For what value of $c$, we have $bar{theta}$ is unbiased?
$begingroup$
Given a population with the density function
$$f(x)=frac{2(theta-x)}{theta^2}$$ for $0<x<theta$, consider the estimator of $theta$ of the form $bar{theta}=cbar{x}$ where $c$ is constant.
For what value of $c$, do we have that $bar{theta}$ is unbiased?
My attempt:
By definition, a estimator is unbiased if $E[bar{theta}]=theta$
But here i'm a little confused because i have a density function. Can someone help me?
probability
edited Nov 25 '18 at 21:56
Clement C.
49.8k33886
asked Nov 25 '18 at 21:43
Bvss12Bvss12
1,790618
$endgroup$
add a comment |
$begingroup$
Given a population with the density function
$$f(x)=frac{2(theta-x)}{theta^2}$$ for $0<x<theta$, consider the estimator of $theta$ of the form $bar{theta}=cbar{x}$ where $c$ is constant.
For what value of $c$, do we have that $bar{theta}$ is unbiased?
My attempt:
By definition, a estimator is unbiased if $E[bar{theta}]=theta$
But here i'm a little confused because i have a density function. Can someone help me?
probability
edited Nov 25 '18 at 21:56
Clement C.
49.8k33886
asked Nov 25 '18 at 21:43
Bvss12Bvss12
1,790618
$endgroup$
add a comment |
$begingroup$
Given a population with the density function
$$f(x)=frac{2(theta-x)}{theta^2}$$ for $0<x<theta$, consider the estimator of $theta$ of the form $bar{theta}=cbar{x}$ where $c$ is constant.
For what value of $c$, do we have that $bar{theta}$ is unbiased?
My attempt:
By definition, a estimator is unbiased if $E[bar{theta}]=theta$
But here i'm a little confused because i have a density function. Can someone help me?
probability
edited Nov 25 '18 at 21:56
Clement C.
49.8k33886
asked Nov 25 '18 at 21:43
Bvss12Bvss12
1,790618
$endgroup$
Given a population with the density function
$$f(x)=frac{2(theta-x)}{theta^2}$$ for $0<x<theta$, consider the estimator of $theta$ of the form $bar{theta}=cbar{x}$ where $c$ is constant.
For what value of $c$, do we have that $bar{theta}$ is unbiased?
My attempt:
By definition, a estimator is unbiased if $E[bar{theta}]=theta$
But here i'm a little confused because i have a density function. Can someone help me?
probability
probability
edited Nov 25 '18 at 21:56
Clement C.
49.8k33886
asked Nov 25 '18 at 21:43
Bvss12Bvss12
1,790618
edited Nov 25 '18 at 21:56
Clement C.
49.8k33886
asked Nov 25 '18 at 21:43
Bvss12Bvss12
1,790618
edited Nov 25 '18 at 21:56
Clement C.
49.8k33886
edited Nov 25 '18 at 21:56
Clement C.
49.8k33886
edited Nov 25 '18 at 21:56
Clement C.
49.8k33886
49.8k33886
asked Nov 25 '18 at 21:43
Bvss12Bvss12
1,790618
asked Nov 25 '18 at 21:43
Bvss12Bvss12
1,790618
asked Nov 25 '18 at 21:43
Bvss12Bvss12
1,790618
1,790618
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
This may get you started:
$$E[bartheta]=E[cbar X]=cE[bar X]=frac{c}{n}sum_{i=1}^nE[X_i]$$
Then using the density function, you can find the expected value inside the sum.
answered Nov 25 '18 at 21:50
gd1035gd1035
4571210
$endgroup$
1
$begingroup$
Thanks for all!
$endgroup$
– Bvss12
Nov 25 '18 at 22:17
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This may get you started:
$$E[bartheta]=E[cbar X]=cE[bar X]=frac{c}{n}sum_{i=1}^nE[X_i]$$
Then using the density function, you can find the expected value inside the sum.
answered Nov 25 '18 at 21:50
gd1035gd1035
4571210
$endgroup$
1
$begingroup$
Thanks for all!
$endgroup$
– Bvss12
Nov 25 '18 at 22:17
add a comment |
$begingroup$
This may get you started:
$$E[bartheta]=E[cbar X]=cE[bar X]=frac{c}{n}sum_{i=1}^nE[X_i]$$
Then using the density function, you can find the expected value inside the sum.
answered Nov 25 '18 at 21:50
gd1035gd1035
4571210
$endgroup$
1
$begingroup$
Thanks for all!
$endgroup$
– Bvss12
Nov 25 '18 at 22:17
add a comment |
$begingroup$
This may get you started:
$$E[bartheta]=E[cbar X]=cE[bar X]=frac{c}{n}sum_{i=1}^nE[X_i]$$
Then using the density function, you can find the expected value inside the sum.
answered Nov 25 '18 at 21:50
gd1035gd1035
4571210
$endgroup$
This may get you started:
$$E[bartheta]=E[cbar X]=cE[bar X]=frac{c}{n}sum_{i=1}^nE[X_i]$$
Then using the density function, you can find the expected value inside the sum.
answered Nov 25 '18 at 21:50
gd1035gd1035
4571210
answered Nov 25 '18 at 21:50
gd1035gd1035
4571210
answered Nov 25 '18 at 21:50
gd1035gd1035
4571210
answered Nov 25 '18 at 21:50
gd1035gd1035
4571210
4571210
1
$begingroup$
Thanks for all!
$endgroup$
– Bvss12
Nov 25 '18 at 22:17
add a comment |
1
$begingroup$
Thanks for all!
$endgroup$
– Bvss12
Nov 25 '18 at 22:17
1
1
$begingroup$
Thanks for all!
$endgroup$
– Bvss12
Nov 25 '18 at 22:17
$begingroup$
Thanks for all!
$endgroup$
– Bvss12
Nov 25 '18 at 22:17
add a comment |
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