Mixing
Method of moments estimator for $theta^2$ when $X_isim p_{theta}(x)=frac{2x}{theta^2}1_{0{leq}x{leq}theta}$
$begingroup$
Let $X_1, ..., X_n$ be from a sample from a probability distribution:
$$p_{theta}(x) = frac{2x}{theta^2}1_{0{leq}x{leq}theta}$$
, where $theta > 0$ is an unknown parameter.
I have found from (i) and (ii) of the question that the unbiased method of moments estimator for $theta$ is $frac{3bar{X}}{2}$.
(iii) asks for an estimator for $theta^2$, which I found to be $2bar{X^2}$ using the second moment of the distribution. However, this is inconsistent with the answer in my textbook which says that it is $(frac{3bar{x}}{2})^2$.
I am confused as to the method my textbook used to arrive at this answer. It seems that the estimator for $theta$ is simply squared, which I assumed is only possible for Maximum Likelihood Estimators when the function of $theta$ is bijective.
(iv) asks for an unbiased estimator for $theta^2$, which if I were to use the estimator as given by my textbook, I would require the variance of the distribution to compute, which I am unable to obtain.
Would really appreciate some help on this. Thanks!
statistics probability-distributions parameter-estimation
edited Jan 27 at 20:21
StubbornAtom
6,29831339
asked Jan 27 at 10:05
Shi TianShi Tian
62
$endgroup$
add a comment |
$begingroup$
Let $X_1, ..., X_n$ be from a sample from a probability distribution:
$$p_{theta}(x) = frac{2x}{theta^2}1_{0{leq}x{leq}theta}$$
, where $theta > 0$ is an unknown parameter.
I have found from (i) and (ii) of the question that the unbiased method of moments estimator for $theta$ is $frac{3bar{X}}{2}$.
(iii) asks for an estimator for $theta^2$, which I found to be $2bar{X^2}$ using the second moment of the distribution. However, this is inconsistent with the answer in my textbook which says that it is $(frac{3bar{x}}{2})^2$.
I am confused as to the method my textbook used to arrive at this answer. It seems that the estimator for $theta$ is simply squared, which I assumed is only possible for Maximum Likelihood Estimators when the function of $theta$ is bijective.
(iv) asks for an unbiased estimator for $theta^2$, which if I were to use the estimator as given by my textbook, I would require the variance of the distribution to compute, which I am unable to obtain.
Would really appreciate some help on this. Thanks!
statistics probability-distributions parameter-estimation
edited Jan 27 at 20:21
StubbornAtom
6,29831339
asked Jan 27 at 10:05
Shi TianShi Tian
62
$endgroup$
$begingroup$
Are you asked for only a method of moments estimator of $theta^2$ ?
$endgroup$
– StubbornAtom
Jan 27 at 10:13
$begingroup$
The exact phrasing of the question is simply: Give the method of moments estimator for $theta^2$
$endgroup$
– Shi Tian
Jan 27 at 10:17
$begingroup$
If you are using sample raw moments to equate with population raw moments, then shouldn't you have $frac{theta^2}{2}=frac{1}{n}sumlimits_{k=1}^n X_k^2implies hattheta^2(X_1,ldots,X_n)=frac{2}{n}sumlimits_{k=1}^n X_k^2$ ? Your book's answer is different though.
$endgroup$
– StubbornAtom
Jan 27 at 20:48
1
$begingroup$
Possible duplicate of Method of moments estimator for $theta^{2}$.
$endgroup$
– StubbornAtom
Jan 27 at 20:51
add a comment |
$begingroup$
Let $X_1, ..., X_n$ be from a sample from a probability distribution:
$$p_{theta}(x) = frac{2x}{theta^2}1_{0{leq}x{leq}theta}$$
, where $theta > 0$ is an unknown parameter.
I have found from (i) and (ii) of the question that the unbiased method of moments estimator for $theta$ is $frac{3bar{X}}{2}$.
(iii) asks for an estimator for $theta^2$, which I found to be $2bar{X^2}$ using the second moment of the distribution. However, this is inconsistent with the answer in my textbook which says that it is $(frac{3bar{x}}{2})^2$.
I am confused as to the method my textbook used to arrive at this answer. It seems that the estimator for $theta$ is simply squared, which I assumed is only possible for Maximum Likelihood Estimators when the function of $theta$ is bijective.
(iv) asks for an unbiased estimator for $theta^2$, which if I were to use the estimator as given by my textbook, I would require the variance of the distribution to compute, which I am unable to obtain.
Would really appreciate some help on this. Thanks!
statistics probability-distributions parameter-estimation
edited Jan 27 at 20:21
StubbornAtom
6,29831339
asked Jan 27 at 10:05
Shi TianShi Tian
62
$endgroup$
Let $X_1, ..., X_n$ be from a sample from a probability distribution:
$$p_{theta}(x) = frac{2x}{theta^2}1_{0{leq}x{leq}theta}$$
, where $theta > 0$ is an unknown parameter.
I have found from (i) and (ii) of the question that the unbiased method of moments estimator for $theta$ is $frac{3bar{X}}{2}$.
(iii) asks for an estimator for $theta^2$, which I found to be $2bar{X^2}$ using the second moment of the distribution. However, this is inconsistent with the answer in my textbook which says that it is $(frac{3bar{x}}{2})^2$.
I am confused as to the method my textbook used to arrive at this answer. It seems that the estimator for $theta$ is simply squared, which I assumed is only possible for Maximum Likelihood Estimators when the function of $theta$ is bijective.
(iv) asks for an unbiased estimator for $theta^2$, which if I were to use the estimator as given by my textbook, I would require the variance of the distribution to compute, which I am unable to obtain.
Would really appreciate some help on this. Thanks!
statistics probability-distributions parameter-estimation
statistics probability-distributions parameter-estimation
edited Jan 27 at 20:21
StubbornAtom
6,29831339
asked Jan 27 at 10:05
Shi TianShi Tian
62
edited Jan 27 at 20:21
StubbornAtom
6,29831339
asked Jan 27 at 10:05
Shi TianShi Tian
62
edited Jan 27 at 20:21
StubbornAtom
6,29831339
edited Jan 27 at 20:21
StubbornAtom
6,29831339
edited Jan 27 at 20:21
StubbornAtom
6,29831339
6,29831339
asked Jan 27 at 10:05
Shi TianShi Tian
62
asked Jan 27 at 10:05
Shi TianShi Tian
62
asked Jan 27 at 10:05
Shi TianShi Tian
62
62
$begingroup$
Are you asked for only a method of moments estimator of $theta^2$ ?
$endgroup$
– StubbornAtom
Jan 27 at 10:13
$begingroup$
The exact phrasing of the question is simply: Give the method of moments estimator for $theta^2$
$endgroup$
– Shi Tian
Jan 27 at 10:17
$begingroup$
If you are using sample raw moments to equate with population raw moments, then shouldn't you have $frac{theta^2}{2}=frac{1}{n}sumlimits_{k=1}^n X_k^2implies hattheta^2(X_1,ldots,X_n)=frac{2}{n}sumlimits_{k=1}^n X_k^2$ ? Your book's answer is different though.
$endgroup$
– StubbornAtom
Jan 27 at 20:48
1
$begingroup$
Possible duplicate of Method of moments estimator for $theta^{2}$.
$endgroup$
– StubbornAtom
Jan 27 at 20:51
add a comment |
$begingroup$
Are you asked for only a method of moments estimator of $theta^2$ ?
$endgroup$
– StubbornAtom
Jan 27 at 10:13
$begingroup$
The exact phrasing of the question is simply: Give the method of moments estimator for $theta^2$
$endgroup$
– Shi Tian
Jan 27 at 10:17
$begingroup$
If you are using sample raw moments to equate with population raw moments, then shouldn't you have $frac{theta^2}{2}=frac{1}{n}sumlimits_{k=1}^n X_k^2implies hattheta^2(X_1,ldots,X_n)=frac{2}{n}sumlimits_{k=1}^n X_k^2$ ? Your book's answer is different though.
$endgroup$
– StubbornAtom
Jan 27 at 20:48
1
$begingroup$
Possible duplicate of Method of moments estimator for $theta^{2}$.
$endgroup$
– StubbornAtom
Jan 27 at 20:51
$begingroup$
Are you asked for only a method of moments estimator of $theta^2$ ?
$endgroup$
– StubbornAtom
Jan 27 at 10:13
$begingroup$
Are you asked for only a method of moments estimator of $theta^2$ ?
$endgroup$
– StubbornAtom
Jan 27 at 10:13
$begingroup$
The exact phrasing of the question is simply: Give the method of moments estimator for $theta^2$
$endgroup$
– Shi Tian
Jan 27 at 10:17
$begingroup$
The exact phrasing of the question is simply: Give the method of moments estimator for $theta^2$
$endgroup$
– Shi Tian
Jan 27 at 10:17
$begingroup$
If you are using sample raw moments to equate with population raw moments, then shouldn't you have $frac{theta^2}{2}=frac{1}{n}sumlimits_{k=1}^n X_k^2implies hattheta^2(X_1,ldots,X_n)=frac{2}{n}sumlimits_{k=1}^n X_k^2$ ? Your book's answer is different though.
$endgroup$
– StubbornAtom
Jan 27 at 20:48
$begingroup$
If you are using sample raw moments to equate with population raw moments, then shouldn't you have $frac{theta^2}{2}=frac{1}{n}sumlimits_{k=1}^n X_k^2implies hattheta^2(X_1,ldots,X_n)=frac{2}{n}sumlimits_{k=1}^n X_k^2$ ? Your book's answer is different though.
$endgroup$
– StubbornAtom
Jan 27 at 20:48
1
1
$begingroup$
Possible duplicate of Method of moments estimator for $theta^{2}$.
$endgroup$
– StubbornAtom
Jan 27 at 20:51
$begingroup$
Possible duplicate of Method of moments estimator for $theta^{2}$.
$endgroup$
– StubbornAtom
Jan 27 at 20:51
add a comment |
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$begingroup$
Are you asked for only a method of moments estimator of $theta^2$ ?
$endgroup$
– StubbornAtom
Jan 27 at 10:13
$begingroup$
The exact phrasing of the question is simply: Give the method of moments estimator for $theta^2$
$endgroup$
– Shi Tian
Jan 27 at 10:17
$begingroup$
If you are using sample raw moments to equate with population raw moments, then shouldn't you have $frac{theta^2}{2}=frac{1}{n}sumlimits_{k=1}^n X_k^2implies hattheta^2(X_1,ldots,X_n)=frac{2}{n}sumlimits_{k=1}^n X_k^2$ ? Your book's answer is different though.
$endgroup$
– StubbornAtom
Jan 27 at 20:48
1
$begingroup$
Possible duplicate of Method of moments estimator for $theta^{2}$.
$endgroup$
– StubbornAtom
Jan 27 at 20:51