Finding Best Unbiased Estimator of Uniform Distribution












2












$begingroup$


Let $X_i$, $i=1,...,n$ be iid with $f(x,theta) = frac{1}{2theta}$ for $-theta<x<theta$. Find the best unbiased estimator of $theta$ if one exists.



So I first tried $T(X)=X_{(n)}$, which gave me $$mathbb{E}[T]=frac{n-1}{n+1}theta.$$ I thus concluded that $T(X)=frac{n+1}{n-1}X_{(n)}$ should be an unbiased estimator.
Next, I wanted to check whether it is the best unbiased estimator using the Cramer Rao Lower Bound. It states
$$operatorname{Var}(T)geq frac{1}{I_X(theta)},$$
so I first calculated the Fisher information for a single random variable. The calculation yields
$$I_X(theta)=mathbb{E}_theta[V(theta,x)^2] = int_{-theta}^thetaleft(partial_thetalogfrac{1}{2theta}right)^{!2}frac{1}{2theta},dx = frac{1}{theta^2}.$$
Thus the lower bound is $dfrac{theta^2}{n}$. Next, I computed the variance of the estimator
$$operatorname{Var}[T] = int_{-theta}^theta left(frac{n+1}{n-1}right)^{!2}x^2nfrac{(x+theta)^{n-1}}{2^ntheta^n},dx-theta^2 = ... = theta^2frac{4n}{(n-1)^2(n-2)},$$
which is obviously lower than the lower bound. However, I am really unsure where I made a mistake. Any help is greatly appreciated.






statistical-inference estimation parameter-estimation







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  • $begingroup$
    First of all, the information inequality does not apply to pdfs where support depends on parameter. And if by best unbiased estimator you mean in the sense of having minimum variance, then your proposed unbiased estimator is not the UMVUE.
    $endgroup$
    – StubbornAtom
    Dec 7 '18 at 17:37












  • $begingroup$
    I am not mistaken, the UMVUE of $theta$ is $frac{n+1}{n}max_{1le ile n} |X_i|$, where $max |X_i|$ is a complete sufficient statistic.
    $endgroup$
    – StubbornAtom
    Dec 7 '18 at 17:43












  • $begingroup$
    Thank you, but how would one go about proving that it is the UMVUE without this theorem?
    $endgroup$
    – Analysis801
    Dec 7 '18 at 17:44










  • $begingroup$
    What theorem have you used here? UMVUE in this case is shown by using the Lehmann-Scheffe theorem.
    $endgroup$
    – StubbornAtom
    Dec 7 '18 at 17:47
















2












$begingroup$


Let $X_i$, $i=1,...,n$ be iid with $f(x,theta) = frac{1}{2theta}$ for $-theta<x<theta$. Find the best unbiased estimator of $theta$ if one exists.



So I first tried $T(X)=X_{(n)}$, which gave me $$mathbb{E}[T]=frac{n-1}{n+1}theta.$$ I thus concluded that $T(X)=frac{n+1}{n-1}X_{(n)}$ should be an unbiased estimator.
Next, I wanted to check whether it is the best unbiased estimator using the Cramer Rao Lower Bound. It states
$$operatorname{Var}(T)geq frac{1}{I_X(theta)},$$
so I first calculated the Fisher information for a single random variable. The calculation yields
$$I_X(theta)=mathbb{E}_theta[V(theta,x)^2] = int_{-theta}^thetaleft(partial_thetalogfrac{1}{2theta}right)^{!2}frac{1}{2theta},dx = frac{1}{theta^2}.$$
Thus the lower bound is $dfrac{theta^2}{n}$. Next, I computed the variance of the estimator
$$operatorname{Var}[T] = int_{-theta}^theta left(frac{n+1}{n-1}right)^{!2}x^2nfrac{(x+theta)^{n-1}}{2^ntheta^n},dx-theta^2 = ... = theta^2frac{4n}{(n-1)^2(n-2)},$$
which is obviously lower than the lower bound. However, I am really unsure where I made a mistake. Any help is greatly appreciated.






statistical-inference estimation parameter-estimation







share|cite|improve this question











$endgroup$












  • $begingroup$
    First of all, the information inequality does not apply to pdfs where support depends on parameter. And if by best unbiased estimator you mean in the sense of having minimum variance, then your proposed unbiased estimator is not the UMVUE.
    $endgroup$
    – StubbornAtom
    Dec 7 '18 at 17:37












  • $begingroup$
    I am not mistaken, the UMVUE of $theta$ is $frac{n+1}{n}max_{1le ile n} |X_i|$, where $max |X_i|$ is a complete sufficient statistic.
    $endgroup$
    – StubbornAtom
    Dec 7 '18 at 17:43












  • $begingroup$
    Thank you, but how would one go about proving that it is the UMVUE without this theorem?
    $endgroup$
    – Analysis801
    Dec 7 '18 at 17:44










  • $begingroup$
    What theorem have you used here? UMVUE in this case is shown by using the Lehmann-Scheffe theorem.
    $endgroup$
    – StubbornAtom
    Dec 7 '18 at 17:47














2












2








2





$begingroup$


Let $X_i$, $i=1,...,n$ be iid with $f(x,theta) = frac{1}{2theta}$ for $-theta<x<theta$. Find the best unbiased estimator of $theta$ if one exists.



So I first tried $T(X)=X_{(n)}$, which gave me $$mathbb{E}[T]=frac{n-1}{n+1}theta.$$ I thus concluded that $T(X)=frac{n+1}{n-1}X_{(n)}$ should be an unbiased estimator.
Next, I wanted to check whether it is the best unbiased estimator using the Cramer Rao Lower Bound. It states
$$operatorname{Var}(T)geq frac{1}{I_X(theta)},$$
so I first calculated the Fisher information for a single random variable. The calculation yields
$$I_X(theta)=mathbb{E}_theta[V(theta,x)^2] = int_{-theta}^thetaleft(partial_thetalogfrac{1}{2theta}right)^{!2}frac{1}{2theta},dx = frac{1}{theta^2}.$$
Thus the lower bound is $dfrac{theta^2}{n}$. Next, I computed the variance of the estimator
$$operatorname{Var}[T] = int_{-theta}^theta left(frac{n+1}{n-1}right)^{!2}x^2nfrac{(x+theta)^{n-1}}{2^ntheta^n},dx-theta^2 = ... = theta^2frac{4n}{(n-1)^2(n-2)},$$
which is obviously lower than the lower bound. However, I am really unsure where I made a mistake. Any help is greatly appreciated.






statistical-inference estimation parameter-estimation







share|cite|improve this question











$endgroup$




Let $X_i$, $i=1,...,n$ be iid with $f(x,theta) = frac{1}{2theta}$ for $-theta<x<theta$. Find the best unbiased estimator of $theta$ if one exists.



So I first tried $T(X)=X_{(n)}$, which gave me $$mathbb{E}[T]=frac{n-1}{n+1}theta.$$ I thus concluded that $T(X)=frac{n+1}{n-1}X_{(n)}$ should be an unbiased estimator.
Next, I wanted to check whether it is the best unbiased estimator using the Cramer Rao Lower Bound. It states
$$operatorname{Var}(T)geq frac{1}{I_X(theta)},$$
so I first calculated the Fisher information for a single random variable. The calculation yields
$$I_X(theta)=mathbb{E}_theta[V(theta,x)^2] = int_{-theta}^thetaleft(partial_thetalogfrac{1}{2theta}right)^{!2}frac{1}{2theta},dx = frac{1}{theta^2}.$$
Thus the lower bound is $dfrac{theta^2}{n}$. Next, I computed the variance of the estimator
$$operatorname{Var}[T] = int_{-theta}^theta left(frac{n+1}{n-1}right)^{!2}x^2nfrac{(x+theta)^{n-1}}{2^ntheta^n},dx-theta^2 = ... = theta^2frac{4n}{(n-1)^2(n-2)},$$
which is obviously lower than the lower bound. However, I am really unsure where I made a mistake. Any help is greatly appreciated.






statistical-inference estimation parameter-estimation




statistical-inference estimation parameter-estimation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 17:03









Adrian Keister

4,84561933




4,84561933










asked Dec 7 '18 at 16:51









Analysis801Analysis801

1196




1196












  • $begingroup$
    First of all, the information inequality does not apply to pdfs where support depends on parameter. And if by best unbiased estimator you mean in the sense of having minimum variance, then your proposed unbiased estimator is not the UMVUE.
    $endgroup$
    – StubbornAtom
    Dec 7 '18 at 17:37












  • $begingroup$
    I am not mistaken, the UMVUE of $theta$ is $frac{n+1}{n}max_{1le ile n} |X_i|$, where $max |X_i|$ is a complete sufficient statistic.
    $endgroup$
    – StubbornAtom
    Dec 7 '18 at 17:43












  • $begingroup$
    Thank you, but how would one go about proving that it is the UMVUE without this theorem?
    $endgroup$
    – Analysis801
    Dec 7 '18 at 17:44










  • $begingroup$
    What theorem have you used here? UMVUE in this case is shown by using the Lehmann-Scheffe theorem.
    $endgroup$
    – StubbornAtom
    Dec 7 '18 at 17:47


















  • $begingroup$
    First of all, the information inequality does not apply to pdfs where support depends on parameter. And if by best unbiased estimator you mean in the sense of having minimum variance, then your proposed unbiased estimator is not the UMVUE.
    $endgroup$
    – StubbornAtom
    Dec 7 '18 at 17:37












  • $begingroup$
    I am not mistaken, the UMVUE of $theta$ is $frac{n+1}{n}max_{1le ile n} |X_i|$, where $max |X_i|$ is a complete sufficient statistic.
    $endgroup$
    – StubbornAtom
    Dec 7 '18 at 17:43












  • $begingroup$
    Thank you, but how would one go about proving that it is the UMVUE without this theorem?
    $endgroup$
    – Analysis801
    Dec 7 '18 at 17:44










  • $begingroup$
    What theorem have you used here? UMVUE in this case is shown by using the Lehmann-Scheffe theorem.
    $endgroup$
    – StubbornAtom
    Dec 7 '18 at 17:47
















$begingroup$
First of all, the information inequality does not apply to pdfs where support depends on parameter. And if by best unbiased estimator you mean in the sense of having minimum variance, then your proposed unbiased estimator is not the UMVUE.
$endgroup$
– StubbornAtom
Dec 7 '18 at 17:37






$begingroup$
First of all, the information inequality does not apply to pdfs where support depends on parameter. And if by best unbiased estimator you mean in the sense of having minimum variance, then your proposed unbiased estimator is not the UMVUE.
$endgroup$
– StubbornAtom
Dec 7 '18 at 17:37














$begingroup$
I am not mistaken, the UMVUE of $theta$ is $frac{n+1}{n}max_{1le ile n} |X_i|$, where $max |X_i|$ is a complete sufficient statistic.
$endgroup$
– StubbornAtom
Dec 7 '18 at 17:43






$begingroup$
I am not mistaken, the UMVUE of $theta$ is $frac{n+1}{n}max_{1le ile n} |X_i|$, where $max |X_i|$ is a complete sufficient statistic.
$endgroup$
– StubbornAtom
Dec 7 '18 at 17:43














$begingroup$
Thank you, but how would one go about proving that it is the UMVUE without this theorem?
$endgroup$
– Analysis801
Dec 7 '18 at 17:44




$begingroup$
Thank you, but how would one go about proving that it is the UMVUE without this theorem?
$endgroup$
– Analysis801
Dec 7 '18 at 17:44












$begingroup$
What theorem have you used here? UMVUE in this case is shown by using the Lehmann-Scheffe theorem.
$endgroup$
– StubbornAtom
Dec 7 '18 at 17:47




$begingroup$
What theorem have you used here? UMVUE in this case is shown by using the Lehmann-Scheffe theorem.
$endgroup$
– StubbornAtom
Dec 7 '18 at 17:47










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