Statistics: how to prove efficiency of a given estimator?
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So the question gave $D_1$ and $D_2$ as unbiased, efficient and consistent estimators of $delta$ . $D_3$ is a new estimator which is obtained by taking a weighted average of $D_1$ and $D_2$ with one quarter of the weight placed on $D_1$. Now, the question is can I prove that $D_3$ is an efficient estimator of $delta$?
I know for that for $D_3$ to be efficient, $Var(D_3)$ has to be less than $Var(D_1)$ and $Var(D_3)$ has to be less than $Var(D_2)$.
I also know that $$Var(D_3)=Var left(frac{1}{4D_1}+frac{3}{4D_2}right)$$
However, I'm not sure how to prove (or if it is even possible to prove) that $D_3$ is an efficient estimator.
statistics
edited Jun 3 '16 at 21:19
M47145
3,21131130
asked May 3 '15 at 22:54
Nikitau
600513
add a comment |
up vote
1
down vote
favorite
So the question gave $D_1$ and $D_2$ as unbiased, efficient and consistent estimators of $delta$ . $D_3$ is a new estimator which is obtained by taking a weighted average of $D_1$ and $D_2$ with one quarter of the weight placed on $D_1$. Now, the question is can I prove that $D_3$ is an efficient estimator of $delta$?
I know for that for $D_3$ to be efficient, $Var(D_3)$ has to be less than $Var(D_1)$ and $Var(D_3)$ has to be less than $Var(D_2)$.
I also know that $$Var(D_3)=Var left(frac{1}{4D_1}+frac{3}{4D_2}right)$$
However, I'm not sure how to prove (or if it is even possible to prove) that $D_3$ is an efficient estimator.
statistics
edited Jun 3 '16 at 21:19
M47145
3,21131130
asked May 3 '15 at 22:54
Nikitau
600513
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
So the question gave $D_1$ and $D_2$ as unbiased, efficient and consistent estimators of $delta$ . $D_3$ is a new estimator which is obtained by taking a weighted average of $D_1$ and $D_2$ with one quarter of the weight placed on $D_1$. Now, the question is can I prove that $D_3$ is an efficient estimator of $delta$?
I know for that for $D_3$ to be efficient, $Var(D_3)$ has to be less than $Var(D_1)$ and $Var(D_3)$ has to be less than $Var(D_2)$.
I also know that $$Var(D_3)=Var left(frac{1}{4D_1}+frac{3}{4D_2}right)$$
However, I'm not sure how to prove (or if it is even possible to prove) that $D_3$ is an efficient estimator.
statistics
edited Jun 3 '16 at 21:19
M47145
3,21131130
asked May 3 '15 at 22:54
Nikitau
600513
So the question gave $D_1$ and $D_2$ as unbiased, efficient and consistent estimators of $delta$ . $D_3$ is a new estimator which is obtained by taking a weighted average of $D_1$ and $D_2$ with one quarter of the weight placed on $D_1$. Now, the question is can I prove that $D_3$ is an efficient estimator of $delta$?
I know for that for $D_3$ to be efficient, $Var(D_3)$ has to be less than $Var(D_1)$ and $Var(D_3)$ has to be less than $Var(D_2)$.
I also know that $$Var(D_3)=Var left(frac{1}{4D_1}+frac{3}{4D_2}right)$$
However, I'm not sure how to prove (or if it is even possible to prove) that $D_3$ is an efficient estimator.
statistics
statistics
edited Jun 3 '16 at 21:19
M47145
3,21131130
asked May 3 '15 at 22:54
Nikitau
600513
edited Jun 3 '16 at 21:19
M47145
3,21131130
asked May 3 '15 at 22:54
Nikitau
600513
edited Jun 3 '16 at 21:19
M47145
3,21131130
edited Jun 3 '16 at 21:19
M47145
3,21131130
edited Jun 3 '16 at 21:19
M47145
3,21131130
3,21131130
asked May 3 '15 at 22:54
Nikitau
600513
asked May 3 '15 at 22:54
Nikitau
600513
asked May 3 '15 at 22:54
Nikitau
600513
600513
add a comment |
add a comment |
1 Answer
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You can use the properties of variance so that Var(D3)=(1/16)Var(D1)+(9/16)Var(D2). To prove the efficiency, you could perhaps use the Cramer-Rao bound and fisher information.
answered May 3 '15 at 23:07
user153009
361211
Ah, I see. However, my syllabus doesn't cover the Cramer-Rao bound or the fisher information. Would there be any other way to prove efficiency?
– Nikitau
May 3 '15 at 23:52
Consider: (1) Variances don't add in this way unless $D_1$ and $D_2$ are independent, which is not necessarily so. (2) If they are independent, then $D_3$ has a smaller variance than $D_1$ or $D_2$. (3) No need to prove $D_1$ and $D_2$ are efficient because that is assumed. (4) If $D_1$ and $D_2$ are both efficient, they must have the same variance.
– BruceET
May 4 '15 at 2:46
Ah, thanks so much! So would it be alright if I concluded that D3 is efficient if and only if D1 and D2 are independent? Thanks so much!
– Nikitau
May 4 '15 at 14:46
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
You can use the properties of variance so that Var(D3)=(1/16)Var(D1)+(9/16)Var(D2). To prove the efficiency, you could perhaps use the Cramer-Rao bound and fisher information.
answered May 3 '15 at 23:07
user153009
361211
Ah, I see. However, my syllabus doesn't cover the Cramer-Rao bound or the fisher information. Would there be any other way to prove efficiency?
– Nikitau
May 3 '15 at 23:52
Consider: (1) Variances don't add in this way unless $D_1$ and $D_2$ are independent, which is not necessarily so. (2) If they are independent, then $D_3$ has a smaller variance than $D_1$ or $D_2$. (3) No need to prove $D_1$ and $D_2$ are efficient because that is assumed. (4) If $D_1$ and $D_2$ are both efficient, they must have the same variance.
– BruceET
May 4 '15 at 2:46
Ah, thanks so much! So would it be alright if I concluded that D3 is efficient if and only if D1 and D2 are independent? Thanks so much!
– Nikitau
May 4 '15 at 14:46
add a comment |
up vote
0
down vote
You can use the properties of variance so that Var(D3)=(1/16)Var(D1)+(9/16)Var(D2). To prove the efficiency, you could perhaps use the Cramer-Rao bound and fisher information.
answered May 3 '15 at 23:07
user153009
361211
Ah, I see. However, my syllabus doesn't cover the Cramer-Rao bound or the fisher information. Would there be any other way to prove efficiency?
– Nikitau
May 3 '15 at 23:52
Consider: (1) Variances don't add in this way unless $D_1$ and $D_2$ are independent, which is not necessarily so. (2) If they are independent, then $D_3$ has a smaller variance than $D_1$ or $D_2$. (3) No need to prove $D_1$ and $D_2$ are efficient because that is assumed. (4) If $D_1$ and $D_2$ are both efficient, they must have the same variance.
– BruceET
May 4 '15 at 2:46
Ah, thanks so much! So would it be alright if I concluded that D3 is efficient if and only if D1 and D2 are independent? Thanks so much!
– Nikitau
May 4 '15 at 14:46
add a comment |
up vote
0
down vote
up vote
0
down vote
You can use the properties of variance so that Var(D3)=(1/16)Var(D1)+(9/16)Var(D2). To prove the efficiency, you could perhaps use the Cramer-Rao bound and fisher information.
answered May 3 '15 at 23:07
user153009
361211
You can use the properties of variance so that Var(D3)=(1/16)Var(D1)+(9/16)Var(D2). To prove the efficiency, you could perhaps use the Cramer-Rao bound and fisher information.
answered May 3 '15 at 23:07
user153009
361211
answered May 3 '15 at 23:07
user153009
361211
answered May 3 '15 at 23:07
user153009
361211
answered May 3 '15 at 23:07
user153009
361211
361211
Ah, I see. However, my syllabus doesn't cover the Cramer-Rao bound or the fisher information. Would there be any other way to prove efficiency?
– Nikitau
May 3 '15 at 23:52
Consider: (1) Variances don't add in this way unless $D_1$ and $D_2$ are independent, which is not necessarily so. (2) If they are independent, then $D_3$ has a smaller variance than $D_1$ or $D_2$. (3) No need to prove $D_1$ and $D_2$ are efficient because that is assumed. (4) If $D_1$ and $D_2$ are both efficient, they must have the same variance.
– BruceET
May 4 '15 at 2:46
Ah, thanks so much! So would it be alright if I concluded that D3 is efficient if and only if D1 and D2 are independent? Thanks so much!
– Nikitau
May 4 '15 at 14:46
add a comment |
Ah, I see. However, my syllabus doesn't cover the Cramer-Rao bound or the fisher information. Would there be any other way to prove efficiency?
– Nikitau
May 3 '15 at 23:52
Consider: (1) Variances don't add in this way unless $D_1$ and $D_2$ are independent, which is not necessarily so. (2) If they are independent, then $D_3$ has a smaller variance than $D_1$ or $D_2$. (3) No need to prove $D_1$ and $D_2$ are efficient because that is assumed. (4) If $D_1$ and $D_2$ are both efficient, they must have the same variance.
– BruceET
May 4 '15 at 2:46
Ah, thanks so much! So would it be alright if I concluded that D3 is efficient if and only if D1 and D2 are independent? Thanks so much!
– Nikitau
May 4 '15 at 14:46
Ah, I see. However, my syllabus doesn't cover the Cramer-Rao bound or the fisher information. Would there be any other way to prove efficiency?
– Nikitau
May 3 '15 at 23:52
Ah, I see. However, my syllabus doesn't cover the Cramer-Rao bound or the fisher information. Would there be any other way to prove efficiency?
– Nikitau
May 3 '15 at 23:52
Consider: (1) Variances don't add in this way unless $D_1$ and $D_2$ are independent, which is not necessarily so. (2) If they are independent, then $D_3$ has a smaller variance than $D_1$ or $D_2$. (3) No need to prove $D_1$ and $D_2$ are efficient because that is assumed. (4) If $D_1$ and $D_2$ are both efficient, they must have the same variance.
– BruceET
May 4 '15 at 2:46
Consider: (1) Variances don't add in this way unless $D_1$ and $D_2$ are independent, which is not necessarily so. (2) If they are independent, then $D_3$ has a smaller variance than $D_1$ or $D_2$. (3) No need to prove $D_1$ and $D_2$ are efficient because that is assumed. (4) If $D_1$ and $D_2$ are both efficient, they must have the same variance.
– BruceET
May 4 '15 at 2:46
Ah, thanks so much! So would it be alright if I concluded that D3 is efficient if and only if D1 and D2 are independent? Thanks so much!
– Nikitau
May 4 '15 at 14:46
Ah, thanks so much! So would it be alright if I concluded that D3 is efficient if and only if D1 and D2 are independent? Thanks so much!
– Nikitau
May 4 '15 at 14:46
add a comment |
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