Error when trying to derive variance of sample mean












0














Assume that $X$ is a random variable with mean $E[X]$ variance $sigma^2$. Let $mu = frac{1}{N}sum_{n=1}^N X_n$, where $X_n$ is i.i.d with respect to $n$ having mean $E[X]$ and variance $sigma^2$, be an estimator of $E[X]$. I tried to show that $Variance(E[X] - mu) = frac{sigma^2}{N}$ but I have made some error which I'm blind to... An approach which yields the correct answer can be found at
http://sepwww.stanford.edu/sep/prof/pvi/rand/paper_html/node16.html.



I'm proceeding as follows:
Let $e = E[X]-mu$ then
begin{align}
Variance(e) &= E[(e -E[e])^2] \
&= E[e^2] - E[e]^2 \
&= E[e^2] & text{unbiased estimator} \
&= E[(E[X]-mu)^2] \
&= E[E[X]^2 - E[X]mu + mu^2] \
&= E[mu^2] - E[X]^2 & text{since }E[mu] = E[X] \
&= Eleft[left(frac{1}{N}sum_{n=1}^N X_nright)^2right] - E[X]^2 \
&= Eleft[frac{1}{N^2}sum_{m=1}^Nsum_{n=1}^N X_m X_nright] - E[X]^2 \
&= left(frac{1}{N^2}sum_{m=1}^Nsum_{n=1}^N E[X_m X_n]right) - E[X]^2 \
&= left(frac{1}{N^2}sum_{n=1}^N E[X_n^2]right) - E[X]^2 & text{due to independence} \
&= left(frac{1}{N^2}sum_{n=1}^N E[X^2]right) - E[X]^2 & text{from assumption on distr. of $X_n$} \
&= frac{1}{N} E[X^2] - E[X]^2 \
&= frac{1}{N} E[X^2] - E[X]^2 \
&= frac{1}{N} (sigma^2 + E[X]^2) - E[X]^2 \
&= frac{sigma^2}{N} + (frac{1}{N} - 1) E[X]^2 \
end{align}



EDIT:
Correcting the error pointed out by NCh gives the correct answer. Continuing from the 9th row
begin{align}
&= left(frac{1}{N^2}sum_{m=1}^Nsum_{n=1}^N E[X_m X_n]right) - E[X]^2 \
&= frac{1}{N^2}left(left(sum_{n=1}^N E[X_n^2]right) + left(sum_{m=1}^Nsum_{nin{1,dots,N}-{m}} E[X_m]E[X_n]right)right) - E[X]^2 & text{due to independence} \
&= frac{1}{N^2}left(left(sum_{n=1}^N E[X^2]right) + left(sum_{m=1}^Nsum_{nin{1,dots,N}-{m}} E[X]^2right)right) - E[X]^2 & text{from assumption on distr. of $X_n$} \
&= frac{1}{N^2}(NE[X^2] + N(N-1)E[X]^2) - E[X] \
&= frac{1}{N}(E[X^2] - E[X]^2) \
&= frac{sigma^2}{N} \
end{align}






probability statistics self-learning estimation expected-value







share|cite|improve this question





























    0














    Assume that $X$ is a random variable with mean $E[X]$ variance $sigma^2$. Let $mu = frac{1}{N}sum_{n=1}^N X_n$, where $X_n$ is i.i.d with respect to $n$ having mean $E[X]$ and variance $sigma^2$, be an estimator of $E[X]$. I tried to show that $Variance(E[X] - mu) = frac{sigma^2}{N}$ but I have made some error which I'm blind to... An approach which yields the correct answer can be found at
    http://sepwww.stanford.edu/sep/prof/pvi/rand/paper_html/node16.html.



    I'm proceeding as follows:
    Let $e = E[X]-mu$ then
    begin{align}
    Variance(e) &= E[(e -E[e])^2] \
    &= E[e^2] - E[e]^2 \
    &= E[e^2] & text{unbiased estimator} \
    &= E[(E[X]-mu)^2] \
    &= E[E[X]^2 - E[X]mu + mu^2] \
    &= E[mu^2] - E[X]^2 & text{since }E[mu] = E[X] \
    &= Eleft[left(frac{1}{N}sum_{n=1}^N X_nright)^2right] - E[X]^2 \
    &= Eleft[frac{1}{N^2}sum_{m=1}^Nsum_{n=1}^N X_m X_nright] - E[X]^2 \
    &= left(frac{1}{N^2}sum_{m=1}^Nsum_{n=1}^N E[X_m X_n]right) - E[X]^2 \
    &= left(frac{1}{N^2}sum_{n=1}^N E[X_n^2]right) - E[X]^2 & text{due to independence} \
    &= left(frac{1}{N^2}sum_{n=1}^N E[X^2]right) - E[X]^2 & text{from assumption on distr. of $X_n$} \
    &= frac{1}{N} E[X^2] - E[X]^2 \
    &= frac{1}{N} E[X^2] - E[X]^2 \
    &= frac{1}{N} (sigma^2 + E[X]^2) - E[X]^2 \
    &= frac{sigma^2}{N} + (frac{1}{N} - 1) E[X]^2 \
    end{align}



    EDIT:
    Correcting the error pointed out by NCh gives the correct answer. Continuing from the 9th row
    begin{align}
    &= left(frac{1}{N^2}sum_{m=1}^Nsum_{n=1}^N E[X_m X_n]right) - E[X]^2 \
    &= frac{1}{N^2}left(left(sum_{n=1}^N E[X_n^2]right) + left(sum_{m=1}^Nsum_{nin{1,dots,N}-{m}} E[X_m]E[X_n]right)right) - E[X]^2 & text{due to independence} \
    &= frac{1}{N^2}left(left(sum_{n=1}^N E[X^2]right) + left(sum_{m=1}^Nsum_{nin{1,dots,N}-{m}} E[X]^2right)right) - E[X]^2 & text{from assumption on distr. of $X_n$} \
    &= frac{1}{N^2}(NE[X^2] + N(N-1)E[X]^2) - E[X] \
    &= frac{1}{N}(E[X^2] - E[X]^2) \
    &= frac{sigma^2}{N} \
    end{align}






    probability statistics self-learning estimation expected-value







    share|cite|improve this question



























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      Assume that $X$ is a random variable with mean $E[X]$ variance $sigma^2$. Let $mu = frac{1}{N}sum_{n=1}^N X_n$, where $X_n$ is i.i.d with respect to $n$ having mean $E[X]$ and variance $sigma^2$, be an estimator of $E[X]$. I tried to show that $Variance(E[X] - mu) = frac{sigma^2}{N}$ but I have made some error which I'm blind to... An approach which yields the correct answer can be found at
      http://sepwww.stanford.edu/sep/prof/pvi/rand/paper_html/node16.html.



      I'm proceeding as follows:
      Let $e = E[X]-mu$ then
      begin{align}
      Variance(e) &= E[(e -E[e])^2] \
      &= E[e^2] - E[e]^2 \
      &= E[e^2] & text{unbiased estimator} \
      &= E[(E[X]-mu)^2] \
      &= E[E[X]^2 - E[X]mu + mu^2] \
      &= E[mu^2] - E[X]^2 & text{since }E[mu] = E[X] \
      &= Eleft[left(frac{1}{N}sum_{n=1}^N X_nright)^2right] - E[X]^2 \
      &= Eleft[frac{1}{N^2}sum_{m=1}^Nsum_{n=1}^N X_m X_nright] - E[X]^2 \
      &= left(frac{1}{N^2}sum_{m=1}^Nsum_{n=1}^N E[X_m X_n]right) - E[X]^2 \
      &= left(frac{1}{N^2}sum_{n=1}^N E[X_n^2]right) - E[X]^2 & text{due to independence} \
      &= left(frac{1}{N^2}sum_{n=1}^N E[X^2]right) - E[X]^2 & text{from assumption on distr. of $X_n$} \
      &= frac{1}{N} E[X^2] - E[X]^2 \
      &= frac{1}{N} E[X^2] - E[X]^2 \
      &= frac{1}{N} (sigma^2 + E[X]^2) - E[X]^2 \
      &= frac{sigma^2}{N} + (frac{1}{N} - 1) E[X]^2 \
      end{align}



      EDIT:
      Correcting the error pointed out by NCh gives the correct answer. Continuing from the 9th row
      begin{align}
      &= left(frac{1}{N^2}sum_{m=1}^Nsum_{n=1}^N E[X_m X_n]right) - E[X]^2 \
      &= frac{1}{N^2}left(left(sum_{n=1}^N E[X_n^2]right) + left(sum_{m=1}^Nsum_{nin{1,dots,N}-{m}} E[X_m]E[X_n]right)right) - E[X]^2 & text{due to independence} \
      &= frac{1}{N^2}left(left(sum_{n=1}^N E[X^2]right) + left(sum_{m=1}^Nsum_{nin{1,dots,N}-{m}} E[X]^2right)right) - E[X]^2 & text{from assumption on distr. of $X_n$} \
      &= frac{1}{N^2}(NE[X^2] + N(N-1)E[X]^2) - E[X] \
      &= frac{1}{N}(E[X^2] - E[X]^2) \
      &= frac{sigma^2}{N} \
      end{align}






      probability statistics self-learning estimation expected-value







      share|cite|improve this question















      Assume that $X$ is a random variable with mean $E[X]$ variance $sigma^2$. Let $mu = frac{1}{N}sum_{n=1}^N X_n$, where $X_n$ is i.i.d with respect to $n$ having mean $E[X]$ and variance $sigma^2$, be an estimator of $E[X]$. I tried to show that $Variance(E[X] - mu) = frac{sigma^2}{N}$ but I have made some error which I'm blind to... An approach which yields the correct answer can be found at
      http://sepwww.stanford.edu/sep/prof/pvi/rand/paper_html/node16.html.



      I'm proceeding as follows:
      Let $e = E[X]-mu$ then
      begin{align}
      Variance(e) &= E[(e -E[e])^2] \
      &= E[e^2] - E[e]^2 \
      &= E[e^2] & text{unbiased estimator} \
      &= E[(E[X]-mu)^2] \
      &= E[E[X]^2 - E[X]mu + mu^2] \
      &= E[mu^2] - E[X]^2 & text{since }E[mu] = E[X] \
      &= Eleft[left(frac{1}{N}sum_{n=1}^N X_nright)^2right] - E[X]^2 \
      &= Eleft[frac{1}{N^2}sum_{m=1}^Nsum_{n=1}^N X_m X_nright] - E[X]^2 \
      &= left(frac{1}{N^2}sum_{m=1}^Nsum_{n=1}^N E[X_m X_n]right) - E[X]^2 \
      &= left(frac{1}{N^2}sum_{n=1}^N E[X_n^2]right) - E[X]^2 & text{due to independence} \
      &= left(frac{1}{N^2}sum_{n=1}^N E[X^2]right) - E[X]^2 & text{from assumption on distr. of $X_n$} \
      &= frac{1}{N} E[X^2] - E[X]^2 \
      &= frac{1}{N} E[X^2] - E[X]^2 \
      &= frac{1}{N} (sigma^2 + E[X]^2) - E[X]^2 \
      &= frac{sigma^2}{N} + (frac{1}{N} - 1) E[X]^2 \
      end{align}



      EDIT:
      Correcting the error pointed out by NCh gives the correct answer. Continuing from the 9th row
      begin{align}
      &= left(frac{1}{N^2}sum_{m=1}^Nsum_{n=1}^N E[X_m X_n]right) - E[X]^2 \
      &= frac{1}{N^2}left(left(sum_{n=1}^N E[X_n^2]right) + left(sum_{m=1}^Nsum_{nin{1,dots,N}-{m}} E[X_m]E[X_n]right)right) - E[X]^2 & text{due to independence} \
      &= frac{1}{N^2}left(left(sum_{n=1}^N E[X^2]right) + left(sum_{m=1}^Nsum_{nin{1,dots,N}-{m}} E[X]^2right)right) - E[X]^2 & text{from assumption on distr. of $X_n$} \
      &= frac{1}{N^2}(NE[X^2] + N(N-1)E[X]^2) - E[X] \
      &= frac{1}{N}(E[X^2] - E[X]^2) \
      &= frac{sigma^2}{N} \
      end{align}






      probability statistics self-learning estimation expected-value




      probability statistics self-learning estimation expected-value






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      edited Dec 4 '18 at 8:04

























      asked Dec 3 '18 at 14:10









      Angelos

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          An error is when you went from 9th to 10th raw:
          $$mathbb E[X_m X_n]neq mathbb E[X_n^2]$$
          for $mneq n$.



          By independence,
          $$mathbb E[X_m X_n]=mathbb E[X_m]mathbb E[X_n]=mathbb E[X]^2$$



          Note also that by properties of variance,
          $$
          Var(E[X]-mu) = Var(mu) = Varleft(frac{sum_{i=1}^N X_i}{N}right) = frac{1}{N^2} Varleft(sum_{i=1}^N X_iright) = frac{1}{N^2} left(sum_{i=1}^N Var(X_i)right) = frac{N sigma^2}{N^2} = frac{sigma^2}{N}
          $$






          share|cite|improve this answer





















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            An error is when you went from 9th to 10th raw:
            $$mathbb E[X_m X_n]neq mathbb E[X_n^2]$$
            for $mneq n$.



            By independence,
            $$mathbb E[X_m X_n]=mathbb E[X_m]mathbb E[X_n]=mathbb E[X]^2$$



            Note also that by properties of variance,
            $$
            Var(E[X]-mu) = Var(mu) = Varleft(frac{sum_{i=1}^N X_i}{N}right) = frac{1}{N^2} Varleft(sum_{i=1}^N X_iright) = frac{1}{N^2} left(sum_{i=1}^N Var(X_i)right) = frac{N sigma^2}{N^2} = frac{sigma^2}{N}
            $$






            share|cite|improve this answer


























              0














              An error is when you went from 9th to 10th raw:
              $$mathbb E[X_m X_n]neq mathbb E[X_n^2]$$
              for $mneq n$.



              By independence,
              $$mathbb E[X_m X_n]=mathbb E[X_m]mathbb E[X_n]=mathbb E[X]^2$$



              Note also that by properties of variance,
              $$
              Var(E[X]-mu) = Var(mu) = Varleft(frac{sum_{i=1}^N X_i}{N}right) = frac{1}{N^2} Varleft(sum_{i=1}^N X_iright) = frac{1}{N^2} left(sum_{i=1}^N Var(X_i)right) = frac{N sigma^2}{N^2} = frac{sigma^2}{N}
              $$






              share|cite|improve this answer
























                0












                0








                0






                An error is when you went from 9th to 10th raw:
                $$mathbb E[X_m X_n]neq mathbb E[X_n^2]$$
                for $mneq n$.



                By independence,
                $$mathbb E[X_m X_n]=mathbb E[X_m]mathbb E[X_n]=mathbb E[X]^2$$



                Note also that by properties of variance,
                $$
                Var(E[X]-mu) = Var(mu) = Varleft(frac{sum_{i=1}^N X_i}{N}right) = frac{1}{N^2} Varleft(sum_{i=1}^N X_iright) = frac{1}{N^2} left(sum_{i=1}^N Var(X_i)right) = frac{N sigma^2}{N^2} = frac{sigma^2}{N}
                $$






                share|cite|improve this answer












                An error is when you went from 9th to 10th raw:
                $$mathbb E[X_m X_n]neq mathbb E[X_n^2]$$
                for $mneq n$.



                By independence,
                $$mathbb E[X_m X_n]=mathbb E[X_m]mathbb E[X_n]=mathbb E[X]^2$$



                Note also that by properties of variance,
                $$
                Var(E[X]-mu) = Var(mu) = Varleft(frac{sum_{i=1}^N X_i}{N}right) = frac{1}{N^2} Varleft(sum_{i=1}^N X_iright) = frac{1}{N^2} left(sum_{i=1}^N Var(X_i)right) = frac{N sigma^2}{N^2} = frac{sigma^2}{N}
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 4 '18 at 3:07









                NCh

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                6,2682723






























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