Unbiased estimate for a parameter
$begingroup$
They ask me to estimate any parameter and I do not if the solution is correct:
The life time X of a battery is considered to be a random variable with density function
$f (x; Θ) =frac{ 2 }{ Θ²} (Θ - x)$ if 0 < x <Θ
Be X1, X2, ..., Xn a simple random sample of X. Knowing that $E(X) = frac{Θ}{ 3}$ and $E(X²) = frac{Θ²}{6}$, determine:
(a) An unbiased estimator for Θ, Θ ^
(b) The variance of the estimator Θ ^
For (A) I have
$mu= frac{Θ}{3}$
$ Θ=3 mu$
$ Θ= 3 X$
$E(Θ)= E(3X) $
$ 3 E(X)= 3 *mu$
$3 * mu =3 * frac{Θ}{3} = Θ$
(that if it is unbiased)
For (B) I have:
$Var(Θ)= Var(3X) = 3^2 Var(X) = 9 σ^2 /n$
statistics statistical-inference estimation parameter-estimation
asked Dec 16 '18 at 13:45
FernandoFernando
495
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add a comment |
$begingroup$
They ask me to estimate any parameter and I do not if the solution is correct:
The life time X of a battery is considered to be a random variable with density function
$f (x; Θ) =frac{ 2 }{ Θ²} (Θ - x)$ if 0 < x <Θ
Be X1, X2, ..., Xn a simple random sample of X. Knowing that $E(X) = frac{Θ}{ 3}$ and $E(X²) = frac{Θ²}{6}$, determine:
(a) An unbiased estimator for Θ, Θ ^
(b) The variance of the estimator Θ ^
For (A) I have
$mu= frac{Θ}{3}$
$ Θ=3 mu$
$ Θ= 3 X$
$E(Θ)= E(3X) $
$ 3 E(X)= 3 *mu$
$3 * mu =3 * frac{Θ}{3} = Θ$
(that if it is unbiased)
For (B) I have:
$Var(Θ)= Var(3X) = 3^2 Var(X) = 9 σ^2 /n$
statistics statistical-inference estimation parameter-estimation
asked Dec 16 '18 at 13:45
FernandoFernando
495
$endgroup$
$begingroup$
Similar to math.stackexchange.com/questions/3038477/…
$endgroup$
– Henry
Dec 16 '18 at 15:54
add a comment |
$begingroup$
They ask me to estimate any parameter and I do not if the solution is correct:
The life time X of a battery is considered to be a random variable with density function
$f (x; Θ) =frac{ 2 }{ Θ²} (Θ - x)$ if 0 < x <Θ
Be X1, X2, ..., Xn a simple random sample of X. Knowing that $E(X) = frac{Θ}{ 3}$ and $E(X²) = frac{Θ²}{6}$, determine:
(a) An unbiased estimator for Θ, Θ ^
(b) The variance of the estimator Θ ^
For (A) I have
$mu= frac{Θ}{3}$
$ Θ=3 mu$
$ Θ= 3 X$
$E(Θ)= E(3X) $
$ 3 E(X)= 3 *mu$
$3 * mu =3 * frac{Θ}{3} = Θ$
(that if it is unbiased)
For (B) I have:
$Var(Θ)= Var(3X) = 3^2 Var(X) = 9 σ^2 /n$
statistics statistical-inference estimation parameter-estimation
asked Dec 16 '18 at 13:45
FernandoFernando
495
$endgroup$
They ask me to estimate any parameter and I do not if the solution is correct:
The life time X of a battery is considered to be a random variable with density function
$f (x; Θ) =frac{ 2 }{ Θ²} (Θ - x)$ if 0 < x <Θ
Be X1, X2, ..., Xn a simple random sample of X. Knowing that $E(X) = frac{Θ}{ 3}$ and $E(X²) = frac{Θ²}{6}$, determine:
(a) An unbiased estimator for Θ, Θ ^
(b) The variance of the estimator Θ ^
For (A) I have
$mu= frac{Θ}{3}$
$ Θ=3 mu$
$ Θ= 3 X$
$E(Θ)= E(3X) $
$ 3 E(X)= 3 *mu$
$3 * mu =3 * frac{Θ}{3} = Θ$
(that if it is unbiased)
For (B) I have:
$Var(Θ)= Var(3X) = 3^2 Var(X) = 9 σ^2 /n$
statistics statistical-inference estimation parameter-estimation
statistics statistical-inference estimation parameter-estimation
asked Dec 16 '18 at 13:45
FernandoFernando
495
asked Dec 16 '18 at 13:45
FernandoFernando
495
asked Dec 16 '18 at 13:45
FernandoFernando
495
asked Dec 16 '18 at 13:45
FernandoFernando
495
asked Dec 16 '18 at 13:45
FernandoFernando
495
495
$begingroup$
Similar to math.stackexchange.com/questions/3038477/…
$endgroup$
– Henry
Dec 16 '18 at 15:54
add a comment |
$begingroup$
Similar to math.stackexchange.com/questions/3038477/…
$endgroup$
– Henry
Dec 16 '18 at 15:54
$begingroup$
Similar to math.stackexchange.com/questions/3038477/…
$endgroup$
– Henry
Dec 16 '18 at 15:54
$begingroup$
Similar to math.stackexchange.com/questions/3038477/…
$endgroup$
– Henry
Dec 16 '18 at 15:54
add a comment |
1 Answer
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Strictly speaking, you have a sample from $X$, so your chosen estimator should be based on the sample $X1,X_2,ldots,X_n$ rather than on the underlying $X$
Although you did not say so, you seem to have chosen $3bar{X}=frac3nsumlimits_{i=1}^n X_i$ as your estimator of $Theta$. This does indeed have an expectation of $3 E[bar{X}] =3 E[X] = Theta$ so is unbiased.
You may have miscalculated the variance of $X$ which is $text{Var}(X) = Eleft[X^2right]-left(E[X]right)^2 = frac{Theta^2}{6}-left(frac{Theta}{3}right)^2 = frac{Theta^2}{18}$ so the variance of your estimator is $text{Var}(3bar{X}) = 9text{Var}(bar{X}) = frac{9}{n}text{Var}(X) = frac{Theta^2}{2n}$
Alternatively you might have chosen $3X_1$ or something similar as your estimator. That would have mean $3 E[X_1] =3 E[X] = Theta$ so is also unbiased. Its variance is $text{Var}(3X_1) = 9text{Var}(X_1) = 9text{Var}(X) = frac{Theta^2}{2}$
answered Dec 16 '18 at 15:53
HenryHenry
99.2k478164
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add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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active
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votes
$begingroup$
Strictly speaking, you have a sample from $X$, so your chosen estimator should be based on the sample $X1,X_2,ldots,X_n$ rather than on the underlying $X$
Although you did not say so, you seem to have chosen $3bar{X}=frac3nsumlimits_{i=1}^n X_i$ as your estimator of $Theta$. This does indeed have an expectation of $3 E[bar{X}] =3 E[X] = Theta$ so is unbiased.
You may have miscalculated the variance of $X$ which is $text{Var}(X) = Eleft[X^2right]-left(E[X]right)^2 = frac{Theta^2}{6}-left(frac{Theta}{3}right)^2 = frac{Theta^2}{18}$ so the variance of your estimator is $text{Var}(3bar{X}) = 9text{Var}(bar{X}) = frac{9}{n}text{Var}(X) = frac{Theta^2}{2n}$
Alternatively you might have chosen $3X_1$ or something similar as your estimator. That would have mean $3 E[X_1] =3 E[X] = Theta$ so is also unbiased. Its variance is $text{Var}(3X_1) = 9text{Var}(X_1) = 9text{Var}(X) = frac{Theta^2}{2}$
answered Dec 16 '18 at 15:53
HenryHenry
99.2k478164
$endgroup$
add a comment |
$begingroup$
Strictly speaking, you have a sample from $X$, so your chosen estimator should be based on the sample $X1,X_2,ldots,X_n$ rather than on the underlying $X$
Although you did not say so, you seem to have chosen $3bar{X}=frac3nsumlimits_{i=1}^n X_i$ as your estimator of $Theta$. This does indeed have an expectation of $3 E[bar{X}] =3 E[X] = Theta$ so is unbiased.
You may have miscalculated the variance of $X$ which is $text{Var}(X) = Eleft[X^2right]-left(E[X]right)^2 = frac{Theta^2}{6}-left(frac{Theta}{3}right)^2 = frac{Theta^2}{18}$ so the variance of your estimator is $text{Var}(3bar{X}) = 9text{Var}(bar{X}) = frac{9}{n}text{Var}(X) = frac{Theta^2}{2n}$
Alternatively you might have chosen $3X_1$ or something similar as your estimator. That would have mean $3 E[X_1] =3 E[X] = Theta$ so is also unbiased. Its variance is $text{Var}(3X_1) = 9text{Var}(X_1) = 9text{Var}(X) = frac{Theta^2}{2}$
answered Dec 16 '18 at 15:53
HenryHenry
99.2k478164
$endgroup$
add a comment |
$begingroup$
Strictly speaking, you have a sample from $X$, so your chosen estimator should be based on the sample $X1,X_2,ldots,X_n$ rather than on the underlying $X$
Although you did not say so, you seem to have chosen $3bar{X}=frac3nsumlimits_{i=1}^n X_i$ as your estimator of $Theta$. This does indeed have an expectation of $3 E[bar{X}] =3 E[X] = Theta$ so is unbiased.
You may have miscalculated the variance of $X$ which is $text{Var}(X) = Eleft[X^2right]-left(E[X]right)^2 = frac{Theta^2}{6}-left(frac{Theta}{3}right)^2 = frac{Theta^2}{18}$ so the variance of your estimator is $text{Var}(3bar{X}) = 9text{Var}(bar{X}) = frac{9}{n}text{Var}(X) = frac{Theta^2}{2n}$
Alternatively you might have chosen $3X_1$ or something similar as your estimator. That would have mean $3 E[X_1] =3 E[X] = Theta$ so is also unbiased. Its variance is $text{Var}(3X_1) = 9text{Var}(X_1) = 9text{Var}(X) = frac{Theta^2}{2}$
answered Dec 16 '18 at 15:53
HenryHenry
99.2k478164
$endgroup$
Strictly speaking, you have a sample from $X$, so your chosen estimator should be based on the sample $X1,X_2,ldots,X_n$ rather than on the underlying $X$
Although you did not say so, you seem to have chosen $3bar{X}=frac3nsumlimits_{i=1}^n X_i$ as your estimator of $Theta$. This does indeed have an expectation of $3 E[bar{X}] =3 E[X] = Theta$ so is unbiased.
You may have miscalculated the variance of $X$ which is $text{Var}(X) = Eleft[X^2right]-left(E[X]right)^2 = frac{Theta^2}{6}-left(frac{Theta}{3}right)^2 = frac{Theta^2}{18}$ so the variance of your estimator is $text{Var}(3bar{X}) = 9text{Var}(bar{X}) = frac{9}{n}text{Var}(X) = frac{Theta^2}{2n}$
Alternatively you might have chosen $3X_1$ or something similar as your estimator. That would have mean $3 E[X_1] =3 E[X] = Theta$ so is also unbiased. Its variance is $text{Var}(3X_1) = 9text{Var}(X_1) = 9text{Var}(X) = frac{Theta^2}{2}$
answered Dec 16 '18 at 15:53
HenryHenry
99.2k478164
answered Dec 16 '18 at 15:53
HenryHenry
99.2k478164
answered Dec 16 '18 at 15:53
HenryHenry
99.2k478164
answered Dec 16 '18 at 15:53
HenryHenry
99.2k478164
99.2k478164
add a comment |
add a comment |
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$begingroup$
Similar to math.stackexchange.com/questions/3038477/…
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– Henry
Dec 16 '18 at 15:54