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Thread: I need some help, please

  1. #16
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    Re: I need some help, please




    k[2σ2+2σ2]=σ2
    k[4σ2]=σ2
    k=1/4

    is this correct or have i completely lost it???
    Looks fine with me.

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    Re: I need some help, please

    So than how would i conclude it?
    could i say that T is the unbiased estimator of σ2. im confused because normally you have to work out E(T) and than your answer should equal σ2.

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    Re: I need some help, please

    once you found the k value how should i prove its an unbiased estimator of sigma squared?

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    Re: I need some help, please

    Now the question is asking you to choose the value of k
    such that T is an unbiased estimator of \sigma^2

    So once you found the corresponding value of k,
    you have answered your question.

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    Re: I need some help, please

    could i say

    k[2σ2+2σ2]=σ2
    k[4σ2]=σ2
    k=1/4

    therefore T= 1/4[2σ2+2σ2]= σ2
    T=σ2
    therefore T is unbiased estimator of σ2.

    Also i just want to say thank you for the help i really appreciate it. only if i could be a help to you.

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    Re: I need some help, please

    Originally, from the definition of T given in your question,
    k is not known yet; however, once we fix it,
    we can determine whether it is an unbiased estimator for \sigma^2 or not.

    Similarly, we can solve for k such that it is an unbiased estimator.

    The thing I want to say is that the question want you to show

    E[T] = \sigma^2 \Rightarrow k = ...

    Of course the converse is also true:

    k = ... \Rightarrow E[T] = \sigma^2

    However, the question is given the condition "T is an unbiased estimator"
    It will be pretty strange and weird to present like we guess for a value of k
    and then check whether it is an unbiased estimator or not.



    Your argument is fine. But please note that the equation is
    E[T] = \sigma^2 but not T = \sigma^2.

    T is an estimator, which is random before you take the
    realization (the estimates).

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    Re: I need some help, please

    ok thanks i will just write the whole solution out and i just want you to confirm if it right or not.

    my answer;

    X1, X2...X4~N( ,σ2)

    E(Xi) =
    V(Xi) = σ2

    E[(x1-x2)^2] = X1^2 - 2X1X2 +X2^2
    E[X1^2-2X1X2+X2^2] = E[X1^2]-2E[X1]E[X2]+E[X2^2]
    E[X1] = , E[X1^2] = Var(X1) + (E[X1])^2 =σ2 +2
    E[X1^2]-2E[X1]E[X2]+E[X2^2] = σ2 +2 - 2 + σ2 +2 = 2 σ2

    for E[(X3-X4)^2] = 2 σ2

    T is an unbiased estimator for σ2

    if and only if E(T)= σ2

    T= k[2σ2+2σ2]
    σ2= k[4σ2]

    σ2/4σ2=k
    k= 1/4

    then i concluded it by writing T is an unbiased estimator of σ2.

    i hope all my question is answered and there is nothing left.

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    Re: I need some help, please

    Many thanks
    I got it

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    Re: I need some help, please

    X1, X2, .Xn is a random sample from a uniform distribution U(0,α). Suppose that we consider the following three statistics as estimators of the parameter α.

    Tn= sample maximum Un=aTn Vn=b ,

    where a, b are constants. You are given the following results.

    E(Tn) = nα/(n+1)

    V(Tn)= nα^2 /(n+1)^2 (n+2)

    Also given is that Un and Vn are both unbiased estimators of the parameter α.


    (i) Find the values of the constants a and b.

    (ii) Show that Un and Vn are consistent estimators of α.


    ive tried this question and ive uploaded the solutions, please can anyone chekc that is it correct or not if not please guide me thanks.



    X1....Xn ~ U(0,α)

    E(Xi) = α/2
    V(Xi) = α^2 / 12

    (i)

    Un= aTn
    Vn = bXbar


    E(Un) = E(aTn)= aE(Tn)
    = a x nα/(n+1)
    a= n+1/n

    E(Vn) = E(bXbar) = bE(Xbar)

    = b x α/2 = α(Vn=Un) = b =2
    b=2

    please check if part (i) is correct.

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    Re: I need some help, please

    Quote Originally Posted by vinux View Post
    Hi Fahad,
    This is easy one.

    E[ (cT-\sigma^2)^2 ] =\sigma^4 \times E[ (\frac{cT}{ \sigma^2 }-1)^2 ]
    =\sigma^4 \times E[ (c \frac{T}{\sigma^2}-1)^2 ] =\sigma^4 \times E[ (c W-1)^2] where W follows chi-square with n-1 df
    =\sigma^4 \times {c^2 E[W^2] -2c E[W] -1}

    Now you know the moments of the chi-square distribution. substitute it. you will get the answer
    Thank you for your help
    I think there is a mistake

    E[ (cT-\sigma^2)^2 ] =\sigma^4 \times {c^2 E[W^2] -2c E[W] +1}

    I have done the exercise
    The final answer is
    E[ (cT-\sigma^2)^2 ]= \sigma^4 [c^2(n^2-1)-2c(n-1)+1]

    Many thanks

  11. #26
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    Re: I need some help, please

    Quote Originally Posted by Smith194 View Post
    (i) Find the values of the constants a and b.

    (ii) Show that Un and Vn are consistent estimators of α.


    ive tried this question and ive uploaded the solutions, please can anyone chekc that is it correct or not if not please guide me thanks.



    X1....Xn ~ U(0,α)

    E(Xi) = α/2
    V(Xi) = α^2 / 12

    (i)

    Un= aTn
    Vn = bXbar


    E(Un) = E(aTn)= aE(Tn)
    = a x nα/(n+1)
    a= n+1/n

    E(Vn) = E(bXbar) = bE(Xbar)

    = b x α/2 = α(Vn=Un) = b =2
    b=2

    please check if part (i) is correct.
    Part (i) looks good.

    Part (ii)
    Since V_n is an unbiased estimator for \alpha,
    you just need to show that

    \lim_{n\to\infty} Var[V_n] 
= \lim_{n\to\infty} E[|V_n - \alpha|^2] = 0

    and the convergence in \mathcal{L}^2 implies the
    convergence in probability.

    Similar for U_n

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    Re: I need some help, please

    Could somebody continue with the second part?
    Show that the mean squared error is minimized at c=(n+1)^(-1). This estimator is known as the Pitman estimator for σ^2 in the Gaussian model. Show that the unbiased estimator has c=(n-1)^(-1) and compare its mean squared error to that of the Pitman estimator.]

  13. #28
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    Re: I need some help, please

    Wow pushing an old post

    Once you show the given expression in part 1) as the MSE, then you can immediately figure out it is just a quadratic expression in c, which is just a secondary school problem to find its minimum point.

    THe expectation of the sample variance is pretty standard too.

  14. The Following User Says Thank You to BGM For This Useful Post:

    arimya (11-13-2013)

  15. #29
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    Re: I need some help, please


    Thanks! it was so easy that I didn't see it .

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