Mixing
Showing weighted average is consistent estimate
$begingroup$
Here's the problem statement: Let $X_1$, . . . , $X_n$ be independent random variables with common mean $mu$
and variances $σ_i^2$ . To estimate $μ$, we use the weighted average
$T_n$ = $sum_{i=1}^n w_iX_i$
with weights
$w_i$ = $frac{sigma_i^{−2}}{sum_{i=1}^n sigma_j^{-2}}$
Show that the estimate $T_n$ is consistent (in probability) if $sum_{i=1}^n sigma_j^{-2} Rightarrow infty$.
End Problem.
So I know that a consistent estimator is one that asymptotically approaches the parameter as the sample size goes to infinity. So by the problem statement, if the denominator of each of the weightings goes to infinity, then each of the $w_iX_i$ would just go to 0, yes? I'm a little confused on what this problem is trying to show. Thanks for any help!
probability-theory estimation law-of-large-numbers
asked Jan 31 at 18:48
psunpsun
211
$endgroup$
add a comment |
$begingroup$
Here's the problem statement: Let $X_1$, . . . , $X_n$ be independent random variables with common mean $mu$
and variances $σ_i^2$ . To estimate $μ$, we use the weighted average
$T_n$ = $sum_{i=1}^n w_iX_i$
with weights
$w_i$ = $frac{sigma_i^{−2}}{sum_{i=1}^n sigma_j^{-2}}$
Show that the estimate $T_n$ is consistent (in probability) if $sum_{i=1}^n sigma_j^{-2} Rightarrow infty$.
End Problem.
So I know that a consistent estimator is one that asymptotically approaches the parameter as the sample size goes to infinity. So by the problem statement, if the denominator of each of the weightings goes to infinity, then each of the $w_iX_i$ would just go to 0, yes? I'm a little confused on what this problem is trying to show. Thanks for any help!
probability-theory estimation law-of-large-numbers
asked Jan 31 at 18:48
psunpsun
211
$endgroup$
add a comment |
$begingroup$
Here's the problem statement: Let $X_1$, . . . , $X_n$ be independent random variables with common mean $mu$
and variances $σ_i^2$ . To estimate $μ$, we use the weighted average
$T_n$ = $sum_{i=1}^n w_iX_i$
with weights
$w_i$ = $frac{sigma_i^{−2}}{sum_{i=1}^n sigma_j^{-2}}$
Show that the estimate $T_n$ is consistent (in probability) if $sum_{i=1}^n sigma_j^{-2} Rightarrow infty$.
End Problem.
So I know that a consistent estimator is one that asymptotically approaches the parameter as the sample size goes to infinity. So by the problem statement, if the denominator of each of the weightings goes to infinity, then each of the $w_iX_i$ would just go to 0, yes? I'm a little confused on what this problem is trying to show. Thanks for any help!
probability-theory estimation law-of-large-numbers
asked Jan 31 at 18:48
psunpsun
211
$endgroup$
Here's the problem statement: Let $X_1$, . . . , $X_n$ be independent random variables with common mean $mu$
and variances $σ_i^2$ . To estimate $μ$, we use the weighted average
$T_n$ = $sum_{i=1}^n w_iX_i$
with weights
$w_i$ = $frac{sigma_i^{−2}}{sum_{i=1}^n sigma_j^{-2}}$
Show that the estimate $T_n$ is consistent (in probability) if $sum_{i=1}^n sigma_j^{-2} Rightarrow infty$.
End Problem.
So I know that a consistent estimator is one that asymptotically approaches the parameter as the sample size goes to infinity. So by the problem statement, if the denominator of each of the weightings goes to infinity, then each of the $w_iX_i$ would just go to 0, yes? I'm a little confused on what this problem is trying to show. Thanks for any help!
probability-theory estimation law-of-large-numbers
probability-theory estimation law-of-large-numbers
asked Jan 31 at 18:48
psunpsun
211
asked Jan 31 at 18:48
psunpsun
211
asked Jan 31 at 18:48
psunpsun
211
asked Jan 31 at 18:48
psunpsun
211
asked Jan 31 at 18:48
psunpsun
211
211
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes, $w_{i}X_{i}$ go to 0, but you are interested in the behaviour of $T_{n}$ . Convergence in probability means .
begin{equation}
lim_{n rightarrow infty }P(|T_{n} - mu | ge epsilon) = 0 , forall epsilon > 0
end{equation}
Try using some concentration inequality involving the variance .
answered Jan 31 at 19:33
Popescu ClaudiuPopescu Claudiu
4019
$endgroup$
$begingroup$
It seems that the Bernstein inequality you want to use require finite moments of any order, which is not specified in the question and not needed to reach the conclusion with the assumptions of the problem.
$endgroup$
– Davide Giraudo
Feb 1 at 14:25
$begingroup$
You are right , I see the variables can be unbounded .I will edit the answare . Thanks !.
$endgroup$
– Popescu Claudiu
Feb 1 at 17:42
add a comment |
$begingroup$
Here are some steps:
- Without loss of generality, $mu=0$.
- Compute $operatorname{Var}left(T_nright)$. Using independence this reduces to $sum_{i=1}^noperatorname{Var}left(omega_iX_iright)$. Since $$operatorname{Var}left(omega_iX_iright)=omega_i^2sigma_i^2,$$
it follows that
$$
operatorname{Var}left(T_nright)=sum_{i=1}^nleft(frac{sigma_i^{-2}}{sum_{j=1}^nsigma_j^{-2}}right)^2sigma_i^2=sum_{i=1}^nsigma_i^{-2}frac 1{left(sum_{j=1}^nsigma_j^{-2}right)^2}=frac{1}{sum_{j=1}^nsigma_j^{-2}}.
$$
- Since $mathbb Eleft[T_nright]=mu$, we get that $mathbb Eleft[left(T_n-muright)^2right]to 0$ from which the convergence in probability follows.
answered Feb 1 at 11:30
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
$begingroup$
I'm a little confused by how setting $mu$ = 0 helps. When I calculate Var($T_n$), I get the following: Var($T_n$) = $sum w_i Var(sum(X_i)) = sum w_i sum Var((X_i)) = sum frac{sigma_i^{-2}}{sum sigma_j^{-2}}sum sigma_i^2$, and I'm not sure how this variance shows that the sum $T_n$ converges to the mean.
$endgroup$
– psun
Feb 1 at 20:53
$begingroup$
It have edited. Actually this computation works also if $muneq 0$.
$endgroup$
– Davide Giraudo
Feb 1 at 21:25
$begingroup$
So by showing this is the variance of $T_n$, the variance of this estimate goes to 0. That much I understand. I guess what I don't quite get is how this necessarily shows that $T_n$ converges to the population mean $mu$.
$endgroup$
– psun
Feb 1 at 23:05
$begingroup$
@psun I have edited.
$endgroup$
– Davide Giraudo
Feb 2 at 18:27
$begingroup$
Thanks for everything! This question is clear once you see all the moving parts :)
$endgroup$
– psun
Feb 4 at 17:37
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, $w_{i}X_{i}$ go to 0, but you are interested in the behaviour of $T_{n}$ . Convergence in probability means .
begin{equation}
lim_{n rightarrow infty }P(|T_{n} - mu | ge epsilon) = 0 , forall epsilon > 0
end{equation}
Try using some concentration inequality involving the variance .
answered Jan 31 at 19:33
Popescu ClaudiuPopescu Claudiu
4019
$endgroup$
$begingroup$
It seems that the Bernstein inequality you want to use require finite moments of any order, which is not specified in the question and not needed to reach the conclusion with the assumptions of the problem.
$endgroup$
– Davide Giraudo
Feb 1 at 14:25
$begingroup$
You are right , I see the variables can be unbounded .I will edit the answare . Thanks !.
$endgroup$
– Popescu Claudiu
Feb 1 at 17:42
add a comment |
$begingroup$
Yes, $w_{i}X_{i}$ go to 0, but you are interested in the behaviour of $T_{n}$ . Convergence in probability means .
begin{equation}
lim_{n rightarrow infty }P(|T_{n} - mu | ge epsilon) = 0 , forall epsilon > 0
end{equation}
Try using some concentration inequality involving the variance .
answered Jan 31 at 19:33
Popescu ClaudiuPopescu Claudiu
4019
$endgroup$
$begingroup$
It seems that the Bernstein inequality you want to use require finite moments of any order, which is not specified in the question and not needed to reach the conclusion with the assumptions of the problem.
$endgroup$
– Davide Giraudo
Feb 1 at 14:25
$begingroup$
You are right , I see the variables can be unbounded .I will edit the answare . Thanks !.
$endgroup$
– Popescu Claudiu
Feb 1 at 17:42
add a comment |
$begingroup$
Yes, $w_{i}X_{i}$ go to 0, but you are interested in the behaviour of $T_{n}$ . Convergence in probability means .
begin{equation}
lim_{n rightarrow infty }P(|T_{n} - mu | ge epsilon) = 0 , forall epsilon > 0
end{equation}
Try using some concentration inequality involving the variance .
answered Jan 31 at 19:33
Popescu ClaudiuPopescu Claudiu
4019
$endgroup$
Yes, $w_{i}X_{i}$ go to 0, but you are interested in the behaviour of $T_{n}$ . Convergence in probability means .
begin{equation}
lim_{n rightarrow infty }P(|T_{n} - mu | ge epsilon) = 0 , forall epsilon > 0
end{equation}
Try using some concentration inequality involving the variance .
answered Jan 31 at 19:33
Popescu ClaudiuPopescu Claudiu
4019
edited Feb 1 at 17:44
answered Jan 31 at 19:33
Popescu ClaudiuPopescu Claudiu
4019
answered Jan 31 at 19:33
Popescu ClaudiuPopescu Claudiu
4019
answered Jan 31 at 19:33
Popescu ClaudiuPopescu Claudiu
4019
4019
$begingroup$
It seems that the Bernstein inequality you want to use require finite moments of any order, which is not specified in the question and not needed to reach the conclusion with the assumptions of the problem.
$endgroup$
– Davide Giraudo
Feb 1 at 14:25
$begingroup$
You are right , I see the variables can be unbounded .I will edit the answare . Thanks !.
$endgroup$
– Popescu Claudiu
Feb 1 at 17:42
add a comment |
$begingroup$
It seems that the Bernstein inequality you want to use require finite moments of any order, which is not specified in the question and not needed to reach the conclusion with the assumptions of the problem.
$endgroup$
– Davide Giraudo
Feb 1 at 14:25
$begingroup$
You are right , I see the variables can be unbounded .I will edit the answare . Thanks !.
$endgroup$
– Popescu Claudiu
Feb 1 at 17:42
$begingroup$
It seems that the Bernstein inequality you want to use require finite moments of any order, which is not specified in the question and not needed to reach the conclusion with the assumptions of the problem.
$endgroup$
– Davide Giraudo
Feb 1 at 14:25
$begingroup$
It seems that the Bernstein inequality you want to use require finite moments of any order, which is not specified in the question and not needed to reach the conclusion with the assumptions of the problem.
$endgroup$
– Davide Giraudo
Feb 1 at 14:25
$begingroup$
You are right , I see the variables can be unbounded .I will edit the answare . Thanks !.
$endgroup$
– Popescu Claudiu
Feb 1 at 17:42
$begingroup$
You are right , I see the variables can be unbounded .I will edit the answare . Thanks !.
$endgroup$
– Popescu Claudiu
Feb 1 at 17:42
add a comment |
$begingroup$
Here are some steps:
- Without loss of generality, $mu=0$.
- Compute $operatorname{Var}left(T_nright)$. Using independence this reduces to $sum_{i=1}^noperatorname{Var}left(omega_iX_iright)$. Since $$operatorname{Var}left(omega_iX_iright)=omega_i^2sigma_i^2,$$
it follows that
$$
operatorname{Var}left(T_nright)=sum_{i=1}^nleft(frac{sigma_i^{-2}}{sum_{j=1}^nsigma_j^{-2}}right)^2sigma_i^2=sum_{i=1}^nsigma_i^{-2}frac 1{left(sum_{j=1}^nsigma_j^{-2}right)^2}=frac{1}{sum_{j=1}^nsigma_j^{-2}}.
$$
- Since $mathbb Eleft[T_nright]=mu$, we get that $mathbb Eleft[left(T_n-muright)^2right]to 0$ from which the convergence in probability follows.
answered Feb 1 at 11:30
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
$begingroup$
I'm a little confused by how setting $mu$ = 0 helps. When I calculate Var($T_n$), I get the following: Var($T_n$) = $sum w_i Var(sum(X_i)) = sum w_i sum Var((X_i)) = sum frac{sigma_i^{-2}}{sum sigma_j^{-2}}sum sigma_i^2$, and I'm not sure how this variance shows that the sum $T_n$ converges to the mean.
$endgroup$
– psun
Feb 1 at 20:53
$begingroup$
It have edited. Actually this computation works also if $muneq 0$.
$endgroup$
– Davide Giraudo
Feb 1 at 21:25
$begingroup$
So by showing this is the variance of $T_n$, the variance of this estimate goes to 0. That much I understand. I guess what I don't quite get is how this necessarily shows that $T_n$ converges to the population mean $mu$.
$endgroup$
– psun
Feb 1 at 23:05
$begingroup$
@psun I have edited.
$endgroup$
– Davide Giraudo
Feb 2 at 18:27
$begingroup$
Thanks for everything! This question is clear once you see all the moving parts :)
$endgroup$
– psun
Feb 4 at 17:37
add a comment |
$begingroup$
Here are some steps:
- Without loss of generality, $mu=0$.
- Compute $operatorname{Var}left(T_nright)$. Using independence this reduces to $sum_{i=1}^noperatorname{Var}left(omega_iX_iright)$. Since $$operatorname{Var}left(omega_iX_iright)=omega_i^2sigma_i^2,$$
it follows that
$$
operatorname{Var}left(T_nright)=sum_{i=1}^nleft(frac{sigma_i^{-2}}{sum_{j=1}^nsigma_j^{-2}}right)^2sigma_i^2=sum_{i=1}^nsigma_i^{-2}frac 1{left(sum_{j=1}^nsigma_j^{-2}right)^2}=frac{1}{sum_{j=1}^nsigma_j^{-2}}.
$$
- Since $mathbb Eleft[T_nright]=mu$, we get that $mathbb Eleft[left(T_n-muright)^2right]to 0$ from which the convergence in probability follows.
answered Feb 1 at 11:30
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
$begingroup$
I'm a little confused by how setting $mu$ = 0 helps. When I calculate Var($T_n$), I get the following: Var($T_n$) = $sum w_i Var(sum(X_i)) = sum w_i sum Var((X_i)) = sum frac{sigma_i^{-2}}{sum sigma_j^{-2}}sum sigma_i^2$, and I'm not sure how this variance shows that the sum $T_n$ converges to the mean.
$endgroup$
– psun
Feb 1 at 20:53
$begingroup$
It have edited. Actually this computation works also if $muneq 0$.
$endgroup$
– Davide Giraudo
Feb 1 at 21:25
$begingroup$
So by showing this is the variance of $T_n$, the variance of this estimate goes to 0. That much I understand. I guess what I don't quite get is how this necessarily shows that $T_n$ converges to the population mean $mu$.
$endgroup$
– psun
Feb 1 at 23:05
$begingroup$
@psun I have edited.
$endgroup$
– Davide Giraudo
Feb 2 at 18:27
$begingroup$
Thanks for everything! This question is clear once you see all the moving parts :)
$endgroup$
– psun
Feb 4 at 17:37
add a comment |
$begingroup$
Here are some steps:
- Without loss of generality, $mu=0$.
- Compute $operatorname{Var}left(T_nright)$. Using independence this reduces to $sum_{i=1}^noperatorname{Var}left(omega_iX_iright)$. Since $$operatorname{Var}left(omega_iX_iright)=omega_i^2sigma_i^2,$$
it follows that
$$
operatorname{Var}left(T_nright)=sum_{i=1}^nleft(frac{sigma_i^{-2}}{sum_{j=1}^nsigma_j^{-2}}right)^2sigma_i^2=sum_{i=1}^nsigma_i^{-2}frac 1{left(sum_{j=1}^nsigma_j^{-2}right)^2}=frac{1}{sum_{j=1}^nsigma_j^{-2}}.
$$
- Since $mathbb Eleft[T_nright]=mu$, we get that $mathbb Eleft[left(T_n-muright)^2right]to 0$ from which the convergence in probability follows.
answered Feb 1 at 11:30
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
Here are some steps:
- Without loss of generality, $mu=0$.
- Compute $operatorname{Var}left(T_nright)$. Using independence this reduces to $sum_{i=1}^noperatorname{Var}left(omega_iX_iright)$. Since $$operatorname{Var}left(omega_iX_iright)=omega_i^2sigma_i^2,$$
it follows that
$$
operatorname{Var}left(T_nright)=sum_{i=1}^nleft(frac{sigma_i^{-2}}{sum_{j=1}^nsigma_j^{-2}}right)^2sigma_i^2=sum_{i=1}^nsigma_i^{-2}frac 1{left(sum_{j=1}^nsigma_j^{-2}right)^2}=frac{1}{sum_{j=1}^nsigma_j^{-2}}.
$$
- Since $mathbb Eleft[T_nright]=mu$, we get that $mathbb Eleft[left(T_n-muright)^2right]to 0$ from which the convergence in probability follows.
answered Feb 1 at 11:30
Davide GiraudoDavide Giraudo
128k17156268
edited Feb 1 at 23:12
answered Feb 1 at 11:30
Davide GiraudoDavide Giraudo
128k17156268
answered Feb 1 at 11:30
Davide GiraudoDavide Giraudo
128k17156268
answered Feb 1 at 11:30
Davide GiraudoDavide Giraudo
128k17156268
128k17156268
$begingroup$
I'm a little confused by how setting $mu$ = 0 helps. When I calculate Var($T_n$), I get the following: Var($T_n$) = $sum w_i Var(sum(X_i)) = sum w_i sum Var((X_i)) = sum frac{sigma_i^{-2}}{sum sigma_j^{-2}}sum sigma_i^2$, and I'm not sure how this variance shows that the sum $T_n$ converges to the mean.
$endgroup$
– psun
Feb 1 at 20:53
$begingroup$
It have edited. Actually this computation works also if $muneq 0$.
$endgroup$
– Davide Giraudo
Feb 1 at 21:25
$begingroup$
So by showing this is the variance of $T_n$, the variance of this estimate goes to 0. That much I understand. I guess what I don't quite get is how this necessarily shows that $T_n$ converges to the population mean $mu$.
$endgroup$
– psun
Feb 1 at 23:05
$begingroup$
@psun I have edited.
$endgroup$
– Davide Giraudo
Feb 2 at 18:27
$begingroup$
Thanks for everything! This question is clear once you see all the moving parts :)
$endgroup$
– psun
Feb 4 at 17:37
add a comment |
$begingroup$
I'm a little confused by how setting $mu$ = 0 helps. When I calculate Var($T_n$), I get the following: Var($T_n$) = $sum w_i Var(sum(X_i)) = sum w_i sum Var((X_i)) = sum frac{sigma_i^{-2}}{sum sigma_j^{-2}}sum sigma_i^2$, and I'm not sure how this variance shows that the sum $T_n$ converges to the mean.
$endgroup$
– psun
Feb 1 at 20:53
$begingroup$
It have edited. Actually this computation works also if $muneq 0$.
$endgroup$
– Davide Giraudo
Feb 1 at 21:25
$begingroup$
So by showing this is the variance of $T_n$, the variance of this estimate goes to 0. That much I understand. I guess what I don't quite get is how this necessarily shows that $T_n$ converges to the population mean $mu$.
$endgroup$
– psun
Feb 1 at 23:05
$begingroup$
@psun I have edited.
$endgroup$
– Davide Giraudo
Feb 2 at 18:27
$begingroup$
Thanks for everything! This question is clear once you see all the moving parts :)
$endgroup$
– psun
Feb 4 at 17:37
$begingroup$
I'm a little confused by how setting $mu$ = 0 helps. When I calculate Var($T_n$), I get the following: Var($T_n$) = $sum w_i Var(sum(X_i)) = sum w_i sum Var((X_i)) = sum frac{sigma_i^{-2}}{sum sigma_j^{-2}}sum sigma_i^2$, and I'm not sure how this variance shows that the sum $T_n$ converges to the mean.
$endgroup$
– psun
Feb 1 at 20:53
$begingroup$
I'm a little confused by how setting $mu$ = 0 helps. When I calculate Var($T_n$), I get the following: Var($T_n$) = $sum w_i Var(sum(X_i)) = sum w_i sum Var((X_i)) = sum frac{sigma_i^{-2}}{sum sigma_j^{-2}}sum sigma_i^2$, and I'm not sure how this variance shows that the sum $T_n$ converges to the mean.
$endgroup$
– psun
Feb 1 at 20:53
$begingroup$
It have edited. Actually this computation works also if $muneq 0$.
$endgroup$
– Davide Giraudo
Feb 1 at 21:25
$begingroup$
It have edited. Actually this computation works also if $muneq 0$.
$endgroup$
– Davide Giraudo
Feb 1 at 21:25
$begingroup$
So by showing this is the variance of $T_n$, the variance of this estimate goes to 0. That much I understand. I guess what I don't quite get is how this necessarily shows that $T_n$ converges to the population mean $mu$.
$endgroup$
– psun
Feb 1 at 23:05
$begingroup$
So by showing this is the variance of $T_n$, the variance of this estimate goes to 0. That much I understand. I guess what I don't quite get is how this necessarily shows that $T_n$ converges to the population mean $mu$.
$endgroup$
– psun
Feb 1 at 23:05
$begingroup$
@psun I have edited.
$endgroup$
– Davide Giraudo
Feb 2 at 18:27
$begingroup$
@psun I have edited.
$endgroup$
– Davide Giraudo
Feb 2 at 18:27
$begingroup$
Thanks for everything! This question is clear once you see all the moving parts :)
$endgroup$
– psun
Feb 4 at 17:37
$begingroup$
Thanks for everything! This question is clear once you see all the moving parts :)
$endgroup$
– psun
Feb 4 at 17:37
add a comment |
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Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
