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Math Help - Unbiased Estimators - Uniform Distribution

  1. #1
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    Unbiased Estimators - Uniform Distribution

    Let Y1, Y2, ... Yn denote a random sample from uniform distribution
    on the interval (0, theta) ...

    let YBAR = sample mean, MAX = sample maximum

    Consider:

    estimator1 = 2(YBAR)

    estimator2 = ([n+1]/n)MAX

    Show that both estimators are unbiased estimators of theta.

    Find the efficiency of estimator1 relative to estimator2.

    Which estimator is preferable? Explain.

    I have no idea how to start this. Any help would be appreciated!!
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  2. #2
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    Re: Unbiased Estimators - Uniform Distribution

    do you know what it means for an estimator to be unbiased?

    do you know what the efficiency of an estimator is?

    You should have some idea how to start these problems or they wouldn't give them to you. Right?
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  3. #3
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    Re: Unbiased Estimators - Uniform Distribution

    I know that the Uniform distribution from (0, θ) is:

    f(x) = {1/θ for 0 ≤ x ≤ θ and 0 elsewhere

    E(x) = θ/2 and this would be the 1st population moment

    x̅ would be the first sample moment

    setting E(x) = x̅ you would get: theta hat = 2*x̅

    Note that E(2*x̅ ) = 2*E(x̅ ) = 2*(θ/2) = θ making it an unbiased estimator

    I'm not even sure if that is correct or what I am supposed to be doing.
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    Re: Unbiased Estimators - Uniform Distribution

    Quote Originally Posted by AwesomeHedgehog View Post
    I know that the Uniform distribution from (0, θ) is:

    f(x) = {1/θ for 0 ≤ x ≤ θ and 0 elsewhere

    E(x) = θ/2 and this would be the 1st population moment

    x̅ would be the first sample moment

    setting E(x) = x̅ you would get: theta hat = 2*x̅

    Note that E(2*x̅ ) = 2*E(x̅ ) = 2*(θ/2) = θ making it an unbiased estimator

    I'm not even sure if that is correct or what I am supposed to be doing.
    so much terminology

    You have a bunch of i.i.d. samples $Y_n$ all distributed $U[0, \theta]$

    You take their sample mean. i.e.

    $\bar{Y}=\dfrac 1 N \displaystyle{\sum_{n=1}^N}Y_n$

    The expected value of the sample mean is

    $E[\bar{Y}]=\dfrac 1 N \displaystyle{\sum_{n=1}^N}E[Y_n]$

    $E[\bar{Y}]=\dfrac 1 N \cdot N \left(\dfrac \theta 2\right) = \dfrac \theta 2$

    also

    $E[Y_n] = \dfrac \theta 2$

    so $E[\bar{Y}]=E[Y_n]~\forall n$ and thus the sample mean is an unbiased estimator of $\dfrac \theta 2$ and thus of $\theta$

    see if you can show the 2nd estimator is unbiased.
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    Re: Unbiased Estimators - Uniform Distribution

    For the second estimator, I have:

    Var(x) = (θ^2)/12

    E(x^2) = Var(x) + (E(x))^2 = (θ^2)/12 + θ/2 = [7*θ^2]/12 and this would be the 2nd population moment

    The 2nd sample moment would be: [Σ(Xi)^2]/n

    Setting the 2nd population moment and sample moment equal to each other you get:

    [7*θ^2]/12 = [Σ(Xi)^2]/n

    solving for θ you get: θ = sqrt(12*Σ(Xi)^2]/7n)
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    Re: Unbiased Estimators - Uniform Distribution

    Quote Originally Posted by AwesomeHedgehog View Post
    For the second estimator, I have:

    Var(x) = (θ^2)/12

    E(x^2) = Var(x) + (E(x))^2 = (θ^2)/12 + θ/2 = [7*θ^2]/12 and this would be the 2nd population moment

    The 2nd sample moment would be: [Σ(Xi)^2]/n

    Setting the 2nd population moment and sample moment equal to each other you get:

    [7*θ^2]/12 = [Σ(Xi)^2]/n

    solving for θ you get: θ = sqrt(12*Σ(Xi)^2]/7n)
    you lost me on this. You just need to find the expected value of the estimate $\dfrac{n+1} n \max(Y_n)$

    Can you determine the probability distribution of the max of $n~ U[0,\theta]$ random variables?

    hint: $Pr[\max(Y_n) < y] = Pr[\text{all }Y_n < y] = \left(\dfrac y \theta \right)^n$
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    Re: Unbiased Estimators - Uniform Distribution

    This is what I did:

    P[MAX(Yn) ≤ y] = P[y1, y2, ... , yn ≤ y] = (y/θ)^n

    f(MAX(Yn)) = n/[θ*(y/θ)^(n-1)]

    E[((n+1)/n)*MAX ] = integral from 0 to θ of [(y*(n+1))/n] * f(MAX(Yn)) dy

    => integral from 0 to θ of (n+1)*(y/θ)^n dy

    => (1/θ^n)*θ^(n+1) = θ

    So estimator 2 would be an unbiased estimator
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    Re: Unbiased Estimators - Uniform Distribution

    Quote Originally Posted by AwesomeHedgehog View Post
    This is what I did:

    P[MAX(Yn) ≤ y] = P[y1, y2, ... , yn ≤ y] = (y/θ)^n

    f(MAX(Yn)) = n/[θ*(y/θ)^(n-1)]

    E[((n+1)/n)*MAX ] = integral from 0 to θ of [(y*(n+1))/n] * f(MAX(Yn)) dy

    => integral from 0 to θ of (n+1)*(y/θ)^n dy

    => (1/θ^n)*θ^(n+1) = θ

    So estimator 2 would be an unbiased estimator
    perfect.

    relative efficiency of two different estimators $T_1, T_2$ is given as

    $e(T_1,T_2)=\dfrac{E[(T_2-\theta)^2]}{E[(T_1-\theta)^2]}$

    now go ahead and compute the relative efficiency of these two estimators.
    Last edited by romsek; April 13th 2014 at 06:25 AM.
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    Re: Unbiased Estimators - Uniform Distribution

    To calculate var(y), we need to calculate E(y^2)

    Integral[0, θ] (1/θ)* y^2 dy = (1/(3θ)) y^3 * E[0,θ] = 1/(3θ)( θ^3 - 0) = θ^(2/3)

    Then, var(y) = E(y^2) - (E(y))^2 = theta^2/3 - (theta/2)^2 = theta^2/3 - theta^2/4 = theta^2/12

    var(y-bar) = theta^2/12n

    Then, as estimator 1 = 2 y-bar, var(2 y-bar) = 4 theta^2/12 n = theta^2/3n

    For the var(estimator2), first we calculate E(estimator^2) =

    I[0, theta] y^2((n+1)/n)^2 n/theta(y/theta)^n-1 dy =

    (n+1)^2/(n theta^n) I[0, theta] y^n+1 dy =

    (n+1)^2/(n(n+2) theta^n) y^(n+2) E[0, theta] =

    (n+1)^2/(n(n+2) theta^n)(theta^(n+2)-0) =

    (n+1)^2/(n(n+2) theta^2

    Then, var(estimator 2) = (n+1)^2/(n(n+2) theta^2 - theta^2 = theta^2/(n(n+2))

    Then, eff estimator 1/eff estimator 2 for unbiased estimators = var estimator 2/var estimator 1 =

    theta^2/(n(n+2))/(theta^2/3n) =

    3/(n+2)

    3/(n+2) < 1 for n > 1

    As the efficiency of estimator 1 relative to estimator 2 is less than 1 for all n > 1, estimator 2 is preferred for n > 1
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    Re: Unbiased Estimators - Uniform Distribution

    ??
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  11. #11
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    Re: Unbiased Estimators - Uniform Distribution

    Quote Originally Posted by AwesomeHedgehog View Post
    ??
    it's so hard to read your math.

    I get

    $E[\left(2\bar{Y}-\theta\right)^2] = \dfrac {n }{3}\theta^2$

    $E[\left(\max(Y_n)-\theta\right)^2] =\left(\dfrac{n+1}{n}\right)^2 \dfrac{n \theta^2}{n+2}=\dfrac{(n+1)^2}{n(n+2)}\theta^2$

    $e=\dfrac{\dfrac{(n+1)^2}{n(n+2)}}{\dfrac {n}{3}}= \dfrac{3(n+1)^2}{n^2(n+2)}$

    $e < 1~\forall n>4$

    So as $n>4$ the statistic $\dfrac{n+1}{n}\max(Y_n)$ becomes more efficient than $2\bar{Y}$.

    You should probably double check this.
    Thanks from AwesomeHedgehog
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