check if estimation is unbiased?
$begingroup$
Assume we that we calculate the expected value of some measurements $x=dfrac {x_1 + x_2 + x_3 + x_4} 4$. what if we dont include $x_3$ and $x_4$, but instead we use $x_2$ as $x_3$ and $x_4$. Then We get the following expression $v=dfrac {x_1 + x_2 + x_2 + x_2} 4$.
How do I know if $v$ is a unbiased estimation of $x$?
I am not sure how to approach this problem, any ideas are appreciated!
statistics estimation-theory parameter-estimation
asked Dec 6 '15 at 19:55
dumble24dumble24
1016
$endgroup$
add a comment |
$begingroup$
Assume we that we calculate the expected value of some measurements $x=dfrac {x_1 + x_2 + x_3 + x_4} 4$. what if we dont include $x_3$ and $x_4$, but instead we use $x_2$ as $x_3$ and $x_4$. Then We get the following expression $v=dfrac {x_1 + x_2 + x_2 + x_2} 4$.
How do I know if $v$ is a unbiased estimation of $x$?
I am not sure how to approach this problem, any ideas are appreciated!
statistics estimation-theory parameter-estimation
asked Dec 6 '15 at 19:55
dumble24dumble24
1016
$endgroup$
1
$begingroup$
So $v$ is the same thing as $x$? If that's not what you meant, then you need to clarify your question.
$endgroup$
– Michael Hardy
Dec 6 '15 at 20:07
$begingroup$
I assumed here that $x_k$ are random variables.
$endgroup$
– manofbear
Dec 6 '15 at 20:35
$begingroup$
x is expected value of a random variable
$endgroup$
– dumble24
Dec 6 '15 at 21:05
$begingroup$
(x1+x2+x3+x4)/4 calculates the expected value if a random variable has exactly four, equiprobable possible outcomes x1, x2, ..., x4. Alternatively, if x_1, x_2, ... x_4 denote 4 independent draws form some probability distribution, then (x1+x2+x3+x4)/4 is an estimator of the expected value, but it is not the actual expected value! To be precise, this is an important distinction.
$endgroup$
– Matthew Gunn
Dec 7 '15 at 23:12
add a comment |
$begingroup$
Assume we that we calculate the expected value of some measurements $x=dfrac {x_1 + x_2 + x_3 + x_4} 4$. what if we dont include $x_3$ and $x_4$, but instead we use $x_2$ as $x_3$ and $x_4$. Then We get the following expression $v=dfrac {x_1 + x_2 + x_2 + x_2} 4$.
How do I know if $v$ is a unbiased estimation of $x$?
I am not sure how to approach this problem, any ideas are appreciated!
statistics estimation-theory parameter-estimation
asked Dec 6 '15 at 19:55
dumble24dumble24
1016
$endgroup$
Assume we that we calculate the expected value of some measurements $x=dfrac {x_1 + x_2 + x_3 + x_4} 4$. what if we dont include $x_3$ and $x_4$, but instead we use $x_2$ as $x_3$ and $x_4$. Then We get the following expression $v=dfrac {x_1 + x_2 + x_2 + x_2} 4$.
How do I know if $v$ is a unbiased estimation of $x$?
I am not sure how to approach this problem, any ideas are appreciated!
statistics estimation-theory parameter-estimation
statistics estimation-theory parameter-estimation
asked Dec 6 '15 at 19:55
dumble24dumble24
1016
asked Dec 6 '15 at 19:55
dumble24dumble24
1016
edited Dec 7 '15 at 11:55
dumble24
asked Dec 6 '15 at 19:55
dumble24dumble24
1016
asked Dec 6 '15 at 19:55
dumble24dumble24
1016
asked Dec 6 '15 at 19:55
dumble24dumble24
1016
1016
1
$begingroup$
So $v$ is the same thing as $x$? If that's not what you meant, then you need to clarify your question.
$endgroup$
– Michael Hardy
Dec 6 '15 at 20:07
$begingroup$
I assumed here that $x_k$ are random variables.
$endgroup$
– manofbear
Dec 6 '15 at 20:35
$begingroup$
x is expected value of a random variable
$endgroup$
– dumble24
Dec 6 '15 at 21:05
$begingroup$
(x1+x2+x3+x4)/4 calculates the expected value if a random variable has exactly four, equiprobable possible outcomes x1, x2, ..., x4. Alternatively, if x_1, x_2, ... x_4 denote 4 independent draws form some probability distribution, then (x1+x2+x3+x4)/4 is an estimator of the expected value, but it is not the actual expected value! To be precise, this is an important distinction.
$endgroup$
– Matthew Gunn
Dec 7 '15 at 23:12
add a comment |
1
$begingroup$
So $v$ is the same thing as $x$? If that's not what you meant, then you need to clarify your question.
$endgroup$
– Michael Hardy
Dec 6 '15 at 20:07
$begingroup$
I assumed here that $x_k$ are random variables.
$endgroup$
– manofbear
Dec 6 '15 at 20:35
$begingroup$
x is expected value of a random variable
$endgroup$
– dumble24
Dec 6 '15 at 21:05
$begingroup$
(x1+x2+x3+x4)/4 calculates the expected value if a random variable has exactly four, equiprobable possible outcomes x1, x2, ..., x4. Alternatively, if x_1, x_2, ... x_4 denote 4 independent draws form some probability distribution, then (x1+x2+x3+x4)/4 is an estimator of the expected value, but it is not the actual expected value! To be precise, this is an important distinction.
$endgroup$
– Matthew Gunn
Dec 7 '15 at 23:12
1
1
$begingroup$
So $v$ is the same thing as $x$? If that's not what you meant, then you need to clarify your question.
$endgroup$
– Michael Hardy
Dec 6 '15 at 20:07
$begingroup$
So $v$ is the same thing as $x$? If that's not what you meant, then you need to clarify your question.
$endgroup$
– Michael Hardy
Dec 6 '15 at 20:07
$begingroup$
I assumed here that $x_k$ are random variables.
$endgroup$
– manofbear
Dec 6 '15 at 20:35
$begingroup$
I assumed here that $x_k$ are random variables.
$endgroup$
– manofbear
Dec 6 '15 at 20:35
$begingroup$
x is expected value of a random variable
$endgroup$
– dumble24
Dec 6 '15 at 21:05
$begingroup$
x is expected value of a random variable
$endgroup$
– dumble24
Dec 6 '15 at 21:05
$begingroup$
(x1+x2+x3+x4)/4 calculates the expected value if a random variable has exactly four, equiprobable possible outcomes x1, x2, ..., x4. Alternatively, if x_1, x_2, ... x_4 denote 4 independent draws form some probability distribution, then (x1+x2+x3+x4)/4 is an estimator of the expected value, but it is not the actual expected value! To be precise, this is an important distinction.
$endgroup$
– Matthew Gunn
Dec 7 '15 at 23:12
$begingroup$
(x1+x2+x3+x4)/4 calculates the expected value if a random variable has exactly four, equiprobable possible outcomes x1, x2, ..., x4. Alternatively, if x_1, x_2, ... x_4 denote 4 independent draws form some probability distribution, then (x1+x2+x3+x4)/4 is an estimator of the expected value, but it is not the actual expected value! To be precise, this is an important distinction.
$endgroup$
– Matthew Gunn
Dec 7 '15 at 23:12
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
[EDIT: Assumed $x_k$ are random variables.]
We say $v$ is an estimator of random variable $x$ if $E[v]=E[x]$, where $E[cdot]$ is expectation of random variables.
Recall that expectation is a linear operator, i.e. if $X$ and $Y$ are random variables and $a,b$ are constants, $E[aX+bY]=aE[X]+bE[Y]$. So we get $E[x]=frac{1}{4}(E[x_1]+E[x_2]+E[x_3]+E[x_4])$, and $E[v]=frac{1}{4}(E[x_1]+3E[x_2])$. Notice that $E[x]=E[v]$ is equivalent to $E[x]-E[v]=0$.
So $v$ is an unbiased estimator if $E[x]-E[v]=0 Leftrightarrow E[x_3]+E[x_4]-2E[x_2]=0.$
answered Dec 6 '15 at 20:02
manofbearmanofbear
1,579515
$endgroup$
$begingroup$
How do we decide if $v$ is unbiased or not unbiased using $ E[x_3]+E[x_4]-2E[x_2]=0.$ as $x_2,x_3$ and $x_4$ are not defined?
$endgroup$
– dumble24
Dec 6 '15 at 21:01
$begingroup$
What is this, if E[v] = E[x]?! Why are you taking expectations on the right hand side?
$endgroup$
– Matthew Gunn
Dec 7 '15 at 10:14
$begingroup$
Almost certainly dumble24 is in a classical statistics environment and is estimating some parameter, not another random variable.
$endgroup$
– Matthew Gunn
Dec 7 '15 at 10:24
$begingroup$
Edited the question, added some more context.
$endgroup$
– dumble24
Dec 7 '15 at 11:58
add a comment |
$begingroup$
Let $theta$ be some parameter. Let $X$ be an estimator.
$X$ is called an unbiased estimator for $theta$ if $E[X] = theta$.
Note that $X$ is a random variable (or random vector) while $theta$ would be a scalar (or vector).
Example
Let's say $x_1$ and $x_2$ are random variables with $E[x_1] = E[x_2] = mu$. Then estimator $y = frac{1}{5} x_1 + frac{4}{5} x_2$ would be an unbiased estimate of $mu$ since $E[y] = mu$.
answered Dec 7 '15 at 9:57
Matthew GunnMatthew Gunn
37618
$endgroup$
$begingroup$
Whoever modded this down was either sloppy or doesn't know what he/she is talking about.
$endgroup$
– Matthew Gunn
Dec 8 '15 at 3:59
add a comment |
Your Answer
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2 Answers
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2 Answers
2
active
oldest
votes
active
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oldest
votes
$begingroup$
[EDIT: Assumed $x_k$ are random variables.]
We say $v$ is an estimator of random variable $x$ if $E[v]=E[x]$, where $E[cdot]$ is expectation of random variables.
Recall that expectation is a linear operator, i.e. if $X$ and $Y$ are random variables and $a,b$ are constants, $E[aX+bY]=aE[X]+bE[Y]$. So we get $E[x]=frac{1}{4}(E[x_1]+E[x_2]+E[x_3]+E[x_4])$, and $E[v]=frac{1}{4}(E[x_1]+3E[x_2])$. Notice that $E[x]=E[v]$ is equivalent to $E[x]-E[v]=0$.
So $v$ is an unbiased estimator if $E[x]-E[v]=0 Leftrightarrow E[x_3]+E[x_4]-2E[x_2]=0.$
answered Dec 6 '15 at 20:02
manofbearmanofbear
1,579515
$endgroup$
$begingroup$
How do we decide if $v$ is unbiased or not unbiased using $ E[x_3]+E[x_4]-2E[x_2]=0.$ as $x_2,x_3$ and $x_4$ are not defined?
$endgroup$
– dumble24
Dec 6 '15 at 21:01
$begingroup$
What is this, if E[v] = E[x]?! Why are you taking expectations on the right hand side?
$endgroup$
– Matthew Gunn
Dec 7 '15 at 10:14
$begingroup$
Almost certainly dumble24 is in a classical statistics environment and is estimating some parameter, not another random variable.
$endgroup$
– Matthew Gunn
Dec 7 '15 at 10:24
$begingroup$
Edited the question, added some more context.
$endgroup$
– dumble24
Dec 7 '15 at 11:58
add a comment |
$begingroup$
[EDIT: Assumed $x_k$ are random variables.]
We say $v$ is an estimator of random variable $x$ if $E[v]=E[x]$, where $E[cdot]$ is expectation of random variables.
Recall that expectation is a linear operator, i.e. if $X$ and $Y$ are random variables and $a,b$ are constants, $E[aX+bY]=aE[X]+bE[Y]$. So we get $E[x]=frac{1}{4}(E[x_1]+E[x_2]+E[x_3]+E[x_4])$, and $E[v]=frac{1}{4}(E[x_1]+3E[x_2])$. Notice that $E[x]=E[v]$ is equivalent to $E[x]-E[v]=0$.
So $v$ is an unbiased estimator if $E[x]-E[v]=0 Leftrightarrow E[x_3]+E[x_4]-2E[x_2]=0.$
answered Dec 6 '15 at 20:02
manofbearmanofbear
1,579515
$endgroup$
$begingroup$
How do we decide if $v$ is unbiased or not unbiased using $ E[x_3]+E[x_4]-2E[x_2]=0.$ as $x_2,x_3$ and $x_4$ are not defined?
$endgroup$
– dumble24
Dec 6 '15 at 21:01
$begingroup$
What is this, if E[v] = E[x]?! Why are you taking expectations on the right hand side?
$endgroup$
– Matthew Gunn
Dec 7 '15 at 10:14
$begingroup$
Almost certainly dumble24 is in a classical statistics environment and is estimating some parameter, not another random variable.
$endgroup$
– Matthew Gunn
Dec 7 '15 at 10:24
$begingroup$
Edited the question, added some more context.
$endgroup$
– dumble24
Dec 7 '15 at 11:58
add a comment |
$begingroup$
[EDIT: Assumed $x_k$ are random variables.]
We say $v$ is an estimator of random variable $x$ if $E[v]=E[x]$, where $E[cdot]$ is expectation of random variables.
Recall that expectation is a linear operator, i.e. if $X$ and $Y$ are random variables and $a,b$ are constants, $E[aX+bY]=aE[X]+bE[Y]$. So we get $E[x]=frac{1}{4}(E[x_1]+E[x_2]+E[x_3]+E[x_4])$, and $E[v]=frac{1}{4}(E[x_1]+3E[x_2])$. Notice that $E[x]=E[v]$ is equivalent to $E[x]-E[v]=0$.
So $v$ is an unbiased estimator if $E[x]-E[v]=0 Leftrightarrow E[x_3]+E[x_4]-2E[x_2]=0.$
answered Dec 6 '15 at 20:02
manofbearmanofbear
1,579515
$endgroup$
[EDIT: Assumed $x_k$ are random variables.]
We say $v$ is an estimator of random variable $x$ if $E[v]=E[x]$, where $E[cdot]$ is expectation of random variables.
Recall that expectation is a linear operator, i.e. if $X$ and $Y$ are random variables and $a,b$ are constants, $E[aX+bY]=aE[X]+bE[Y]$. So we get $E[x]=frac{1}{4}(E[x_1]+E[x_2]+E[x_3]+E[x_4])$, and $E[v]=frac{1}{4}(E[x_1]+3E[x_2])$. Notice that $E[x]=E[v]$ is equivalent to $E[x]-E[v]=0$.
So $v$ is an unbiased estimator if $E[x]-E[v]=0 Leftrightarrow E[x_3]+E[x_4]-2E[x_2]=0.$
answered Dec 6 '15 at 20:02
manofbearmanofbear
1,579515
edited Dec 6 '15 at 20:36
answered Dec 6 '15 at 20:02
manofbearmanofbear
1,579515
answered Dec 6 '15 at 20:02
manofbearmanofbear
1,579515
answered Dec 6 '15 at 20:02
manofbearmanofbear
1,579515
1,579515
$begingroup$
How do we decide if $v$ is unbiased or not unbiased using $ E[x_3]+E[x_4]-2E[x_2]=0.$ as $x_2,x_3$ and $x_4$ are not defined?
$endgroup$
– dumble24
Dec 6 '15 at 21:01
$begingroup$
What is this, if E[v] = E[x]?! Why are you taking expectations on the right hand side?
$endgroup$
– Matthew Gunn
Dec 7 '15 at 10:14
$begingroup$
Almost certainly dumble24 is in a classical statistics environment and is estimating some parameter, not another random variable.
$endgroup$
– Matthew Gunn
Dec 7 '15 at 10:24
$begingroup$
Edited the question, added some more context.
$endgroup$
– dumble24
Dec 7 '15 at 11:58
add a comment |
$begingroup$
How do we decide if $v$ is unbiased or not unbiased using $ E[x_3]+E[x_4]-2E[x_2]=0.$ as $x_2,x_3$ and $x_4$ are not defined?
$endgroup$
– dumble24
Dec 6 '15 at 21:01
$begingroup$
What is this, if E[v] = E[x]?! Why are you taking expectations on the right hand side?
$endgroup$
– Matthew Gunn
Dec 7 '15 at 10:14
$begingroup$
Almost certainly dumble24 is in a classical statistics environment and is estimating some parameter, not another random variable.
$endgroup$
– Matthew Gunn
Dec 7 '15 at 10:24
$begingroup$
Edited the question, added some more context.
$endgroup$
– dumble24
Dec 7 '15 at 11:58
$begingroup$
How do we decide if $v$ is unbiased or not unbiased using $ E[x_3]+E[x_4]-2E[x_2]=0.$ as $x_2,x_3$ and $x_4$ are not defined?
$endgroup$
– dumble24
Dec 6 '15 at 21:01
$begingroup$
How do we decide if $v$ is unbiased or not unbiased using $ E[x_3]+E[x_4]-2E[x_2]=0.$ as $x_2,x_3$ and $x_4$ are not defined?
$endgroup$
– dumble24
Dec 6 '15 at 21:01
$begingroup$
What is this, if E[v] = E[x]?! Why are you taking expectations on the right hand side?
$endgroup$
– Matthew Gunn
Dec 7 '15 at 10:14
$begingroup$
What is this, if E[v] = E[x]?! Why are you taking expectations on the right hand side?
$endgroup$
– Matthew Gunn
Dec 7 '15 at 10:14
$begingroup$
Almost certainly dumble24 is in a classical statistics environment and is estimating some parameter, not another random variable.
$endgroup$
– Matthew Gunn
Dec 7 '15 at 10:24
$begingroup$
Almost certainly dumble24 is in a classical statistics environment and is estimating some parameter, not another random variable.
$endgroup$
– Matthew Gunn
Dec 7 '15 at 10:24
$begingroup$
Edited the question, added some more context.
$endgroup$
– dumble24
Dec 7 '15 at 11:58
$begingroup$
Edited the question, added some more context.
$endgroup$
– dumble24
Dec 7 '15 at 11:58
add a comment |
$begingroup$
Let $theta$ be some parameter. Let $X$ be an estimator.
$X$ is called an unbiased estimator for $theta$ if $E[X] = theta$.
Note that $X$ is a random variable (or random vector) while $theta$ would be a scalar (or vector).
Example
Let's say $x_1$ and $x_2$ are random variables with $E[x_1] = E[x_2] = mu$. Then estimator $y = frac{1}{5} x_1 + frac{4}{5} x_2$ would be an unbiased estimate of $mu$ since $E[y] = mu$.
answered Dec 7 '15 at 9:57
Matthew GunnMatthew Gunn
37618
$endgroup$
$begingroup$
Whoever modded this down was either sloppy or doesn't know what he/she is talking about.
$endgroup$
– Matthew Gunn
Dec 8 '15 at 3:59
add a comment |
$begingroup$
Let $theta$ be some parameter. Let $X$ be an estimator.
$X$ is called an unbiased estimator for $theta$ if $E[X] = theta$.
Note that $X$ is a random variable (or random vector) while $theta$ would be a scalar (or vector).
Example
Let's say $x_1$ and $x_2$ are random variables with $E[x_1] = E[x_2] = mu$. Then estimator $y = frac{1}{5} x_1 + frac{4}{5} x_2$ would be an unbiased estimate of $mu$ since $E[y] = mu$.
answered Dec 7 '15 at 9:57
Matthew GunnMatthew Gunn
37618
$endgroup$
$begingroup$
Whoever modded this down was either sloppy or doesn't know what he/she is talking about.
$endgroup$
– Matthew Gunn
Dec 8 '15 at 3:59
add a comment |
$begingroup$
Let $theta$ be some parameter. Let $X$ be an estimator.
$X$ is called an unbiased estimator for $theta$ if $E[X] = theta$.
Note that $X$ is a random variable (or random vector) while $theta$ would be a scalar (or vector).
Example
Let's say $x_1$ and $x_2$ are random variables with $E[x_1] = E[x_2] = mu$. Then estimator $y = frac{1}{5} x_1 + frac{4}{5} x_2$ would be an unbiased estimate of $mu$ since $E[y] = mu$.
answered Dec 7 '15 at 9:57
Matthew GunnMatthew Gunn
37618
$endgroup$
Let $theta$ be some parameter. Let $X$ be an estimator.
$X$ is called an unbiased estimator for $theta$ if $E[X] = theta$.
Note that $X$ is a random variable (or random vector) while $theta$ would be a scalar (or vector).
Example
Let's say $x_1$ and $x_2$ are random variables with $E[x_1] = E[x_2] = mu$. Then estimator $y = frac{1}{5} x_1 + frac{4}{5} x_2$ would be an unbiased estimate of $mu$ since $E[y] = mu$.
answered Dec 7 '15 at 9:57
Matthew GunnMatthew Gunn
37618
edited Dec 7 '15 at 10:17
answered Dec 7 '15 at 9:57
Matthew GunnMatthew Gunn
37618
answered Dec 7 '15 at 9:57
Matthew GunnMatthew Gunn
37618
answered Dec 7 '15 at 9:57
Matthew GunnMatthew Gunn
37618
37618
$begingroup$
Whoever modded this down was either sloppy or doesn't know what he/she is talking about.
$endgroup$
– Matthew Gunn
Dec 8 '15 at 3:59
add a comment |
$begingroup$
Whoever modded this down was either sloppy or doesn't know what he/she is talking about.
$endgroup$
– Matthew Gunn
Dec 8 '15 at 3:59
$begingroup$
Whoever modded this down was either sloppy or doesn't know what he/she is talking about.
$endgroup$
– Matthew Gunn
Dec 8 '15 at 3:59
$begingroup$
Whoever modded this down was either sloppy or doesn't know what he/she is talking about.
$endgroup$
– Matthew Gunn
Dec 8 '15 at 3:59
add a comment |
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So $v$ is the same thing as $x$? If that's not what you meant, then you need to clarify your question.
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– Michael Hardy
Dec 6 '15 at 20:07
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I assumed here that $x_k$ are random variables.
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– manofbear
Dec 6 '15 at 20:35
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x is expected value of a random variable
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– dumble24
Dec 6 '15 at 21:05
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(x1+x2+x3+x4)/4 calculates the expected value if a random variable has exactly four, equiprobable possible outcomes x1, x2, ..., x4. Alternatively, if x_1, x_2, ... x_4 denote 4 independent draws form some probability distribution, then (x1+x2+x3+x4)/4 is an estimator of the expected value, but it is not the actual expected value! To be precise, this is an important distinction.
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– Matthew Gunn
Dec 7 '15 at 23:12