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Math Help - density for order stat, sufficient stat, MLE question

  1. #1
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    density for order stat, sufficient stat, MLE question

    Pls take a look, this is an exam practice question but no answer is provided.

    Question.

    Let Y_1, ... Y_n be a random sample from a distribution with density function

    f_Y(y;\theta)=3\theta^{-3}y^2, 0\leq{y}\leq{\theta}
    f_Y(y;\theta)=0 otherwise

    where \theta>0 is an unknown parameter.

    (a) Let X=max_{1\leq{i}\leq{n}}Y_i. Find the density function of X.

    (b) find a simple sufficient statistic for \theta.

    (c) find the maximum likelihood estimator \hat{\theta} for \theta.

    (d) compute the mean square error of the maximum likelihood estimator \hat{\theta}.


    Answer.

    (a) I take it X is the maximum Y in the sample ie Y_{(n)} so I find the density of it

    P(Y_{(n)}\leq{y})=P(Y_1\leq{y},... Y_n\leq{y})=[P(Y\leq{y})}^n=[F_Y(y)]^n=(3\theta^{-3}\frac{y^3}{3})^n=\theta^{-3n}y^{3n}

    f_X(y)=f_{Y_{(n)}}(y)=\frac{d}{dy}F_{Y_{(n)}}(y)=n  f_Y(y)[F_y(y)]^n=n*3\theta^{-3}y^2*\theta^{-3n}y^{3n}=3n\theta^{-3-3n}y^{2+3n}


    (b) to find sufficient statistics for \theta, I will use Factorisation theorem, ie factorising the joint density function of the sample

    f_Y(y;\theta)=(3\theta^{-3})^n\prod_{i=1}^nY_i^2I_{Y_{(1)}\geq{0}}(y)I_{Y_{  (n)}\leq\theta}(y)=(3\theta^{-3})^nI_{Y_{(n)}\leq\theta}(y)*\prod_{i=1}^nY_i^2I_  {Y_{(1)}\geq{0}}(y)

    the second component depends on the sample only; the first component depends on \theta and on the sample through Y_{(n)}(y) and can be a sufficient statistic for \theta.

    ?? does this makes sense ??

    (c) here is where I am stuck and cannot move on to (d)

    To find MLE for \theta, I will find maximum of log-likelihood of the density of Y

    L_Y(y;\theta)=(3\theta^{-3})^n\prod_{i=1}^nY_i^2 subject to a constraint 0\leq{y}\leq{\theta}

    l_Y(y;\theta)=ln(L_Y)=ln[(3\theta^{-3})^n]+ln[\prod_{i=1}^nY_i^2]=n\ln(3\theta^{-3})+\Sigma_{i=1}^n\ln{Y_i^2}

    \frac{\partial}{\partial\theta}(l_Y)=n\frac{1}{3\t  heta^{-3}}(-9\theta^{-4})=-\frac{3n}{\theta^{-3}}\theta^{-4}=-\frac{3n}{\theta}

    this can be equal to zero only if n=0 and this is not what I am looking for here...

    Can you help me to spot where I've gone wrong?

    thanks!
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  2. #2
    Senior Member Sambit's Avatar
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    Remember one thing: whenever the range of the variable in the pdf depends upon the unknown parameter, MLE can NOT be obtained by just differentiation.
    Here, the range of y depends on \theta (since y\leq\theta)

    MLE is such a value of the unknown parameter which maximizes the pdf. What do you think, for what value of \theta the pdf will be maximum?
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  3. #3
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    Quote Originally Posted by Sambit View Post
    Remember one thing: whenever the range of the variable in the pdf depends upon the unknown parameter, MLE can NOT be obtained by just differentiation.
    Here, the range of y depends on \theta (since y\leq\theta)

    MLE is such a value of the unknown parameter which maximizes the pdf. What do you think, for what value of \theta the pdf will be maximum?
    Oh, good point )))

    So, my likelihood function

    L_Y(\theta;y)=(3\theta^{-3})^n\prod_{i=1}^nY_i^2I_{Y_{(1)}\geq{0}}I_{Y_{(n)  }\leq{\theta}}

    will reach maximum value (given fixed Y and \theta>0) if \theta=Y_{(n)}.

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  4. #4
    Senior Member Sambit's Avatar
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    Further hint: the pdf attains maximum when \theta is minimum. Do you follow that?
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  5. #5
    Senior Member Sambit's Avatar
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    Yeah....you got it right.....
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  6. #6
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    Quote Originally Posted by Sambit View Post
    Further hint: the pdf attains maximum when \theta is minimum. Do you follow that?
    Yes I saw that (negative power of theta) but its lower bound is zero...

    I must admit I can only do straightforward MLEs (and only on a clear day) so I am going blind here ))). I've posted a suggestion above. Appreciate your comment! This could be a good learning question for me.
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  7. #7
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    Quote Originally Posted by Sambit View Post
    Yeah....you got it right.....
    thanks!! will proceed to (d) then.
    Last edited by Volga; March 10th 2011 at 09:50 PM.
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  8. #8
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    (d) find MSE for \hat{\theta}=Y_{(n)}=X. Now I can use the density for X that I worked out in part (a) - truly if there is a rifle on the stage, there must be a moment in the performance when it will shoot ))))

    MSE_{\theta}(X)=E[(X-\theta)^2]=[Bias_{\theta}(X)]^2+Var_{\theta}(X)

    So I am planning to proceed as follows:

    Bias(X)=Bias (Y_{(n)})=\int_0^{\theta}(y-\theta)3n\theta^{-3-3n}y^{2+3n}dy

    Var(X)=E(X^2)-E(X)^2

    E(X)=E(Y_{(n)})=\int_0^{\theta}y3n\theta^{-3-3n}y^{2+3n}dy

    E(X^2)=E(Y_{(n)}^2)=\int_0^{\theta}y^23n\theta^{-3-3n}y^{2+3n}dy

    Can you tell me if I am on the right path??...
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  9. #9
    MHF Contributor matheagle's Avatar
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    To simplify the calculus, you can let W_n=Y_{(n)}/\theta

    Then F_W(w)=F_{Y{(n)}}(\theta w)=w^{3n} on 0<w<1.

    This makes all these calculations easier.
    This is a Beta random variable by the way
    with \alpha=3n and \beta=1
    so you need not do any computation at all.
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  10. #10
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    Quote Originally Posted by matheagle View Post
    This is a Beta random variable by the way
    with \alpha=3n and \beta=1
    so you need not do any computation at all.
    ah! another challenge ))) let me try it, the things you've taught me so far proved to be valuable indeed
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  11. #11
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    So, if W=\frac{Y_{(n)}}{\theta}, F_W(w)=F(\theta{w})=w^{3n}, f_W(w)=3nw^{3n-1}, 0\leq{w}\leq{1} and W\sim{Beta}(3n,1)

    Then we use Beta distribution to find mean and variance of W:


    Mean= E(W)=\frac{\alpha}{\alpha+\beta}=\frac{3n}{3n+1}

    E(Y_{(n)})=E(\theta{W})=\theta{E}(W)=\frac{\theta{  3n}}{3n+1}


    Variance= Var(W)=\frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+  \beta+1)}=\frac{3n}{(3n+1)^2(3n+2)}

    Var(Y_{(n)})=Var(\theta{W})=\theta^2Var(W)=\frac{\  theta^23n}{(3n+1)^2(3n+2)}

    (and I have checked the E and Var above by direct integration for w)


    Bias(Y_{(n)})=E(Y_{(n)}-\theta))=E(W\theta-\theta)=-E(\theta(1-W))=

    =-\int_0^1\theta(1-w)w^{3n}dw=-\theta\int_0^1w^{3n}(1-w)dw=-\theta{B}(3n+1,2)=

    =-\theta\frac{\Gamma(3n+1)\Gamma(2)}{\Gamma(3n+1+2)}  =-\theta\frac{(3n+1-1)!(2-1)!}{(3n+1+2-1)!}=-\theta\frac{(3n)!}{(3n+2)!}=-\frac{\theta}{(3n+1)(3n+2)}

    MSE_{\theta}(Y_{(n)})=\frac{\theta^2}{(3n+1)^2(3n+  2)^2}+\frac{\theta^2}{(3n+1)^2(3n+2)}=\frac{\theta  ^2(3n+3)}{(3n+1)^2(3n+2)}

    How's that?
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  12. #12
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by Volga View Post
    ah! another challenge ))) let me try it, the things you've taught me so far proved to be valuable indeed
    if you keep saying that
    someone may have a cow here
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  13. #13
    MHF Contributor matheagle's Avatar
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    I only just glanced at your calculations and it looks like you're missing a square of the term in the denominator of the MSE.
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  14. #14
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    Quote Originally Posted by matheagle View Post
    I only just glanced at your calculations and it looks like you're missing a square of the term in the denominator of the MSE.
    Thanks! you are right. But I cannot go back to that post and edit it, so this error (I think it is rather a typo) will be forever left on my record on this forum ))) whoever else reads this post, be aware...
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