Choosing the best estimator
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Given yield measurements $X_1,X_2,X_3$ from three independent runs of an experiment with variance $sigma^2$, which is the better of the two estimators:
$hattheta_{1}$= $frac{X_1+X_2+X_3}{3}$, $hattheta_{2}$=$frac{X_1+2X_2+X_3}{4}$
I know that in order to find the best estimator if both are unbiased, we are supposed to choose the one with the smallest variance. I need help just starting this problem. Thank you.
probability statistics
asked Sep 28 '13 at 16:52
glou92glou92
63
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add a comment |
$begingroup$
Given yield measurements $X_1,X_2,X_3$ from three independent runs of an experiment with variance $sigma^2$, which is the better of the two estimators:
$hattheta_{1}$= $frac{X_1+X_2+X_3}{3}$, $hattheta_{2}$=$frac{X_1+2X_2+X_3}{4}$
I know that in order to find the best estimator if both are unbiased, we are supposed to choose the one with the smallest variance. I need help just starting this problem. Thank you.
probability statistics
asked Sep 28 '13 at 16:52
glou92glou92
63
$endgroup$
add a comment |
$begingroup$
Given yield measurements $X_1,X_2,X_3$ from three independent runs of an experiment with variance $sigma^2$, which is the better of the two estimators:
$hattheta_{1}$= $frac{X_1+X_2+X_3}{3}$, $hattheta_{2}$=$frac{X_1+2X_2+X_3}{4}$
I know that in order to find the best estimator if both are unbiased, we are supposed to choose the one with the smallest variance. I need help just starting this problem. Thank you.
probability statistics
asked Sep 28 '13 at 16:52
glou92glou92
63
$endgroup$
Given yield measurements $X_1,X_2,X_3$ from three independent runs of an experiment with variance $sigma^2$, which is the better of the two estimators:
$hattheta_{1}$= $frac{X_1+X_2+X_3}{3}$, $hattheta_{2}$=$frac{X_1+2X_2+X_3}{4}$
I know that in order to find the best estimator if both are unbiased, we are supposed to choose the one with the smallest variance. I need help just starting this problem. Thank you.
probability statistics
probability statistics
asked Sep 28 '13 at 16:52
glou92glou92
63
asked Sep 28 '13 at 16:52
glou92glou92
63
edited Sep 28 '13 at 17:02
user85362
asked Sep 28 '13 at 16:52
glou92glou92
63
asked Sep 28 '13 at 16:52
glou92glou92
63
asked Sep 28 '13 at 16:52
glou92glou92
63
63
add a comment |
add a comment |
2 Answers
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Variance of the Sample mean estimate is just $$text{VAR}(bar{mu})=text{VAR}left(frac{sum_i X_i}{n}right)$$ for the first case and $$text{VAR}(bar{mu_2})=text{VAR}left(frac{X_1+X_2+X_2+X_3}{4}right)$$ in the second. Now apply scaling property of variances to see which one has a higher variance.
answered Sep 28 '13 at 17:02
SudarsanSudarsan
836718
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Hints:
$text{Var}(X_1+X_2) = text{Var}(X_1)+text{Var}(X_2)$ if they are independent
$text{Var}(k X) = k^2 text{Var}(X)$ and so $text{Var}left(dfrac{X}{k}right) = dfrac{text{Var}(X)}{k^2} $
answered Sep 29 '13 at 13:21
HenryHenry
100k480166
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
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active
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$begingroup$
Variance of the Sample mean estimate is just $$text{VAR}(bar{mu})=text{VAR}left(frac{sum_i X_i}{n}right)$$ for the first case and $$text{VAR}(bar{mu_2})=text{VAR}left(frac{X_1+X_2+X_2+X_3}{4}right)$$ in the second. Now apply scaling property of variances to see which one has a higher variance.
answered Sep 28 '13 at 17:02
SudarsanSudarsan
836718
$endgroup$
add a comment |
$begingroup$
Variance of the Sample mean estimate is just $$text{VAR}(bar{mu})=text{VAR}left(frac{sum_i X_i}{n}right)$$ for the first case and $$text{VAR}(bar{mu_2})=text{VAR}left(frac{X_1+X_2+X_2+X_3}{4}right)$$ in the second. Now apply scaling property of variances to see which one has a higher variance.
answered Sep 28 '13 at 17:02
SudarsanSudarsan
836718
$endgroup$
add a comment |
$begingroup$
Variance of the Sample mean estimate is just $$text{VAR}(bar{mu})=text{VAR}left(frac{sum_i X_i}{n}right)$$ for the first case and $$text{VAR}(bar{mu_2})=text{VAR}left(frac{X_1+X_2+X_2+X_3}{4}right)$$ in the second. Now apply scaling property of variances to see which one has a higher variance.
answered Sep 28 '13 at 17:02
SudarsanSudarsan
836718
$endgroup$
Variance of the Sample mean estimate is just $$text{VAR}(bar{mu})=text{VAR}left(frac{sum_i X_i}{n}right)$$ for the first case and $$text{VAR}(bar{mu_2})=text{VAR}left(frac{X_1+X_2+X_2+X_3}{4}right)$$ in the second. Now apply scaling property of variances to see which one has a higher variance.
answered Sep 28 '13 at 17:02
SudarsanSudarsan
836718
answered Sep 28 '13 at 17:02
SudarsanSudarsan
836718
answered Sep 28 '13 at 17:02
SudarsanSudarsan
836718
answered Sep 28 '13 at 17:02
SudarsanSudarsan
836718
836718
add a comment |
add a comment |
$begingroup$
Hints:
$text{Var}(X_1+X_2) = text{Var}(X_1)+text{Var}(X_2)$ if they are independent
$text{Var}(k X) = k^2 text{Var}(X)$ and so $text{Var}left(dfrac{X}{k}right) = dfrac{text{Var}(X)}{k^2} $
answered Sep 29 '13 at 13:21
HenryHenry
100k480166
$endgroup$
add a comment |
$begingroup$
Hints:
$text{Var}(X_1+X_2) = text{Var}(X_1)+text{Var}(X_2)$ if they are independent
$text{Var}(k X) = k^2 text{Var}(X)$ and so $text{Var}left(dfrac{X}{k}right) = dfrac{text{Var}(X)}{k^2} $
answered Sep 29 '13 at 13:21
HenryHenry
100k480166
$endgroup$
add a comment |
$begingroup$
Hints:
$text{Var}(X_1+X_2) = text{Var}(X_1)+text{Var}(X_2)$ if they are independent
$text{Var}(k X) = k^2 text{Var}(X)$ and so $text{Var}left(dfrac{X}{k}right) = dfrac{text{Var}(X)}{k^2} $
answered Sep 29 '13 at 13:21
HenryHenry
100k480166
$endgroup$
Hints:
$text{Var}(X_1+X_2) = text{Var}(X_1)+text{Var}(X_2)$ if they are independent
$text{Var}(k X) = k^2 text{Var}(X)$ and so $text{Var}left(dfrac{X}{k}right) = dfrac{text{Var}(X)}{k^2} $
answered Sep 29 '13 at 13:21
HenryHenry
100k480166
answered Sep 29 '13 at 13:21
HenryHenry
100k480166
answered Sep 29 '13 at 13:21
HenryHenry
100k480166
answered Sep 29 '13 at 13:21
HenryHenry
100k480166
100k480166
add a comment |
add a comment |
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