Let $Y_n = X_n + X_{n+1}$ and $T_n = frac{1}{n} sum_{i=1}^n Y_i$. I want to find $Var[T_n]$
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 yesterday
qcc101
479113
add a comment |
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 yesterday
qcc101
479113
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.
– saz
yesterday
add a comment |
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 yesterday
qcc101
479113
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 yesterday
qcc101
479113
asked yesterday
qcc101
479113
asked yesterday
qcc101
479113
asked yesterday
qcc101
479113
asked yesterday
qcc101
479113
479113
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.
– saz
yesterday
add a comment |
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.
– saz
yesterday
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.
– saz
yesterday
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.
– saz
yesterday
add a comment |
1 Answer
1
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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 yesterday
Henry
98.3k475162
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 yesterday
Henry
98.3k475162
add a comment |
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 yesterday
Henry
98.3k475162
add a comment |
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 yesterday
Henry
98.3k475162
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 yesterday
Henry
98.3k475162
answered yesterday
Henry
98.3k475162
answered yesterday
Henry
98.3k475162
answered yesterday
Henry
98.3k475162
98.3k475162
add a comment |
add a comment |
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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.
– saz
yesterday