Unbiased estimate for a parameter












0












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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







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  • $begingroup$
    Similar to math.stackexchange.com/questions/3038477/…
    $endgroup$
    – Henry
    Dec 16 '18 at 15:54
















0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    Similar to math.stackexchange.com/questions/3038477/…
    $endgroup$
    – Henry
    Dec 16 '18 at 15:54














0












0








0





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













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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


















  • $begingroup$
    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










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$begingroup$

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 $text{Var}(3X_1) = 9text{Var}(X_1) = 9text{Var}(X) = frac{Theta^2}{2}$






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    $begingroup$

    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 $text{Var}(3X_1) = 9text{Var}(X_1) = 9text{Var}(X) = frac{Theta^2}{2}$






    share|cite|improve this answer









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      0












      $begingroup$

      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 $text{Var}(3X_1) = 9text{Var}(X_1) = 9text{Var}(X) = frac{Theta^2}{2}$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        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 $text{Var}(3X_1) = 9text{Var}(X_1) = 9text{Var}(X) = frac{Theta^2}{2}$






        share|cite|improve this answer









        $endgroup$



        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 $text{Var}(3X_1) = 9text{Var}(X_1) = 9text{Var}(X) = frac{Theta^2}{2}$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 16 '18 at 15:53









        HenryHenry

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