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Math Help - Order Stat and unbiased estimator

  1. #1
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    Order Stat and unbiased estimator

    f(y) = \left\{ \begin{array}{rcl}<br />
\alpha y^{\alpha-1}/\theta^{\alpha} & \mbox{if} & 0 \leq y \leq \theta \\<br />
0 & \mbox{if} & \mbox{otherwise}<br />
\end{array}\right.<br />

    \alpha > 0, which is a fixed known value, and \theta is unknown. Consider the estimator \hat{\theta} =\max(Y_1, Y_2,..., Y_n)

    a) show that \hat{\theta} is a biased estimator for \theta.
    b) Find a multiple of \hat{\theta} that is an unbiased estimator of \theta.

    Solutions:

    a)  E[Y] = \int_0^{\theta} \frac{\alpha y^{\alpha -1}}{\theta^{\alpha}} \cdot y \ dy

    =\frac{\alpha}{\theta^{\alpha}} \int_0^{\theta} y^{\alpha} / dy

    =\frac{\alpha}{\theta^{\alpha}} \times \frac{y^{\alpha+1}}{\alpha+1} \bigg{|}^{\theta}_0

    =\frac{\alpha}{\theta^{\alpha}} \times \frac{\theta^{\alpha+1}}{\alpha+1}= \frac{\alpha \cdot \theta}{\alpha+1}

    now for the \hat{\theta}

    F(y) = \int_0^{\theta} \frac{\alpha y^{\alpha -1}}{\theta^{\alpha}} \ dy

    F(y) =  \frac{\alpha} {\theta^{\alpha}} \int_0^{\theta} y^{\alpha -1} \ dy

    F(y) =  \frac{\alpha} {\theta^{\alpha}} \cdot \frac{y^{\alpha}}{\alpha}\bigg{|}^{\theta}_{0} =\frac{\alpha} {\theta^{\alpha}} \cdot \frac{\theta^{\alpha}}{\alpha} =1

    f_{(n)}(y) = n[F(y)]^{n-1}f(y)

    f_{(n)}(y) = n[1]^{n-1} \times \alpha y^{\alpha-1}/\theta^{\alpha}

    E[Y] = \int^{\theta}_0 n \times \alpha y^{\alpha-1}/\theta^{\alpha} \times y \ dy

    = \frac{n\alpha}{\theta^{\alpha}} \int^{\theta}_0 y^{\alpha} \ dy

    = \frac{n\alpha}{\theta^{\alpha}} \times \frac{y^{\alpha+1}}{\alpha+1} \bigg{|}^{\theta}_0

    = \frac{n\alpha}{\theta^{\alpha}} \times \frac{\theta^{\alpha+1}}{\alpha+1}  = \frac{n\alpha \theta}{\alpha+1}

    now \frac{n\alpha \theta}{\alpha+1} \neq \frac{\alpha \theta}{\alpha+1} therefore it's biased.

    b) n =1 for it to be an unbiased estimator.

    is this correct?
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  2. #2
    Grand Panjandrum
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    Cumulative distribution for y:

    F(y) =\begin{cases} 0& y \le 0 \\ <br />
\int_0^{y} \frac{\alpha \xi^{\alpha -1}}{\theta^{\alpha}} \ d \xi & 0<y<\theta \\<br />
1 & y\ge \theta \end{cases}

    (you will find it advantageous if you describe what you are trying to do in words before embarking on the symbolic manipulations)

    CB
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  3. #3
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    Quote Originally Posted by lllll View Post
    f(y) = \left\{ \begin{array}{rcl}<br />
\alpha y^{\alpha-1}/\theta^{\alpha} & \mbox{if} & 0 \leq y \leq \theta \\<br />
0 & \mbox{if} & \mbox{otherwise}<br />
\end{array}\right.<br />

    \alpha > 0, which is a fixed known value, and \theta is unknown. Consider the estimator \hat{\theta} =\max(Y_1, Y_2,..., Y_n)

    a) show that \hat{\theta} is a biased estimator for \theta.
    b) Find a multiple of \hat{\theta} that is an unbiased estimator of \theta.

    Solutions:

    a)  E[Y] = \int_0^{\theta} \frac{\alpha y^{\alpha -1}}{\theta^{\alpha}} \cdot y \ dy

    =\frac{\alpha}{\theta^{\alpha}} \int_0^{\theta} y^{\alpha} / dy

    =\frac{\alpha}{\theta^{\alpha}} \times \frac{y^{\alpha+1}}{\alpha+1} \bigg{|}^{\theta}_0

    =\frac{\alpha}{\theta^{\alpha}} \times \frac{\theta^{\alpha+1}}{\alpha+1}= \frac{\alpha \cdot \theta}{\alpha+1}

    Mr F says: The above calculation is not necessary for answering the question.

    now for the \hat{\theta}

    F(y) = \int_0^{\theta} \frac{\alpha y^{\alpha -1}}{\theta^{\alpha}} \ dy Mr F says: All you will do here is prove that the area under the curve is equal to 1, ie. f(y) satisfies one of the two criteria for being a pdf ......

    F(y) = \frac{\alpha} {\theta^{\alpha}} \int_0^{\theta} y^{\alpha -1} \ dy

    F(y) = \frac{\alpha} {\theta^{\alpha}} \cdot \frac{y^{\alpha}}{\alpha}\bigg{|}^{\theta}_{0} =\frac{\alpha} {\theta^{\alpha}} \cdot \frac{\theta^{\alpha}}{\alpha} =1 Mr F say: An unsurprising result!

    f_{(n)}(y) = n[F(y)]^{n-1}f(y)

    f_{(n)}(y) = n[1]^{n-1} \times \alpha y^{\alpha-1}/\theta^{\alpha}

    E[Y] = \int^{\theta}_0 n \times \alpha y^{\alpha-1}/\theta^{\alpha} \times y \ dy Mr F says: Should be {\color{red}E({\color{blue}\hat{\theta}})}.

    = \frac{n\alpha}{\theta^{\alpha}} \int^{\theta}_0 y^{\alpha} \ dy

    = \frac{n\alpha}{\theta^{\alpha}} \times \frac{y^{\alpha+1}}{\alpha+1} \bigg{|}^{\theta}_0

    = \frac{n\alpha}{\theta^{\alpha}} \times \frac{\theta^{\alpha+1}}{\alpha+1} = \frac{n\alpha \theta}{\alpha+1}

    now \frac{n\alpha \theta}{\alpha+1} \neq \frac{\alpha \theta}{\alpha+1} therefore it's biased.

    b) n =1 for it to be an unbiased estimator.

    is this correct?
    Key results (details left for you to fill in):

    F(y) = \frac{y^{\alpha}}{\theta^{\alpha}}.

    f_{(n)}(y) = \frac{n \alpha y^{n \alpha - 1}}{\theta^{n \alpha}}.

    E(\hat{\theta}) = \int_{0}^{\theta} y \, \frac{n \alpha y^{n \alpha - 1}}{\theta^{n \alpha}} \, dy = \frac{n \alpha}{\theta^{n \alpha}} \, \int_{0}^{\theta} y^{n \alpha} \, dy = \frac{n \alpha}{n \alpha + 1} \, \theta \neq \theta.


    An unbiased estimator is therefore \hat{\theta} = \left( \frac{n \alpha + 1}{\alpha} \right)\max(Y_1, Y_2,..., Y_n).
    Last edited by mr fantastic; November 3rd 2008 at 01:31 AM.
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    E(\hat{\theta}) = \int_{0}^{\theta} y \, \frac{n \alpha y^{n \alpha - 1}}{\theta^{n \alpha}} \, dy = \frac{n \alpha}{\theta^{n \alpha}} \, \int_{0}^{\theta} y^{n \alpha} \, dy = \frac{n \alpha}{n \alpha + 1} \, \theta \neq \theta.
    When I integrate this out I get one. I get the stated answer for E[\hat{\theta^2}]

    E[\hat{\theta}] = \int_{0}^{\theta} \, n \left(\frac{y^{\alpha}}{\theta^{\alpha}} \right)^{n-1} \times \frac{\alpha y^{\alpha - 1}}{\theta^{\alpha}} \times y \  dy

    E[\hat{\theta}] = \int_{0}^{\theta} \alpha n \frac{y^{\alpha n -\alpha}}{\theta^{\alpha n- \alpha}} \times \frac{y^{\alpha}}{\theta^{\alpha}} \ dy

    E[\hat{\theta}] = \alpha n \int_{0}^{\theta} \frac{y^{\alpha n}}{\theta^{\alpha n}} \ dy

    E[\hat{\theta}] = \frac{\alpha n}{\theta^{\alpha n}} \int_{0}^{\theta} y^{\alpha n} \ dy

    E[\hat{\theta}] = \frac{\rlap{\color{red}\vrule height3.5pt depth-2pt width 1.2 em}<br />
\alpha n}{\theta^{\alpha n}} \frac{y^{\alpha n}}{\rlap{\color{red}\vrule height3.5pt depth-2pt width 1.2 em}\alpha n} \bigg{|}^{\theta}_0

    E[\hat{\theta}] = \frac{\theta^{\alpha n}}{\theta^{\alpha n}} - \frac{0^{\alpha n}}{\theta^{\alpha n}}= 1


    now if I do E[\hat{\theta^2}] I get:

    E[\hat{\theta^2}]= \int_{0}^{\theta} \, n \left(\frac{y^{\alpha}}{\theta^{\alpha}} \right)^{n-1} \times \frac{\alpha y^{\alpha - 1}}{\theta^{\alpha}} \times y^2 \  dy

    E[\hat{\theta^2}] = \int_{0}^{\theta} \alpha n \frac{y^{\alpha n -\alpha}}{\theta^{\alpha n- \alpha}} \times \frac{y^{\alpha+1}}{\theta^{\alpha}} \ dy

    E[\hat{\theta^2}] = \alpha n \int_{0}^{\theta} \frac{y^{\alpha n+1}}{\theta^{\alpha n}} \ dy

    E[\hat{\theta^2}] = \frac{\alpha n}{\theta^{\alpha n}} \int_{0}^{\theta} y^{\alpha n+1} \ dy

    E[\hat{\theta^2}] = \frac{\alpha n}{\theta^{\alpha n}} \frac{y^{\alpha n+1}}{n+1} \bigg{|}^{\theta}_0

    E[\hat{\theta^2}] = \frac{\alpha n}{\theta^{\alpha n}} \frac{\theta^{\alpha n+1}}{n+1}-\frac{\alpha n}{\theta^{\alpha n}} \frac{0^{\alpha n+1}}{n+1} = \frac{\alpha n }{n+1} \theta

    so now for an unbiased estimator E[\hat{\theta}] \neq \theta \longrightarrow 1=\theta since my E[\hat{\theta}] is simply a constant in this case would it have to be equal to \theta?
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  5. #5
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    Quote Originally Posted by lllll View Post
    When I integrate this out I get one. I get the stated answer for E[\hat{\theta^2}]

    E[\hat{\theta}] = \int_{0}^{\theta} \, n \left(\frac{y^{\alpha}}{\theta^{\alpha}} \right)^{n-1} \times \frac{\alpha y^{\alpha - 1}}{\theta^{\alpha}} \times y \ dy

    E[\hat{\theta}] = \int_{0}^{\theta} \alpha n \frac{y^{\alpha n -\alpha}}{\theta^{\alpha n- \alpha}} \times \frac{y^{\alpha}}{\theta^{\alpha}} \ dy

    E[\hat{\theta}] = \alpha n \int_{0}^{\theta} \frac{y^{\alpha n}}{\theta^{\alpha n}} \ dy

    E[\hat{\theta}] = \frac{\alpha n}{\theta^{\alpha n}} \int_{0}^{\theta} y^{\alpha n} \ dy

    E[\hat{\theta}] = \frac{\rlap{\color{red}\vrule height3.5pt depth-2pt width 1.2 em}<br />
\alpha n}{\theta^{\alpha n}} \frac{y^{\alpha n}}{\rlap{\color{red}\vrule height3.5pt depth-2pt width 1.2 em}\alpha n} \bigg{|}^{\theta}_0 Mr F says: This is wrong. It should be {\color{red} \frac{\alpha n}{\theta^{\alpha n}} \left[ \frac{y^{\alpha n{\color{blue} + 1}}}{\alpha n{\color{blue} + 1}}\right]_{0}^{\theta}}.

    [snip]
    ..
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  6. #6
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    Quote Originally Posted by lllll View Post
    [snip]
    now if I do E[\hat{\theta^2}] I get:

    E[\hat{\theta^2}]= \int_{0}^{\theta} \, n \left(\frac{y^{\alpha}}{\theta^{\alpha}} \right)^{n-1} \times \frac{\alpha y^{\alpha - 1}}{\theta^{\alpha}} \times y^2 \ dy

    E[\hat{\theta^2}] = \int_{0}^{\theta} \alpha n \frac{y^{\alpha n -\alpha}}{\theta^{\alpha n- \alpha}} \times \frac{y^{\alpha+1}}{\theta^{\alpha}} \ dy

    E[\hat{\theta^2}] = \alpha n \int_{0}^{\theta} \frac{y^{\alpha n+1}}{\theta^{\alpha n}} \ dy

    E[\hat{\theta^2}] = \frac{\alpha n}{\theta^{\alpha n}} \int_{0}^{\theta} y^{\alpha n+1} \ dy

    E[\hat{\theta^2}] = \frac{\alpha n}{\theta^{\alpha n}} \frac{y^{\alpha n+1}}{n+1} \bigg{|}^{\theta}_0 Mr F says: This is wrong. It should be {\color{red}\frac{\alpha n}{\theta^{\alpha n}} ~ \frac{y^{\alpha n + {\color{blue}2}}}{{\color{blue}\alpha} n + {\color{blue}2}} \bigg{|}^{\theta}_{0}}

    [snip]
    ..
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