me. Thanks
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  2. #2
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    The answer as given is 2/3. But I not sure how to get it
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  3. #3
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    Quote Originally Posted by noob mathematician View Post
    Under H0, a random variable has the cumulative distribution function F_0(x)=x^2, 0\leq x\leq 1; and under H1, it has the cumulative distribution function F_1(x)=x^3, 0\leq x\leq 1

    Question: If the two hypotheses have equal prior probability, for what values of x is the posterior probability of H0 greater than that of H1?

    This is what I did:

    Since I know that \frac{P(H_0|x)}{P(H_1|x)}=\frac{P(H_0)}{P(H_1)}\fr  ac{P(x|H_0)}{P(x|H_1)}, Given that the prior probability is equal will yield \frac{P(H_0|x)}{P(H_1|x)}=\frac{P(x|H_0)}{P(x|H_1)  } However, x^2 \geq x^3 given that 0<x<1, so it seems that for all values the statement is true, but the answer is apparent not as above.

    Hope someone can help me. Thanks
    P(x|H_0)=\frac{d}{dx}x^2,\ \  0<x<1

    P(x|H_1)=\frac{d}{dx}x^3, \ \ 0<x<1

    CB
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  4. #4
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    Given a level \alpha test, I managed to find the rejection region as: X>\sqrt{1-\alpha}

    What about the power of the test?
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  5. #5
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    The answer is 1-(1-\alpha)^3 /2. Anyone can help with the explanation?

    My understanding is as followed: to find power of the test, I will use
    P(RejectH_0|H_1)
    = P(X>\sqrt{1-\alpha}|H_1)
    = 1-\int_{0}^{\sqrt{1-\alpha}}3x^2.dx
    = 1-(1-\alpha)^{3/2}

    Can it be that the answer is wrong, which is highly impossible i think.
    Last edited by noob mathematician; October 20th 2009 at 06:21 AM.
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