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Math Help - point estimate question: sufficiency, bias, order stat

  1. #1
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    point estimate question: sufficiency, bias, order stat

    Question.

    Suppose that Y_1, ... Y_n is a random sample from a population with density f_Y(y)=\frac{2y}{\theta^2} for 0<y\leq\theta, where \theta>0.

    (a) is Y_{(n)} a sufficient statistic for \theta?

    (b) Derive the density of the last order statistic Y_{(n)}.

    (c) Find an unbiased estimator of \theta that is a function of Y_{(n)}.

    (d) By considering the condition that a function of Y_{(n)} is unbiased for \theta, determine whether there is a better unbiased estimator for \theta.

    I've just started stat inference and my confidence is so low, I feel like threading on seashells... basically I worked through most of proofs but I don't see how I can apply them. This is a sample exam question (to which I don't have an answer). So I'd appreciate your pointers.

    Answer.

    (a) sufficiency of Y_{(n)}

    Should I apply factorisation theorem here to factorise the density of Y into two parts: one is the function of Y_{(n)} and \theta, another is a function of y only?

    Another thought, given the fact that y is limited by \theta, could it be that y somehow depends on \theta and therefore the highest order statistic Y_{(n)} cannot be a sufficient statistic simply because of this dependency?

    (b) Density for Y_{(n)}

    find anti-derivative of f_Y(y)

    F_Y(y)=\frac{2}{\theta^2}\frac{y^2}{2}, 0<y\leq\theta - for a single observation

    F_{Y_{(n)}}(y)=P(Y_{(n)}{\leq}y)=[F_Y(y)]^n=(\frac{y^2}{\theta^2})^n, 0<y\leq\theta

    f_{Y_{(n)}}(y)=\frac{d}{dy}F_{Y_{(n)}}(y)=n(\frac{  y^2}{\theta^2})^{n-1}\frac{2y}{\theta^2}

    (c) unbiased estimator of \theta that is a function of Y_{(n)}

    I cannot 'see' it just from looking at the distribution, so I was thinking to use the definition of the unbiased estimator, but the integration seems like a puzzle.

    Define a function of Y_{(n)}: g(Y_{(n)}). Then
    E(g(Y_{(n)})=\theta according to the definition of an unbiased estimator.

    Then I would try to find g(Y_{(n)} form the equation

    E(g(Y_{(n)})=\int_0^{\theta}g(Y_{(n)})f_{Y_{(n)}}(  y)dy=\theta

    If I do that, I get an integral of a product of g(Y_{(n)}) and y^{2n-1}, on the left side, and an expression involving theta and n on the right side - is it solvable?

    Any advice? Thanks!
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  2. #2
    MHF Contributor matheagle's Avatar
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    the largest order stat is suff for theta

    I believe that

    E(Y_{(n)})={2n\theta\over 2n+1}

    If that's right, then {2n+1\over 2n}Y_{(n)} is unbiased for theta
    Last edited by matheagle; February 10th 2011 at 08:57 PM.
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  3. #3
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    expected value of i-th order statistic

    One more (general) question related to (c), how do you find an expected value of an i-th order statistics - do you use the usual formula (summation or integration)

    \int_{range of Y}Y_{(i)}f_Y_{(i)}(y)dy

    I can find the density of i-th order stats, but and what do I put as Y_{(i)}? y? or I need to derive a formula for Y_{(i)}, based on the distribution given?
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  4. #4
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    Quote Originally Posted by matheagle View Post
    (that took 1 minute, by the way)
    That took you n years of studying and practising maths, right?
    I wasn't able to integrate AT ALL before last August. I didn't know anything about order statistic three days ago.
    Last edited by Volga; February 11th 2011 at 12:30 AM.
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  5. #5
    MHF Contributor matheagle's Avatar
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    \int yf_{Y_{(i)}}(y)dy

    Little y, Y_{(i)} is the name/label of the rv
    y is the dummy variable we are summing/integrating over.

    \int yf_{Y_{(i)}}(y)dy=\int uf_{Y_{(i)}}(u)du

    I always use f(u) in class
    it's the only time I can say fu without getting in trouble.
    Last edited by matheagle; February 11th 2011 at 01:39 PM.
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  6. #6
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    Quote Originally Posted by matheagle View Post
    the largest order stat is suff for theta
    Can I ask how you made this conclusion? is that so obvious?... or some mathematical manipulations were performed

    Quote Originally Posted by matheagle View Post
    I believe that

    E(Y_{(n)})={2n\theta\over 2n+1}

    If that's right, then {2n+1\over 2n}Y_{(n)} is unbiased for theta
    I see the logic now...
    Last edited by Volga; February 11th 2011 at 05:29 AM.
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  7. #7
    MHF Contributor matheagle's Avatar
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    the likelihood functions factors into the five parts

    {2^n\prod y_i\over \theta^{2n}} I(Y_{(1)}>0) I(Y_{(n)}<\theta)

    the only part involving the data that cannot be separated from theta is the one with the largest order stat.
    that is always the case when dealing with a parameter greater than the random variables.
    When the parameter is a lower bound, the the smallest order stat is suff for that parameter.
    Last edited by matheagle; February 11th 2011 at 06:00 AM. Reason: spelling
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  8. #8
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    Quote Originally Posted by matheagle View Post
    the likelihood functions factors in to the five parts

    {2^n\prod y_i\over \theta^{2n}} I(Y_{(1)}>0) I(Y_{(n)}<\theta)

    the only part involving the data that cannot be separated from theta is the one with the largest order stat.
    that is always the case when dealing with a parameter greater than the random variables.
    When the paramter is a lower bound, the the smallest order stat is suff for that parameter.
    thanks, I got it. I factorised initially, but I missed out the Indicator function.
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  9. #9
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by Volga View Post
    That took you n years of studying and practising maths, right?
    I wasn't able to integrate AT ALL before last August. I didn't know anything about order statistic three days ago.
    According to certain people it's N years
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  10. #10
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    Is that to say N>>n?
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  11. #11
    MHF Contributor matheagle's Avatar
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    according to moo n mr fantasy, way bigger
    maybe N!
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  12. #12
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    Quote Originally Posted by matheagle View Post
    the likelihood functions factors into the five parts

    {2^n\prod y_i\over \theta^{2n}} I(Y_{(1)}>0) I(Y_{(n)}<\theta)
    So, to continue to part (d), I think I need to check 'attaining the Cramer-Rao bound'.

    Given the likelihood function for n observations above,  l(\theta;y)=nln2+\Sigma_{i=1}^nY_i-2ln\theta, 0<y\leq\theta,

    s(\theta;y)=\frac{d}{d\theta}(l(\theta;y))=-\frac{2n}{\theta}

    Now I think I should check the 'attaining Cramer-Rao lower bound' condition, whether or not the score function is linear with the estimator function

    s(\theta;y)=b(\theta)[h(y)-g(\theta)]

    -\frac{2n}{\theta}=b(\theta)[\frac{2n+1}{2n\theta}Y_{(n)}-g(\theta)]

    should I find b and g of theta here?

    I am just copying the formula from the book. How do N! mathematicians do that?...
    Last edited by Volga; February 11th 2011 at 02:57 PM.
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    Another attempt at (d) By considering the condition that a function of Y_{(n)} is unbiased for \theta, determine whether there is a better unbiased estimator for \theta.

    If {2n+1\over 2n}Y_{(n)} is an unbiased etimator for \theta based on a random sample Y_1, ... Y_n, then

    Var({2n+1\over 2n}Y_{(n)})\geq\frac{1}{nI(\theta)} ("I" here denotes information - what it the real Latex code for this letter? is it Greek?...)

    Var({2n+1\over 2n}Y_{(n)})\geq\frac{1}{nI(\theta)}=\frac{\theta^2  }{4n^3}

    Now I find the variance of my unbiased estimator and compare to the above variance threshold

    Var(({2n+1\over 2n}Y_{(n)})=\frac{(2n+1)^2}{4n^2\theta^2}Var({Y_{(  n)})

    E[(Y_{(n)})^2]=\int_0^{\theta}y^2n(\frac{y^2}{{\theta}^2}\frac{2  y}{\theta^2}dy=...=\frac{{\theta^2}2n}{2n+2}

    then Var(({2n+1\over 2n}Y_{(n)})=E[(Y_{(n)})^2]-E(Y_{(n)})^2=\frac{{\theta^2}2n}{2n+2}-(\frac{2n\theta}{2n+1})^2=\theta^2[\frac{2n}{2n+2}-\frac{4n^2}{(2n+1)^2}]

    Var(({2n+1\over 2n}Y_{(n)})=\frac{(2n+1)^2}{4n^2\theta^2}\theta^2[\frac{2n}{2n+2}-\frac{4n^2}{(2n+1)^2}]=...=\frac{1}{4n^2+4n}

    I am concerned that the final Var does not depend on theta - theta squared cancelled out in the computation.

    Now I understand one would see how this variance compares with \frac{\theta^2}{4n^3}

    ???
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  14. #14
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    Seriously, what the name of the letter used as notation for Fisher information?
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