Mixing
Let $Y_n = X_n + X_{n+1}$ and $T_n = frac{1}{n} sum_{i=1}^n Y_i$. I want to find $Var[T_n]$
$begingroup$
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
asked Jan 4 at 19:51
qcc101qcc101
541113
$endgroup$
add a comment |
$begingroup$
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
asked Jan 4 at 19:51
qcc101qcc101
541113
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Jan 4 at 19:51
qcc101qcc101
541113
$endgroup$
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
asked Jan 4 at 19:51
qcc101qcc101
541113
asked Jan 4 at 19:51
qcc101qcc101
541113
asked Jan 4 at 19:51
qcc101qcc101
541113
asked Jan 4 at 19:51
qcc101qcc101
541113
asked Jan 4 at 19:51
qcc101qcc101
541113
541113
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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
answered Jan 4 at 19:57
HenryHenry
99.3k479165
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
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oldest
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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
answered Jan 4 at 19:57
HenryHenry
99.3k479165
$endgroup$
add a comment |
$begingroup$
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
answered Jan 4 at 19:57
HenryHenry
99.3k479165
$endgroup$
add a comment |
$begingroup$
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
answered Jan 4 at 19:57
HenryHenry
99.3k479165
$endgroup$
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
answered Jan 4 at 19:57
HenryHenry
99.3k479165
answered Jan 4 at 19:57
HenryHenry
99.3k479165
answered Jan 4 at 19:57
HenryHenry
99.3k479165
answered Jan 4 at 19:57
HenryHenry
99.3k479165
99.3k479165
add a comment |
add a comment |
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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