check if estimation is unbiased?












0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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


















0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '15 at 11:55







dumble24

















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
















  • 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












2 Answers
2






active

oldest

votes


















0












$begingroup$

[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 $E[x]-E[v]=0 Leftrightarrow E[x_3]+E[x_4]-2E[x_2]=0.$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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



















-1












$begingroup$

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 $E[y] = mu$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

[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 $E[x]-E[v]=0 Leftrightarrow E[x_3]+E[x_4]-2E[x_2]=0.$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















0












$begingroup$

[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 $E[x]-E[v]=0 Leftrightarrow E[x_3]+E[x_4]-2E[x_2]=0.$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














0












0








0





$begingroup$

[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 $E[x]-E[v]=0 Leftrightarrow E[x_3]+E[x_4]-2E[x_2]=0.$






share|cite|improve this answer











$endgroup$



[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 $E[x]-E[v]=0 Leftrightarrow E[x_3]+E[x_4]-2E[x_2]=0.$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 6 '15 at 20:36

























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


















  • $begingroup$
    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











-1












$begingroup$

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 $E[y] = mu$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















-1












$begingroup$

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 $E[y] = mu$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














-1












-1








-1





$begingroup$

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 $E[y] = mu$.






share|cite|improve this answer











$endgroup$



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 $E[y] = mu$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 7 '15 at 10:17

























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


















  • $begingroup$
    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


















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