Calculating the variance of an estimator (unclear on one step)
How can you go from $4V(bar X)$ to $displaystyle frac{4}{n}V(X_1)$? I understand the rest of the steps...
probability statistics estimation-theory
asked Apr 11 '13 at 7:31
SilverSilver
129117
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How can you go from $4V(bar X)$ to $displaystyle frac{4}{n}V(X_1)$? I understand the rest of the steps...
probability statistics estimation-theory
asked Apr 11 '13 at 7:31
SilverSilver
129117
add a comment |
How can you go from $4V(bar X)$ to $displaystyle frac{4}{n}V(X_1)$? I understand the rest of the steps...
probability statistics estimation-theory
asked Apr 11 '13 at 7:31
SilverSilver
129117
How can you go from $4V(bar X)$ to $displaystyle frac{4}{n}V(X_1)$? I understand the rest of the steps...
probability statistics estimation-theory
probability statistics estimation-theory
asked Apr 11 '13 at 7:31
SilverSilver
129117
asked Apr 11 '13 at 7:31
SilverSilver
129117
edited Apr 11 '13 at 8:05
user60610
asked Apr 11 '13 at 7:31
SilverSilver
129117
asked Apr 11 '13 at 7:31
SilverSilver
129117
asked Apr 11 '13 at 7:31
SilverSilver
129117
129117
add a comment |
add a comment |
1 Answer
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Assuming that $X_1,ldots,X_n$ are independent and identically distributed (this is used explicitly in equality 3 and 4) we have
$$
begin{align}
mathrm{Var}(bar{X})&=mathrm{Var}left(frac{1}{n}sum_{i=1}^nX_iright)=frac{1}{n^2}mathrm{Var}left(sum_{i=1}^n X_iright)=frac{1}{n^2}sum_{i=1}^nmathrm{Var}(X_i)\
&=frac{1}{n^2}nmathrm{Var}(X_1)=frac{1}{n}mathrm{Var}(X_1).
end{align}
$$
edited Nov 22 '18 at 19:52
Daniel Shatz
1033
answered Apr 11 '13 at 7:34
Stefan HansenStefan Hansen
20.8k73663
Follow up question: How can you change the Xi to X1? Could you have just as easily have picked X2 or Xn? What's the deal?
– Silver
Apr 11 '13 at 7:42
1
Yes, this is because $X_1,ldots,X_n$ (presumably) are identically distributed. In particular, $mathrm{Var}(X_1)=cdots=mathrm{Var}(X_n)$.
– Stefan Hansen
Apr 11 '13 at 7:44
Ah of course, that makes sense, thanks!
– Silver
Apr 11 '13 at 7:50
add a comment |
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1 Answer
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1 Answer
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Assuming that $X_1,ldots,X_n$ are independent and identically distributed (this is used explicitly in equality 3 and 4) we have
$$
begin{align}
mathrm{Var}(bar{X})&=mathrm{Var}left(frac{1}{n}sum_{i=1}^nX_iright)=frac{1}{n^2}mathrm{Var}left(sum_{i=1}^n X_iright)=frac{1}{n^2}sum_{i=1}^nmathrm{Var}(X_i)\
&=frac{1}{n^2}nmathrm{Var}(X_1)=frac{1}{n}mathrm{Var}(X_1).
end{align}
$$
edited Nov 22 '18 at 19:52
Daniel Shatz
1033
answered Apr 11 '13 at 7:34
Stefan HansenStefan Hansen
20.8k73663
Follow up question: How can you change the Xi to X1? Could you have just as easily have picked X2 or Xn? What's the deal?
– Silver
Apr 11 '13 at 7:42
1
Yes, this is because $X_1,ldots,X_n$ (presumably) are identically distributed. In particular, $mathrm{Var}(X_1)=cdots=mathrm{Var}(X_n)$.
– Stefan Hansen
Apr 11 '13 at 7:44
Ah of course, that makes sense, thanks!
– Silver
Apr 11 '13 at 7:50
add a comment |
Assuming that $X_1,ldots,X_n$ are independent and identically distributed (this is used explicitly in equality 3 and 4) we have
$$
begin{align}
mathrm{Var}(bar{X})&=mathrm{Var}left(frac{1}{n}sum_{i=1}^nX_iright)=frac{1}{n^2}mathrm{Var}left(sum_{i=1}^n X_iright)=frac{1}{n^2}sum_{i=1}^nmathrm{Var}(X_i)\
&=frac{1}{n^2}nmathrm{Var}(X_1)=frac{1}{n}mathrm{Var}(X_1).
end{align}
$$
edited Nov 22 '18 at 19:52
Daniel Shatz
1033
answered Apr 11 '13 at 7:34
Stefan HansenStefan Hansen
20.8k73663
Follow up question: How can you change the Xi to X1? Could you have just as easily have picked X2 or Xn? What's the deal?
– Silver
Apr 11 '13 at 7:42
1
Yes, this is because $X_1,ldots,X_n$ (presumably) are identically distributed. In particular, $mathrm{Var}(X_1)=cdots=mathrm{Var}(X_n)$.
– Stefan Hansen
Apr 11 '13 at 7:44
Ah of course, that makes sense, thanks!
– Silver
Apr 11 '13 at 7:50
add a comment |
Assuming that $X_1,ldots,X_n$ are independent and identically distributed (this is used explicitly in equality 3 and 4) we have
$$
begin{align}
mathrm{Var}(bar{X})&=mathrm{Var}left(frac{1}{n}sum_{i=1}^nX_iright)=frac{1}{n^2}mathrm{Var}left(sum_{i=1}^n X_iright)=frac{1}{n^2}sum_{i=1}^nmathrm{Var}(X_i)\
&=frac{1}{n^2}nmathrm{Var}(X_1)=frac{1}{n}mathrm{Var}(X_1).
end{align}
$$
edited Nov 22 '18 at 19:52
Daniel Shatz
1033
answered Apr 11 '13 at 7:34
Stefan HansenStefan Hansen
20.8k73663
Assuming that $X_1,ldots,X_n$ are independent and identically distributed (this is used explicitly in equality 3 and 4) we have
$$
begin{align}
mathrm{Var}(bar{X})&=mathrm{Var}left(frac{1}{n}sum_{i=1}^nX_iright)=frac{1}{n^2}mathrm{Var}left(sum_{i=1}^n X_iright)=frac{1}{n^2}sum_{i=1}^nmathrm{Var}(X_i)\
&=frac{1}{n^2}nmathrm{Var}(X_1)=frac{1}{n}mathrm{Var}(X_1).
end{align}
$$
edited Nov 22 '18 at 19:52
Daniel Shatz
1033
answered Apr 11 '13 at 7:34
Stefan HansenStefan Hansen
20.8k73663
edited Nov 22 '18 at 19:52
Daniel Shatz
1033
edited Nov 22 '18 at 19:52
Daniel Shatz
1033
edited Nov 22 '18 at 19:52
Daniel Shatz
1033
1033
answered Apr 11 '13 at 7:34
Stefan HansenStefan Hansen
20.8k73663
answered Apr 11 '13 at 7:34
Stefan HansenStefan Hansen
20.8k73663
answered Apr 11 '13 at 7:34
Stefan HansenStefan Hansen
20.8k73663
20.8k73663
Follow up question: How can you change the Xi to X1? Could you have just as easily have picked X2 or Xn? What's the deal?
– Silver
Apr 11 '13 at 7:42
1
Yes, this is because $X_1,ldots,X_n$ (presumably) are identically distributed. In particular, $mathrm{Var}(X_1)=cdots=mathrm{Var}(X_n)$.
– Stefan Hansen
Apr 11 '13 at 7:44
Ah of course, that makes sense, thanks!
– Silver
Apr 11 '13 at 7:50
add a comment |
Follow up question: How can you change the Xi to X1? Could you have just as easily have picked X2 or Xn? What's the deal?
– Silver
Apr 11 '13 at 7:42
1
Yes, this is because $X_1,ldots,X_n$ (presumably) are identically distributed. In particular, $mathrm{Var}(X_1)=cdots=mathrm{Var}(X_n)$.
– Stefan Hansen
Apr 11 '13 at 7:44
Ah of course, that makes sense, thanks!
– Silver
Apr 11 '13 at 7:50
Follow up question: How can you change the Xi to X1? Could you have just as easily have picked X2 or Xn? What's the deal?
– Silver
Apr 11 '13 at 7:42
Follow up question: How can you change the Xi to X1? Could you have just as easily have picked X2 or Xn? What's the deal?
– Silver
Apr 11 '13 at 7:42
1
1
Yes, this is because $X_1,ldots,X_n$ (presumably) are identically distributed. In particular, $mathrm{Var}(X_1)=cdots=mathrm{Var}(X_n)$.
– Stefan Hansen
Apr 11 '13 at 7:44
Yes, this is because $X_1,ldots,X_n$ (presumably) are identically distributed. In particular, $mathrm{Var}(X_1)=cdots=mathrm{Var}(X_n)$.
– Stefan Hansen
Apr 11 '13 at 7:44
Ah of course, that makes sense, thanks!
– Silver
Apr 11 '13 at 7:50
Ah of course, that makes sense, thanks!
– Silver
Apr 11 '13 at 7:50
add a comment |
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