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Calculating variance of an estimator
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Given that $ operatornameVar(x)=frac34theta^2$, I want t find the variance of estimator $hattheta_1 = frac2n3sum_i=1^nX_i$. EDIT: $X_1,...,X_n$ are independent and identically distributed (iid)
I proceed as follows:
$$ operatornameVar(hattheta) = operatornameVar(frac2n3sum_i=1^nX_i)=frac49 frac1n operatornameVar(X_1)= frac49 frac1n frac34 theta^2 = frac13ntheta^2 $$
or
$$ operatornameVar(hattheta) = operatornameVar(frac2n3sum_i=1^nX_i)= operatornameVar(frac2n3nX_1)= operatornameVar(frac23X_1)=frac49 frac34theta^2 = frac13theta^2 $$
Our professor provided us with the solution, so I know the first approach is the correct one, however i do not understand what is wrong with the second approach?
estimation variance estimation-theory mean-square-error
asked Jul 30 at 8:34
user1607
608
 |Â
show 3 more comments
up vote
0
down vote
favorite
Given that $ operatornameVar(x)=frac34theta^2$, I want t find the variance of estimator $hattheta_1 = frac2n3sum_i=1^nX_i$. EDIT: $X_1,...,X_n$ are independent and identically distributed (iid)
I proceed as follows:
$$ operatornameVar(hattheta) = operatornameVar(frac2n3sum_i=1^nX_i)=frac49 frac1n operatornameVar(X_1)= frac49 frac1n frac34 theta^2 = frac13ntheta^2 $$
or
$$ operatornameVar(hattheta) = operatornameVar(frac2n3sum_i=1^nX_i)= operatornameVar(frac2n3nX_1)= operatornameVar(frac23X_1)=frac49 frac34theta^2 = frac13theta^2 $$
Our professor provided us with the solution, so I know the first approach is the correct one, however i do not understand what is wrong with the second approach?
estimation variance estimation-theory mean-square-error
asked Jul 30 at 8:34
user1607
608
Is the first one really correct? It seems to me that he claims $textVar(sum_i X_i) = ntextVar(X_1)$ (2nd =). Then, 3rd and 4th = are clearly wrong.
– Antoine
Jul 30 at 8:38
$X_i, ..., X_n$ are iid...forgot to mention. Then $Var(sum_iX_i )= nVar(X_1)$, right?
– user1607
Jul 30 at 8:44
I count lots of mistakes in this. E.g. do we really have $frac4n^29n=frac4n$?
– drhab
Jul 30 at 8:45
@Antoine Sorry, but you are wrong. And probably the $X_i$ are meant to be independent (which should have been mentioned in the question). The variance of a sum of independent random variables equals the sum of variances.
– drhab
Jul 30 at 8:49
I appologize, i made mistakes when writing the questions and fixed them after they were poited out.
– user1607
Jul 30 at 8:56
 |Â
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given that $ operatornameVar(x)=frac34theta^2$, I want t find the variance of estimator $hattheta_1 = frac2n3sum_i=1^nX_i$. EDIT: $X_1,...,X_n$ are independent and identically distributed (iid)
I proceed as follows:
$$ operatornameVar(hattheta) = operatornameVar(frac2n3sum_i=1^nX_i)=frac49 frac1n operatornameVar(X_1)= frac49 frac1n frac34 theta^2 = frac13ntheta^2 $$
or
$$ operatornameVar(hattheta) = operatornameVar(frac2n3sum_i=1^nX_i)= operatornameVar(frac2n3nX_1)= operatornameVar(frac23X_1)=frac49 frac34theta^2 = frac13theta^2 $$
Our professor provided us with the solution, so I know the first approach is the correct one, however i do not understand what is wrong with the second approach?
estimation variance estimation-theory mean-square-error
asked Jul 30 at 8:34
user1607
608
Given that $ operatornameVar(x)=frac34theta^2$, I want t find the variance of estimator $hattheta_1 = frac2n3sum_i=1^nX_i$. EDIT: $X_1,...,X_n$ are independent and identically distributed (iid)
I proceed as follows:
$$ operatornameVar(hattheta) = operatornameVar(frac2n3sum_i=1^nX_i)=frac49 frac1n operatornameVar(X_1)= frac49 frac1n frac34 theta^2 = frac13ntheta^2 $$
or
$$ operatornameVar(hattheta) = operatornameVar(frac2n3sum_i=1^nX_i)= operatornameVar(frac2n3nX_1)= operatornameVar(frac23X_1)=frac49 frac34theta^2 = frac13theta^2 $$
Our professor provided us with the solution, so I know the first approach is the correct one, however i do not understand what is wrong with the second approach?
estimation variance estimation-theory mean-square-error
asked Jul 30 at 8:34
user1607
608
edited Jul 30 at 8:55
asked Jul 30 at 8:34
user1607
608
asked Jul 30 at 8:34
user1607
608
asked Jul 30 at 8:34
user1607
608
608
Is the first one really correct? It seems to me that he claims $textVar(sum_i X_i) = ntextVar(X_1)$ (2nd =). Then, 3rd and 4th = are clearly wrong.
– Antoine
Jul 30 at 8:38
$X_i, ..., X_n$ are iid...forgot to mention. Then $Var(sum_iX_i )= nVar(X_1)$, right?
– user1607
Jul 30 at 8:44
I count lots of mistakes in this. E.g. do we really have $frac4n^29n=frac4n$?
– drhab
Jul 30 at 8:45
@Antoine Sorry, but you are wrong. And probably the $X_i$ are meant to be independent (which should have been mentioned in the question). The variance of a sum of independent random variables equals the sum of variances.
– drhab
Jul 30 at 8:49
I appologize, i made mistakes when writing the questions and fixed them after they were poited out.
– user1607
Jul 30 at 8:56
 |Â
show 3 more comments
Is the first one really correct? It seems to me that he claims $textVar(sum_i X_i) = ntextVar(X_1)$ (2nd =). Then, 3rd and 4th = are clearly wrong.
– Antoine
Jul 30 at 8:38
$X_i, ..., X_n$ are iid...forgot to mention. Then $Var(sum_iX_i )= nVar(X_1)$, right?
– user1607
Jul 30 at 8:44
I count lots of mistakes in this. E.g. do we really have $frac4n^29n=frac4n$?
– drhab
Jul 30 at 8:45
@Antoine Sorry, but you are wrong. And probably the $X_i$ are meant to be independent (which should have been mentioned in the question). The variance of a sum of independent random variables equals the sum of variances.
– drhab
Jul 30 at 8:49
I appologize, i made mistakes when writing the questions and fixed them after they were poited out.
– user1607
Jul 30 at 8:56
Is the first one really correct? It seems to me that he claims $textVar(sum_i X_i) = ntextVar(X_1)$ (2nd =). Then, 3rd and 4th = are clearly wrong.
– Antoine
Jul 30 at 8:38
Is the first one really correct? It seems to me that he claims $textVar(sum_i X_i) = ntextVar(X_1)$ (2nd =). Then, 3rd and 4th = are clearly wrong.
– Antoine
Jul 30 at 8:38
$X_i, ..., X_n$ are iid...forgot to mention. Then $Var(sum_iX_i )= nVar(X_1)$, right?
– user1607
Jul 30 at 8:44
$X_i, ..., X_n$ are iid...forgot to mention. Then $Var(sum_iX_i )= nVar(X_1)$, right?
– user1607
Jul 30 at 8:44
I count lots of mistakes in this. E.g. do we really have $frac4n^29n=frac4n$?
– drhab
Jul 30 at 8:45
I count lots of mistakes in this. E.g. do we really have $frac4n^29n=frac4n$?
– drhab
Jul 30 at 8:45
@Antoine Sorry, but you are wrong. And probably the $X_i$ are meant to be independent (which should have been mentioned in the question). The variance of a sum of independent random variables equals the sum of variances.
– drhab
Jul 30 at 8:49
@Antoine Sorry, but you are wrong. And probably the $X_i$ are meant to be independent (which should have been mentioned in the question). The variance of a sum of independent random variables equals the sum of variances.
– drhab
Jul 30 at 8:49
I appologize, i made mistakes when writing the questions and fixed them after they were poited out.
– user1607
Jul 30 at 8:56
I appologize, i made mistakes when writing the questions and fixed them after they were poited out.
– user1607
Jul 30 at 8:56
 |Â
show 3 more comments
1 Answer
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oldest
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up vote
2
down vote
accepted
Both approaches seem to be problematic in my view with the second being worse than the first because there is no reason why your equalities should hold. The first approach only has the problem of lacking correct calculations with rational numbers as far as I can tell.
The identity
$$Var(frac2n3sum_i=1^nX_i) = frac4n^29n Var(X_1)$$
is correct (if we assume that the $X_1, dots, X_n$ are independent and identically distributed), since $Var(alpha X) = alpha^2 Var(X)$ for any constant $alpha$ and $Var(X + Y) = Var(X) + Var(Y)$ for independent $X$ and $Y$.
However,
$$frac4n^29n Var(X_1)=frac4 n^39Var(X_1) $$
and not $frac4 nVar(X_1)$ or $frac4 9 nVar(X_1)$ (how are the $n$ supposed to cancel?)
You should now try to plug in $Var(X_1) = frac34 theta^2$.
answered Jul 30 at 8:53
Matthias Klupsch
6,0341127
What is wrong with $Var(frac2n3sum_i=1^nX_i) = frac4n^29 Var (sum_i=1^n X_i) = frac4n^29 Var(nX_1)= frac4n^29 n^2 Var(X_1) $ ?
– user1607
Jul 30 at 9:05
It is not true that $Var(sum_i X_i) = Var(n X_1)$. For example if $n = 2$. Then your equation would read $Var(X_1 + X_2) = Var(2 X_1) = 4 Var(X_1)$ which is not correct. Instead you have $Var(X_1 + X_2) = Var(X_1) + Var(X_2) = 2 Var(X_1)$.
– Matthias Klupsch
Jul 30 at 9:06
thanks for the clarification. So then: $$ Var(hattheta)= frac4n^39Var(X_1)= frac 4n^39 frac34 theta^2 = fracn^33theta^2$$
– user1607
Jul 30 at 12:04
Yes that's correct.
– Matthias Klupsch
Jul 30 at 12:22
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Both approaches seem to be problematic in my view with the second being worse than the first because there is no reason why your equalities should hold. The first approach only has the problem of lacking correct calculations with rational numbers as far as I can tell.
The identity
$$Var(frac2n3sum_i=1^nX_i) = frac4n^29n Var(X_1)$$
is correct (if we assume that the $X_1, dots, X_n$ are independent and identically distributed), since $Var(alpha X) = alpha^2 Var(X)$ for any constant $alpha$ and $Var(X + Y) = Var(X) + Var(Y)$ for independent $X$ and $Y$.
However,
$$frac4n^29n Var(X_1)=frac4 n^39Var(X_1) $$
and not $frac4 nVar(X_1)$ or $frac4 9 nVar(X_1)$ (how are the $n$ supposed to cancel?)
You should now try to plug in $Var(X_1) = frac34 theta^2$.
answered Jul 30 at 8:53
Matthias Klupsch
6,0341127
What is wrong with $Var(frac2n3sum_i=1^nX_i) = frac4n^29 Var (sum_i=1^n X_i) = frac4n^29 Var(nX_1)= frac4n^29 n^2 Var(X_1) $ ?
– user1607
Jul 30 at 9:05
It is not true that $Var(sum_i X_i) = Var(n X_1)$. For example if $n = 2$. Then your equation would read $Var(X_1 + X_2) = Var(2 X_1) = 4 Var(X_1)$ which is not correct. Instead you have $Var(X_1 + X_2) = Var(X_1) + Var(X_2) = 2 Var(X_1)$.
– Matthias Klupsch
Jul 30 at 9:06
thanks for the clarification. So then: $$ Var(hattheta)= frac4n^39Var(X_1)= frac 4n^39 frac34 theta^2 = fracn^33theta^2$$
– user1607
Jul 30 at 12:04
Yes that's correct.
– Matthias Klupsch
Jul 30 at 12:22
add a comment |Â
up vote
2
down vote
accepted
Both approaches seem to be problematic in my view with the second being worse than the first because there is no reason why your equalities should hold. The first approach only has the problem of lacking correct calculations with rational numbers as far as I can tell.
The identity
$$Var(frac2n3sum_i=1^nX_i) = frac4n^29n Var(X_1)$$
is correct (if we assume that the $X_1, dots, X_n$ are independent and identically distributed), since $Var(alpha X) = alpha^2 Var(X)$ for any constant $alpha$ and $Var(X + Y) = Var(X) + Var(Y)$ for independent $X$ and $Y$.
However,
$$frac4n^29n Var(X_1)=frac4 n^39Var(X_1) $$
and not $frac4 nVar(X_1)$ or $frac4 9 nVar(X_1)$ (how are the $n$ supposed to cancel?)
You should now try to plug in $Var(X_1) = frac34 theta^2$.
answered Jul 30 at 8:53
Matthias Klupsch
6,0341127
What is wrong with $Var(frac2n3sum_i=1^nX_i) = frac4n^29 Var (sum_i=1^n X_i) = frac4n^29 Var(nX_1)= frac4n^29 n^2 Var(X_1) $ ?
– user1607
Jul 30 at 9:05
It is not true that $Var(sum_i X_i) = Var(n X_1)$. For example if $n = 2$. Then your equation would read $Var(X_1 + X_2) = Var(2 X_1) = 4 Var(X_1)$ which is not correct. Instead you have $Var(X_1 + X_2) = Var(X_1) + Var(X_2) = 2 Var(X_1)$.
– Matthias Klupsch
Jul 30 at 9:06
thanks for the clarification. So then: $$ Var(hattheta)= frac4n^39Var(X_1)= frac 4n^39 frac34 theta^2 = fracn^33theta^2$$
– user1607
Jul 30 at 12:04
Yes that's correct.
– Matthias Klupsch
Jul 30 at 12:22
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Both approaches seem to be problematic in my view with the second being worse than the first because there is no reason why your equalities should hold. The first approach only has the problem of lacking correct calculations with rational numbers as far as I can tell.
The identity
$$Var(frac2n3sum_i=1^nX_i) = frac4n^29n Var(X_1)$$
is correct (if we assume that the $X_1, dots, X_n$ are independent and identically distributed), since $Var(alpha X) = alpha^2 Var(X)$ for any constant $alpha$ and $Var(X + Y) = Var(X) + Var(Y)$ for independent $X$ and $Y$.
However,
$$frac4n^29n Var(X_1)=frac4 n^39Var(X_1) $$
and not $frac4 nVar(X_1)$ or $frac4 9 nVar(X_1)$ (how are the $n$ supposed to cancel?)
You should now try to plug in $Var(X_1) = frac34 theta^2$.
answered Jul 30 at 8:53
Matthias Klupsch
6,0341127
Both approaches seem to be problematic in my view with the second being worse than the first because there is no reason why your equalities should hold. The first approach only has the problem of lacking correct calculations with rational numbers as far as I can tell.
The identity
$$Var(frac2n3sum_i=1^nX_i) = frac4n^29n Var(X_1)$$
is correct (if we assume that the $X_1, dots, X_n$ are independent and identically distributed), since $Var(alpha X) = alpha^2 Var(X)$ for any constant $alpha$ and $Var(X + Y) = Var(X) + Var(Y)$ for independent $X$ and $Y$.
However,
$$frac4n^29n Var(X_1)=frac4 n^39Var(X_1) $$
and not $frac4 nVar(X_1)$ or $frac4 9 nVar(X_1)$ (how are the $n$ supposed to cancel?)
You should now try to plug in $Var(X_1) = frac34 theta^2$.
answered Jul 30 at 8:53
Matthias Klupsch
6,0341127
answered Jul 30 at 8:53
Matthias Klupsch
6,0341127
answered Jul 30 at 8:53
Matthias Klupsch
6,0341127
answered Jul 30 at 8:53
Matthias Klupsch
6,0341127
6,0341127
What is wrong with $Var(frac2n3sum_i=1^nX_i) = frac4n^29 Var (sum_i=1^n X_i) = frac4n^29 Var(nX_1)= frac4n^29 n^2 Var(X_1) $ ?
– user1607
Jul 30 at 9:05
It is not true that $Var(sum_i X_i) = Var(n X_1)$. For example if $n = 2$. Then your equation would read $Var(X_1 + X_2) = Var(2 X_1) = 4 Var(X_1)$ which is not correct. Instead you have $Var(X_1 + X_2) = Var(X_1) + Var(X_2) = 2 Var(X_1)$.
– Matthias Klupsch
Jul 30 at 9:06
thanks for the clarification. So then: $$ Var(hattheta)= frac4n^39Var(X_1)= frac 4n^39 frac34 theta^2 = fracn^33theta^2$$
– user1607
Jul 30 at 12:04
Yes that's correct.
– Matthias Klupsch
Jul 30 at 12:22
add a comment |Â
What is wrong with $Var(frac2n3sum_i=1^nX_i) = frac4n^29 Var (sum_i=1^n X_i) = frac4n^29 Var(nX_1)= frac4n^29 n^2 Var(X_1) $ ?
– user1607
Jul 30 at 9:05
It is not true that $Var(sum_i X_i) = Var(n X_1)$. For example if $n = 2$. Then your equation would read $Var(X_1 + X_2) = Var(2 X_1) = 4 Var(X_1)$ which is not correct. Instead you have $Var(X_1 + X_2) = Var(X_1) + Var(X_2) = 2 Var(X_1)$.
– Matthias Klupsch
Jul 30 at 9:06
thanks for the clarification. So then: $$ Var(hattheta)= frac4n^39Var(X_1)= frac 4n^39 frac34 theta^2 = fracn^33theta^2$$
– user1607
Jul 30 at 12:04
Yes that's correct.
– Matthias Klupsch
Jul 30 at 12:22
What is wrong with $Var(frac2n3sum_i=1^nX_i) = frac4n^29 Var (sum_i=1^n X_i) = frac4n^29 Var(nX_1)= frac4n^29 n^2 Var(X_1) $ ?
– user1607
Jul 30 at 9:05
What is wrong with $Var(frac2n3sum_i=1^nX_i) = frac4n^29 Var (sum_i=1^n X_i) = frac4n^29 Var(nX_1)= frac4n^29 n^2 Var(X_1) $ ?
– user1607
Jul 30 at 9:05
It is not true that $Var(sum_i X_i) = Var(n X_1)$. For example if $n = 2$. Then your equation would read $Var(X_1 + X_2) = Var(2 X_1) = 4 Var(X_1)$ which is not correct. Instead you have $Var(X_1 + X_2) = Var(X_1) + Var(X_2) = 2 Var(X_1)$.
– Matthias Klupsch
Jul 30 at 9:06
It is not true that $Var(sum_i X_i) = Var(n X_1)$. For example if $n = 2$. Then your equation would read $Var(X_1 + X_2) = Var(2 X_1) = 4 Var(X_1)$ which is not correct. Instead you have $Var(X_1 + X_2) = Var(X_1) + Var(X_2) = 2 Var(X_1)$.
– Matthias Klupsch
Jul 30 at 9:06
thanks for the clarification. So then: $$ Var(hattheta)= frac4n^39Var(X_1)= frac 4n^39 frac34 theta^2 = fracn^33theta^2$$
– user1607
Jul 30 at 12:04
thanks for the clarification. So then: $$ Var(hattheta)= frac4n^39Var(X_1)= frac 4n^39 frac34 theta^2 = fracn^33theta^2$$
– user1607
Jul 30 at 12:04
Yes that's correct.
– Matthias Klupsch
Jul 30 at 12:22
Yes that's correct.
– Matthias Klupsch
Jul 30 at 12:22
add a comment |Â
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Is the first one really correct? It seems to me that he claims $textVar(sum_i X_i) = ntextVar(X_1)$ (2nd =). Then, 3rd and 4th = are clearly wrong.
– Antoine
Jul 30 at 8:38
$X_i, ..., X_n$ are iid...forgot to mention. Then $Var(sum_iX_i )= nVar(X_1)$, right?
– user1607
Jul 30 at 8:44
I count lots of mistakes in this. E.g. do we really have $frac4n^29n=frac4n$?
– drhab
Jul 30 at 8:45
@Antoine Sorry, but you are wrong. And probably the $X_i$ are meant to be independent (which should have been mentioned in the question). The variance of a sum of independent random variables equals the sum of variances.
– drhab
Jul 30 at 8:49
I appologize, i made mistakes when writing the questions and fixed them after they were poited out.
– user1607
Jul 30 at 8:56