Computing the variance of an estimator












0












$begingroup$


I'm having some trouble figuring out the proper estimator for the following problem.



Given three random samples $ X_1 $, $ X_2 $ and $ X_3 $ from a normal distribution $ N(mu, sigma^2) $, compute the variance of the estimator



$$ T = frac{X_1+2X_2-X_3}{2} $$



Here's what I tried:



$$ T = frac{X_1+2X_2-X_3}{2} = frac{1}{2}X_1 + X_2 - frac{1}{2}X_3 $$
$$ Var(T) = Var(frac{1}{2}X_1) + Var(X_2) - Var(frac{1}{2}X_3) $$



Then, since $ Var(rY) = r^2 Var(Y) $



$$ Var(T) = frac{1}{4}sigma^2 + sigma^2 - frac{1}{4}sigma^2 = sigma^2 $$



However, the correct answer should be $ frac{3}{2}sigma^2 $. What am I doing wrong?






probability statistics







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$endgroup$












  • $begingroup$
    View it as $tfrac12 X_1+X_2+(-tfrac12 X_3)$ and try again.
    $endgroup$
    – WimC
    Jan 26 at 19:24
















0












$begingroup$


I'm having some trouble figuring out the proper estimator for the following problem.



Given three random samples $ X_1 $, $ X_2 $ and $ X_3 $ from a normal distribution $ N(mu, sigma^2) $, compute the variance of the estimator



$$ T = frac{X_1+2X_2-X_3}{2} $$



Here's what I tried:



$$ T = frac{X_1+2X_2-X_3}{2} = frac{1}{2}X_1 + X_2 - frac{1}{2}X_3 $$
$$ Var(T) = Var(frac{1}{2}X_1) + Var(X_2) - Var(frac{1}{2}X_3) $$



Then, since $ Var(rY) = r^2 Var(Y) $



$$ Var(T) = frac{1}{4}sigma^2 + sigma^2 - frac{1}{4}sigma^2 = sigma^2 $$



However, the correct answer should be $ frac{3}{2}sigma^2 $. What am I doing wrong?






probability statistics







share|cite|improve this question









$endgroup$












  • $begingroup$
    View it as $tfrac12 X_1+X_2+(-tfrac12 X_3)$ and try again.
    $endgroup$
    – WimC
    Jan 26 at 19:24














0












0








0





$begingroup$


I'm having some trouble figuring out the proper estimator for the following problem.



Given three random samples $ X_1 $, $ X_2 $ and $ X_3 $ from a normal distribution $ N(mu, sigma^2) $, compute the variance of the estimator



$$ T = frac{X_1+2X_2-X_3}{2} $$



Here's what I tried:



$$ T = frac{X_1+2X_2-X_3}{2} = frac{1}{2}X_1 + X_2 - frac{1}{2}X_3 $$
$$ Var(T) = Var(frac{1}{2}X_1) + Var(X_2) - Var(frac{1}{2}X_3) $$



Then, since $ Var(rY) = r^2 Var(Y) $



$$ Var(T) = frac{1}{4}sigma^2 + sigma^2 - frac{1}{4}sigma^2 = sigma^2 $$



However, the correct answer should be $ frac{3}{2}sigma^2 $. What am I doing wrong?






probability statistics







share|cite|improve this question









$endgroup$




I'm having some trouble figuring out the proper estimator for the following problem.



Given three random samples $ X_1 $, $ X_2 $ and $ X_3 $ from a normal distribution $ N(mu, sigma^2) $, compute the variance of the estimator



$$ T = frac{X_1+2X_2-X_3}{2} $$



Here's what I tried:



$$ T = frac{X_1+2X_2-X_3}{2} = frac{1}{2}X_1 + X_2 - frac{1}{2}X_3 $$
$$ Var(T) = Var(frac{1}{2}X_1) + Var(X_2) - Var(frac{1}{2}X_3) $$



Then, since $ Var(rY) = r^2 Var(Y) $



$$ Var(T) = frac{1}{4}sigma^2 + sigma^2 - frac{1}{4}sigma^2 = sigma^2 $$



However, the correct answer should be $ frac{3}{2}sigma^2 $. What am I doing wrong?






probability statistics




probability statistics






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 26 at 19:09









molenzwiebelmolenzwiebel

1153




1153












  • $begingroup$
    View it as $tfrac12 X_1+X_2+(-tfrac12 X_3)$ and try again.
    $endgroup$
    – WimC
    Jan 26 at 19:24


















  • $begingroup$
    View it as $tfrac12 X_1+X_2+(-tfrac12 X_3)$ and try again.
    $endgroup$
    – WimC
    Jan 26 at 19:24
















$begingroup$
View it as $tfrac12 X_1+X_2+(-tfrac12 X_3)$ and try again.
$endgroup$
– WimC
Jan 26 at 19:24




$begingroup$
View it as $tfrac12 X_1+X_2+(-tfrac12 X_3)$ and try again.
$endgroup$
– WimC
Jan 26 at 19:24










1 Answer
1






active

oldest

votes


















1












$begingroup$

You have a sign mistake. For any real random variable $X$, it holds that $Var(X) = Var(-X)$. So your calculation should be



$$ Var(T) = Var(frac{1}{2}X_1) + Var(X_2) + Var(frac{1}{2}X_3) $$



where we assume that the samples are independent.



Therefore,



$$ Var(T) = frac{1}{4}sigma^2 + sigma^2 + frac{1}{4}sigma^2 = 1.5sigma^2 $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ah there's the error. Thanks!
    $endgroup$
    – molenzwiebel
    Jan 26 at 19:52











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1 Answer
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active

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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









1












$begingroup$

You have a sign mistake. For any real random variable $X$, it holds that $Var(X) = Var(-X)$. So your calculation should be



$$ Var(T) = Var(frac{1}{2}X_1) + Var(X_2) + Var(frac{1}{2}X_3) $$



where we assume that the samples are independent.



Therefore,



$$ Var(T) = frac{1}{4}sigma^2 + sigma^2 + frac{1}{4}sigma^2 = 1.5sigma^2 $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ah there's the error. Thanks!
    $endgroup$
    – molenzwiebel
    Jan 26 at 19:52
















1












$begingroup$

You have a sign mistake. For any real random variable $X$, it holds that $Var(X) = Var(-X)$. So your calculation should be



$$ Var(T) = Var(frac{1}{2}X_1) + Var(X_2) + Var(frac{1}{2}X_3) $$



where we assume that the samples are independent.



Therefore,



$$ Var(T) = frac{1}{4}sigma^2 + sigma^2 + frac{1}{4}sigma^2 = 1.5sigma^2 $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ah there's the error. Thanks!
    $endgroup$
    – molenzwiebel
    Jan 26 at 19:52














1












1








1





$begingroup$

You have a sign mistake. For any real random variable $X$, it holds that $Var(X) = Var(-X)$. So your calculation should be



$$ Var(T) = Var(frac{1}{2}X_1) + Var(X_2) + Var(frac{1}{2}X_3) $$



where we assume that the samples are independent.



Therefore,



$$ Var(T) = frac{1}{4}sigma^2 + sigma^2 + frac{1}{4}sigma^2 = 1.5sigma^2 $$






share|cite|improve this answer









$endgroup$



You have a sign mistake. For any real random variable $X$, it holds that $Var(X) = Var(-X)$. So your calculation should be



$$ Var(T) = Var(frac{1}{2}X_1) + Var(X_2) + Var(frac{1}{2}X_3) $$



where we assume that the samples are independent.



Therefore,



$$ Var(T) = frac{1}{4}sigma^2 + sigma^2 + frac{1}{4}sigma^2 = 1.5sigma^2 $$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 26 at 19:20









user144410user144410

1,0412719




1,0412719












  • $begingroup$
    Ah there's the error. Thanks!
    $endgroup$
    – molenzwiebel
    Jan 26 at 19:52


















  • $begingroup$
    Ah there's the error. Thanks!
    $endgroup$
    – molenzwiebel
    Jan 26 at 19:52
















$begingroup$
Ah there's the error. Thanks!
$endgroup$
– molenzwiebel
Jan 26 at 19:52




$begingroup$
Ah there's the error. Thanks!
$endgroup$
– molenzwiebel
Jan 26 at 19:52


















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