Let $Y_n = X_n + X_{n+1}$ and $T_n = frac{1}{n} sum_{i=1}^n Y_i$. I want to find $Var[T_n]$












0












$begingroup$


Let $Y_n = X_n + X_{n+1}$ with $X_n$ ~ Bernoulli$(p)$ independent and identically distributed. Let $T_n = frac{1}{n} sum_{i=1}^n Y_i$. I want to find $Var[T_n]$.



This is what I have done so far:



I know that $Y_n$ ~ Bin(2,p). Then I can write this:



$$ Var[T_n] = Var[frac{1}{n}sum_{i=1}^n Y_i]= frac{1}{n^2}Var[sum_{i=1}^n Y_i] =$$



$$=frac{1}{n^2}sum_{i=1}^nVar[ Y_i] = frac{1}{n^2}*n* 2p(1-p) $$



However, the result should actually contain a $frac{2n-1}{n^2}$. Can somebody point out my mistake?






probability convergence







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  • $begingroup$
    The random variables $Y_i$ are not independent and therefore $text{var}(sum_{i} Y_i) = sum_i text{var}(Y_i)$ fails to hold true.
    $endgroup$
    – saz
    Jan 4 at 19:57
















0












$begingroup$


Let $Y_n = X_n + X_{n+1}$ with $X_n$ ~ Bernoulli$(p)$ independent and identically distributed. Let $T_n = frac{1}{n} sum_{i=1}^n Y_i$. I want to find $Var[T_n]$.



This is what I have done so far:



I know that $Y_n$ ~ Bin(2,p). Then I can write this:



$$ Var[T_n] = Var[frac{1}{n}sum_{i=1}^n Y_i]= frac{1}{n^2}Var[sum_{i=1}^n Y_i] =$$



$$=frac{1}{n^2}sum_{i=1}^nVar[ Y_i] = frac{1}{n^2}*n* 2p(1-p) $$



However, the result should actually contain a $frac{2n-1}{n^2}$. Can somebody point out my mistake?






probability convergence







share|cite|improve this question









$endgroup$












  • $begingroup$
    The random variables $Y_i$ are not independent and therefore $text{var}(sum_{i} Y_i) = sum_i text{var}(Y_i)$ fails to hold true.
    $endgroup$
    – saz
    Jan 4 at 19:57














0












0








0





$begingroup$


Let $Y_n = X_n + X_{n+1}$ with $X_n$ ~ Bernoulli$(p)$ independent and identically distributed. Let $T_n = frac{1}{n} sum_{i=1}^n Y_i$. I want to find $Var[T_n]$.



This is what I have done so far:



I know that $Y_n$ ~ Bin(2,p). Then I can write this:



$$ Var[T_n] = Var[frac{1}{n}sum_{i=1}^n Y_i]= frac{1}{n^2}Var[sum_{i=1}^n Y_i] =$$



$$=frac{1}{n^2}sum_{i=1}^nVar[ Y_i] = frac{1}{n^2}*n* 2p(1-p) $$



However, the result should actually contain a $frac{2n-1}{n^2}$. Can somebody point out my mistake?






probability convergence







share|cite|improve this question









$endgroup$




Let $Y_n = X_n + X_{n+1}$ with $X_n$ ~ Bernoulli$(p)$ independent and identically distributed. Let $T_n = frac{1}{n} sum_{i=1}^n Y_i$. I want to find $Var[T_n]$.



This is what I have done so far:



I know that $Y_n$ ~ Bin(2,p). Then I can write this:



$$ Var[T_n] = Var[frac{1}{n}sum_{i=1}^n Y_i]= frac{1}{n^2}Var[sum_{i=1}^n Y_i] =$$



$$=frac{1}{n^2}sum_{i=1}^nVar[ Y_i] = frac{1}{n^2}*n* 2p(1-p) $$



However, the result should actually contain a $frac{2n-1}{n^2}$. Can somebody point out my mistake?






probability convergence




probability convergence






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share|cite|improve this question











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share|cite|improve this question










asked Jan 4 at 19:51









qcc101qcc101

627213




627213












  • $begingroup$
    The random variables $Y_i$ are not independent and therefore $text{var}(sum_{i} Y_i) = sum_i text{var}(Y_i)$ fails to hold true.
    $endgroup$
    – saz
    Jan 4 at 19:57


















  • $begingroup$
    The random variables $Y_i$ are not independent and therefore $text{var}(sum_{i} Y_i) = sum_i text{var}(Y_i)$ fails to hold true.
    $endgroup$
    – saz
    Jan 4 at 19:57
















$begingroup$
The random variables $Y_i$ are not independent and therefore $text{var}(sum_{i} Y_i) = sum_i text{var}(Y_i)$ fails to hold true.
$endgroup$
– saz
Jan 4 at 19:57




$begingroup$
The random variables $Y_i$ are not independent and therefore $text{var}(sum_{i} Y_i) = sum_i text{var}(Y_i)$ fails to hold true.
$endgroup$
– saz
Jan 4 at 19:57










1 Answer
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$begingroup$

Hints:




  • The $Y_i$s are not independent so you cannot just sum their variances


  • Consider $Var[T_n] = Varleft[frac{1}{n}sumlimits_{i=1}^n Y_iright] = Varleft[frac{1}{n}left(X_1+X_{n+1}+sumlimits_{i=2}^n 2X_iright)right]$ where the $X_i$s are independent







share|cite|improve this answer









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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

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    active

    oldest

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    1












    $begingroup$

    Hints:




    • The $Y_i$s are not independent so you cannot just sum their variances


    • Consider $Var[T_n] = Varleft[frac{1}{n}sumlimits_{i=1}^n Y_iright] = Varleft[frac{1}{n}left(X_1+X_{n+1}+sumlimits_{i=2}^n 2X_iright)right]$ where the $X_i$s are independent







    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Hints:




      • The $Y_i$s are not independent so you cannot just sum their variances


      • Consider $Var[T_n] = Varleft[frac{1}{n}sumlimits_{i=1}^n Y_iright] = Varleft[frac{1}{n}left(X_1+X_{n+1}+sumlimits_{i=2}^n 2X_iright)right]$ where the $X_i$s are independent







      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Hints:




        • The $Y_i$s are not independent so you cannot just sum their variances


        • Consider $Var[T_n] = Varleft[frac{1}{n}sumlimits_{i=1}^n Y_iright] = Varleft[frac{1}{n}left(X_1+X_{n+1}+sumlimits_{i=2}^n 2X_iright)right]$ where the $X_i$s are independent







        share|cite|improve this answer









        $endgroup$



        Hints:




        • The $Y_i$s are not independent so you cannot just sum their variances


        • Consider $Var[T_n] = Varleft[frac{1}{n}sumlimits_{i=1}^n Y_iright] = Varleft[frac{1}{n}left(X_1+X_{n+1}+sumlimits_{i=2}^n 2X_iright)right]$ where the $X_i$s are independent








        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 19:57









        HenryHenry

        101k481168




        101k481168






























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