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Math Help - Order Stats and sufficiency question

  1. #1
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    Order Stats and sufficiency question

    Let Y_1, \ Y_2, \ \dotso,\ Y_n denote a random sample from the prob. density function:

    f(y|\theta) = \left\{ \begin{array}{rcl}<br />
e^{-(y-\theta)} & \mbox{for} & y \geq \theta \\ <br />
0 & \mbox{for} & \mbox{other}<br />
\end{array}\right.

    show that Y_{(1)} =\min(Y_1, \ Y_2, \ \dotso,\ Y_n) is sufficient for \theta

    I asked my prof, and all he told me is that in order for our definition of sufficiency to be valid in this case we simply need to considering the smallest y, where it should be greater then some fixed \theta. If I look at what the book says and go according to http://www.mathhelpforum.com/math-he...-question.html I get something a little closer to what's in the book. Therefore what I get is

    F(y) = \int_{\theta}^u e^{-(y-\theta)} \ dy = 1-e^{\theta-u}

    so the density function for a minimum stat is n[1-F(y)]^{n-1} \cdot f(y) = n[1-(1-e^{\theta-u})]^{n-1} \cdot e^{-(u-\theta)}

    =n[e^{\theta-u}]^{n-1} \cdot e^{\theta-u} = n[e^{\theta-u}]^n

    following the same method as in the link I get:

    L(y_1, \ y_2, \ \dotso, \ y_n|\theta) = f(y_1, \ y_2, \ \dotso, \ y_n|\theta) = f(y_1|\theta) \cdot f(y_2|\theta) \cdot \dotso \cdot f(y_n|\theta)

    =\prod_{i=1}^k n[e^{\theta-u_i}]^n

    =n[e^{\theta-u_1}]^n \times n[e^{\theta-u_2}]^n \times \ \dotso \times n[e^{\theta-u_k}]^n

    =n^k \cdot \exp \bigg{[} \sum_{i=1}^k (\theta -u_i)^n \bigg{]}

    factorizing it out would give

    g(\min (Y_1, \ Y_2, \ \dotso,\ Y_n)|\theta) = \exp \bigg{[} \sum_{i=1}^k (\theta -u_i)^n \bigg{]} \ \mbox{and} \ \ h(y_1,\ y_2,\ \dotso, \ y_n) = n^k

    I'm not quite sure that this is correct.

    Also can it be done using these terms:

    Let X = \sum^n_{i=1} (\theta-u_i)

    =\prod_{i=1}^k n[e^{\theta-u_i}]^n as \prod_{i=1}^k n\cdot \exp\bigg{[}\sum^n_{i=1} (\theta-u_i) \bigg{]} = \prod_{i=1}^k n\cdot e^X

    expanding it out gives:

    =ne^X \times ne^X \times \ \dotso \times ne^X

    =n^k e^{kX}
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  2. #2
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    Quote Originally Posted by lllll View Post
    Let Y_1, \ Y_2, \ \dotso,\ Y_n denote a random sample from the prob. density function:

    f(y|\theta) = \left\{ \begin{array}{rcl}<br />
e^{-(y-\theta)} & \mbox{for} & y \geq \theta \\ <br />
0 & \mbox{for} & \mbox{other}<br />
\end{array}\right.

    show that Y_{(1)} =\min(Y_1, \ Y_2, \ \dotso,\ Y_n) is sufficient for \theta

    I asked my prof, and all he told me is that in order for our definition of sufficiency to be valid in this case we simply need to considering the smallest y, where it should be greater then some fixed \theta. If I look at what the book says and go according to http://www.mathhelpforum.com/math-he...-question.html I get something a little closer to what's in the book. Therefore what I get is

    F(y) = \int_{\theta}^u e^{-(y-\theta)} \ dy = 1-e^{\theta-u}

    so the density function for a minimum stat is n[1-F(y)]^{n-1} \cdot f(y) = n[1-(1-e^{\theta-u})]^{n-1} \cdot e^{-(u-\theta)}

    =n[e^{\theta-u}]^{n-1} \cdot e^{\theta-u} = n[e^{\theta-u}]^n

    following the same method as in the link I get:

    L(y_1, \ y_2, \ \dotso, \ y_n|\theta) = f(y_1, \ y_2, \ \dotso, \ y_n|\theta) = f(y_1|\theta) \cdot f(y_2|\theta) \cdot \dotso \cdot f(y_n|\theta)

    =\prod_{i=1}^k n[e^{\theta-u_i}]^n

    =n[e^{\theta-u_1}]^n \times n[e^{\theta-u_2}]^n \times \ \dotso \times n[e^{\theta-u_k}]^n

    =n^k \cdot \exp \bigg{[} \sum_{i=1}^k (\theta -u_i)^n \bigg{]}

    factorizing it out would give

    g(\min (Y_1, \ Y_2, \ \dotso,\ Y_n)|\theta) = \exp \bigg{[} \sum_{i=1}^k (\theta -u_i)^n \bigg{]} \ \mbox{and} \ \ h(y_1,\ y_2,\ \dotso, \ y_n) = n^k

    I'm not quite sure that this is correct.

    Also can it be done using these terms:

    Let X = \sum^n_{i=1} (\theta-u_i)

    =\prod_{i=1}^k n[e^{\theta-u_i}]^n as \prod_{i=1}^k n\cdot \exp\bigg{[}\sum^n_{i=1} (\theta-u_i) \bigg{]} = \prod_{i=1}^k n\cdot e^X

    expanding it out gives:

    =ne^X \times ne^X \times \ \dotso \times ne^X

    =n^k e^{kX}
    L = e^{-(y_1 - \theta)} \cdot e^{-(y_2 - \theta)} \cdot \, .... \, \cdot e^{-(y_n - \theta)} = e^{-(y_1 + y_2 + \, .... \, + y_n) + n \theta} = e^{-(\sum_{i=1}^{n} y_i) + n \theta}.

    Clearly Y_{(1)} = Y_k for some 1 \leq k \leq n.

    Therefore

    L = e^{-y_k + n \theta} \cdot e^{-(\sum_{i=1, \, i \neq k}^{n} y_i)}

    which has the form that shows sufficiency.
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  3. #3
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    If we changed to to be a maximum value where 0 \leq y \leq \theta ,instead of y\geq \theta, would it be the same procedure yielding the same result?
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  4. #4
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    Quote Originally Posted by lllll View Post
    If we changed to to be a maximum value where 0 \leq y \leq \theta ,instead of y\geq \theta, would it be the same procedure yielding the same result?
    I suppose it would be. In both cases there are still some dotting of i's and crossing of t's you need to do.
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