Calculating the variance of an estimator (unclear on one step)











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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...










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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...










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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...










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      enter image description here



      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






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      edited Apr 11 '13 at 8:05







      user60610

















      asked Apr 11 '13 at 7:31









      Silver

      129117




      129117






















          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
          $$
          begin{align}
          mathrm{Var}(bar{X})&=mathrm{Var}left(frac{1}{n}sum_{i=1}^nX_iright)=frac{1}{n^2}mathrm{Var}left(sum_{i=1}^n X_iright)=frac{1}{n^2}sum_{i=1}^nmathrm{Var}(X_i)\
          &=frac{1}{n^2}nmathrm{Var}(X_1)=frac{1}{n}mathrm{Var}(X_1).
          end{align}
          $$






          share|cite|improve this answer























          • 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











          Your Answer





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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes








          up vote
          3
          down vote



          accepted










          Assuming that $X_1,ldots,X_n$ are independent and identically distributed (this is used explicitly in equality 3 and 4) we have
          $$
          begin{align}
          mathrm{Var}(bar{X})&=mathrm{Var}left(frac{1}{n}sum_{i=1}^nX_iright)=frac{1}{n^2}mathrm{Var}left(sum_{i=1}^n X_iright)=frac{1}{n^2}sum_{i=1}^nmathrm{Var}(X_i)\
          &=frac{1}{n^2}nmathrm{Var}(X_1)=frac{1}{n}mathrm{Var}(X_1).
          end{align}
          $$






          share|cite|improve this answer























          • 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















          up vote
          3
          down vote



          accepted










          Assuming that $X_1,ldots,X_n$ are independent and identically distributed (this is used explicitly in equality 3 and 4) we have
          $$
          begin{align}
          mathrm{Var}(bar{X})&=mathrm{Var}left(frac{1}{n}sum_{i=1}^nX_iright)=frac{1}{n^2}mathrm{Var}left(sum_{i=1}^n X_iright)=frac{1}{n^2}sum_{i=1}^nmathrm{Var}(X_i)\
          &=frac{1}{n^2}nmathrm{Var}(X_1)=frac{1}{n}mathrm{Var}(X_1).
          end{align}
          $$






          share|cite|improve this answer























          • 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













          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          Assuming that $X_1,ldots,X_n$ are independent and identically distributed (this is used explicitly in equality 3 and 4) we have
          $$
          begin{align}
          mathrm{Var}(bar{X})&=mathrm{Var}left(frac{1}{n}sum_{i=1}^nX_iright)=frac{1}{n^2}mathrm{Var}left(sum_{i=1}^n X_iright)=frac{1}{n^2}sum_{i=1}^nmathrm{Var}(X_i)\
          &=frac{1}{n^2}nmathrm{Var}(X_1)=frac{1}{n}mathrm{Var}(X_1).
          end{align}
          $$






          share|cite|improve this answer














          Assuming that $X_1,ldots,X_n$ are independent and identically distributed (this is used explicitly in equality 3 and 4) we have
          $$
          begin{align}
          mathrm{Var}(bar{X})&=mathrm{Var}left(frac{1}{n}sum_{i=1}^nX_iright)=frac{1}{n^2}mathrm{Var}left(sum_{i=1}^n X_iright)=frac{1}{n^2}sum_{i=1}^nmathrm{Var}(X_i)\
          &=frac{1}{n^2}nmathrm{Var}(X_1)=frac{1}{n}mathrm{Var}(X_1).
          end{align}
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 22 at 19:52









          Daniel Shatz

          1033




          1033










          answered Apr 11 '13 at 7:34









          Stefan Hansen

          20.6k73662




          20.6k73662












          • 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






          • 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


















           

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