Variance of an unbiased estimator $L = fracpi4sqrt{X_1X_2}$












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Find variance of unbiased estimator L, where $L = dfracpi4sqrt{X_1X_2}$.



$f(x) = dfrac1theta e^{frac{-x}{theta}}$, $x>0$,



$X_1$ and $X_2$ are independent, and exponentially distributed.



Since $L$ is unbiased so I know $E[L] = theta$, right?



Also, $operatorname{Var}[x] = E[x^2] - E[x]^2$. But I'm struggling with finding $E[x^2]$. Any help will be appreciated. Thank you.






probability integration statistics variance expected-value







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  • 1




    $begingroup$
    How can the "estimator" $L$ depend on the quantity $theta$ being estimated?
    $endgroup$
    – jesterII
    Nov 27 '18 at 11:59










  • $begingroup$
    @jesterll edited.
    $endgroup$
    – OvermanZarathustra
    Nov 27 '18 at 13:20






  • 3




    $begingroup$
    You are given $E(L)$. Now look at $E(L^2)=text{const.}E(X_1X_2)$. Now use the independence of $X_1$ and $X_2$. The variance of $L$ is $E(L^2)-E(L)^2$.
    $endgroup$
    – yurnero
    Nov 27 '18 at 13:50








  • 1




    $begingroup$
    From $operatorname{Var}(L)=frac{pi^2}{16}operatorname{Var}(sqrt{X_1X_2})=frac{pi^2}{16}left[E(X_1X_2)-(E(sqrt{X_1X_2}))^2right]$, the only thing that needs calculating is $E(sqrt{X_1})$. Where is the problem?
    $endgroup$
    – StubbornAtom
    Nov 27 '18 at 16:14
















1












$begingroup$


Find variance of unbiased estimator L, where $L = dfracpi4sqrt{X_1X_2}$.



$f(x) = dfrac1theta e^{frac{-x}{theta}}$, $x>0$,



$X_1$ and $X_2$ are independent, and exponentially distributed.



Since $L$ is unbiased so I know $E[L] = theta$, right?



Also, $operatorname{Var}[x] = E[x^2] - E[x]^2$. But I'm struggling with finding $E[x^2]$. Any help will be appreciated. Thank you.






probability integration statistics variance expected-value







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    How can the "estimator" $L$ depend on the quantity $theta$ being estimated?
    $endgroup$
    – jesterII
    Nov 27 '18 at 11:59










  • $begingroup$
    @jesterll edited.
    $endgroup$
    – OvermanZarathustra
    Nov 27 '18 at 13:20






  • 3




    $begingroup$
    You are given $E(L)$. Now look at $E(L^2)=text{const.}E(X_1X_2)$. Now use the independence of $X_1$ and $X_2$. The variance of $L$ is $E(L^2)-E(L)^2$.
    $endgroup$
    – yurnero
    Nov 27 '18 at 13:50








  • 1




    $begingroup$
    From $operatorname{Var}(L)=frac{pi^2}{16}operatorname{Var}(sqrt{X_1X_2})=frac{pi^2}{16}left[E(X_1X_2)-(E(sqrt{X_1X_2}))^2right]$, the only thing that needs calculating is $E(sqrt{X_1})$. Where is the problem?
    $endgroup$
    – StubbornAtom
    Nov 27 '18 at 16:14














1












1








1





$begingroup$


Find variance of unbiased estimator L, where $L = dfracpi4sqrt{X_1X_2}$.



$f(x) = dfrac1theta e^{frac{-x}{theta}}$, $x>0$,



$X_1$ and $X_2$ are independent, and exponentially distributed.



Since $L$ is unbiased so I know $E[L] = theta$, right?



Also, $operatorname{Var}[x] = E[x^2] - E[x]^2$. But I'm struggling with finding $E[x^2]$. Any help will be appreciated. Thank you.






probability integration statistics variance expected-value







share|cite|improve this question











$endgroup$




Find variance of unbiased estimator L, where $L = dfracpi4sqrt{X_1X_2}$.



$f(x) = dfrac1theta e^{frac{-x}{theta}}$, $x>0$,



$X_1$ and $X_2$ are independent, and exponentially distributed.



Since $L$ is unbiased so I know $E[L] = theta$, right?



Also, $operatorname{Var}[x] = E[x^2] - E[x]^2$. But I'm struggling with finding $E[x^2]$. Any help will be appreciated. Thank you.






probability integration statistics variance expected-value




probability integration statistics variance expected-value






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 13:20







OvermanZarathustra

















asked Nov 27 '18 at 10:41









OvermanZarathustraOvermanZarathustra

156




156








  • 1




    $begingroup$
    How can the "estimator" $L$ depend on the quantity $theta$ being estimated?
    $endgroup$
    – jesterII
    Nov 27 '18 at 11:59










  • $begingroup$
    @jesterll edited.
    $endgroup$
    – OvermanZarathustra
    Nov 27 '18 at 13:20






  • 3




    $begingroup$
    You are given $E(L)$. Now look at $E(L^2)=text{const.}E(X_1X_2)$. Now use the independence of $X_1$ and $X_2$. The variance of $L$ is $E(L^2)-E(L)^2$.
    $endgroup$
    – yurnero
    Nov 27 '18 at 13:50








  • 1




    $begingroup$
    From $operatorname{Var}(L)=frac{pi^2}{16}operatorname{Var}(sqrt{X_1X_2})=frac{pi^2}{16}left[E(X_1X_2)-(E(sqrt{X_1X_2}))^2right]$, the only thing that needs calculating is $E(sqrt{X_1})$. Where is the problem?
    $endgroup$
    – StubbornAtom
    Nov 27 '18 at 16:14














  • 1




    $begingroup$
    How can the "estimator" $L$ depend on the quantity $theta$ being estimated?
    $endgroup$
    – jesterII
    Nov 27 '18 at 11:59










  • $begingroup$
    @jesterll edited.
    $endgroup$
    – OvermanZarathustra
    Nov 27 '18 at 13:20






  • 3




    $begingroup$
    You are given $E(L)$. Now look at $E(L^2)=text{const.}E(X_1X_2)$. Now use the independence of $X_1$ and $X_2$. The variance of $L$ is $E(L^2)-E(L)^2$.
    $endgroup$
    – yurnero
    Nov 27 '18 at 13:50








  • 1




    $begingroup$
    From $operatorname{Var}(L)=frac{pi^2}{16}operatorname{Var}(sqrt{X_1X_2})=frac{pi^2}{16}left[E(X_1X_2)-(E(sqrt{X_1X_2}))^2right]$, the only thing that needs calculating is $E(sqrt{X_1})$. Where is the problem?
    $endgroup$
    – StubbornAtom
    Nov 27 '18 at 16:14








1




1




$begingroup$
How can the "estimator" $L$ depend on the quantity $theta$ being estimated?
$endgroup$
– jesterII
Nov 27 '18 at 11:59




$begingroup$
How can the "estimator" $L$ depend on the quantity $theta$ being estimated?
$endgroup$
– jesterII
Nov 27 '18 at 11:59












$begingroup$
@jesterll edited.
$endgroup$
– OvermanZarathustra
Nov 27 '18 at 13:20




$begingroup$
@jesterll edited.
$endgroup$
– OvermanZarathustra
Nov 27 '18 at 13:20




3




3




$begingroup$
You are given $E(L)$. Now look at $E(L^2)=text{const.}E(X_1X_2)$. Now use the independence of $X_1$ and $X_2$. The variance of $L$ is $E(L^2)-E(L)^2$.
$endgroup$
– yurnero
Nov 27 '18 at 13:50






$begingroup$
You are given $E(L)$. Now look at $E(L^2)=text{const.}E(X_1X_2)$. Now use the independence of $X_1$ and $X_2$. The variance of $L$ is $E(L^2)-E(L)^2$.
$endgroup$
– yurnero
Nov 27 '18 at 13:50






1




1




$begingroup$
From $operatorname{Var}(L)=frac{pi^2}{16}operatorname{Var}(sqrt{X_1X_2})=frac{pi^2}{16}left[E(X_1X_2)-(E(sqrt{X_1X_2}))^2right]$, the only thing that needs calculating is $E(sqrt{X_1})$. Where is the problem?
$endgroup$
– StubbornAtom
Nov 27 '18 at 16:14




$begingroup$
From $operatorname{Var}(L)=frac{pi^2}{16}operatorname{Var}(sqrt{X_1X_2})=frac{pi^2}{16}left[E(X_1X_2)-(E(sqrt{X_1X_2}))^2right]$, the only thing that needs calculating is $E(sqrt{X_1})$. Where is the problem?
$endgroup$
– StubbornAtom
Nov 27 '18 at 16:14










1 Answer
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$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
mathbb{E}bracks{L} & equiv
int_{0}^{infty}{expo{-x_{1}/theta} over theta}
int_{0}^{infty}{expo{-x_{2}/theta} over theta}
pars{{pi over 4}root{x_{1}x_{2}}}dd x_{1},dd x_{2}
\[5mm] & =
{pi over 4}pars{root{theta}
int_{0}^{infty}expo{-x_{1}/theta}
,root{x_{1} over theta},{dd x_{1} over theta}}
\[2mm] & phantom{===,}
pars{root{theta}
int_{0}^{infty}expo{-x_{2}/theta},root{x_{2} over theta}
,{dd x_{2} over theta}}
\[5mm] & =
{pi over 4}
underbrace{pars{int_{0}^{infty}x^{1/2}expo{-x}dd x}^{2}}
_{ds{= Gamma^{2}pars{3/2} = {pi/4}}} theta =
bbx{{pi^{2} over 16},theta}
\[1cm]
mathbb{E}bracks{L^{2}} & equiv
int_{0}^{infty}{expo{-x_{1}/theta} over theta}
int_{0}^{infty}{expo{-x_{2}/theta} over theta}
pars{{pi over 4}root{x_{1}x_{2}}}^{2}dd x_{1},dd x_{2}
\[5mm] & =
{pi^{2} over 16}pars{thetaint_{0}^{infty}
expo{-x_{1}/theta},{x_{1} over theta}
,{dd x_{1} over theta}}
pars{thetaint_{0}^{infty}
expo{-x_{2}/theta},{x_{1} over theta}
,{dd x_{2} over theta}}
\[5mm] & =
bbx{{pi^{2} over 16},theta^{2}}
\[1cm]
mbox{Var}pars{L} & =
bbx{{pi^{2} over 16}pars{1 - {pi^{2} over 16}}theta^{2}}
end{align}






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    1 Answer
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    active

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    1












    $begingroup$

    $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
    newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
    newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
    newcommand{dd}{mathrm{d}}
    newcommand{ds}[1]{displaystyle{#1}}
    newcommand{expo}[1]{,mathrm{e}^{#1},}
    newcommand{ic}{mathrm{i}}
    newcommand{mc}[1]{mathcal{#1}}
    newcommand{mrm}[1]{mathrm{#1}}
    newcommand{pars}[1]{left(,{#1},right)}
    newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
    newcommand{root}[2]{,sqrt[#1]{,{#2},},}
    newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
    newcommand{verts}[1]{leftvert,{#1},rightvert}$
    begin{align}
    mathbb{E}bracks{L} & equiv
    int_{0}^{infty}{expo{-x_{1}/theta} over theta}
    int_{0}^{infty}{expo{-x_{2}/theta} over theta}
    pars{{pi over 4}root{x_{1}x_{2}}}dd x_{1},dd x_{2}
    \[5mm] & =
    {pi over 4}pars{root{theta}
    int_{0}^{infty}expo{-x_{1}/theta}
    ,root{x_{1} over theta},{dd x_{1} over theta}}
    \[2mm] & phantom{===,}
    pars{root{theta}
    int_{0}^{infty}expo{-x_{2}/theta},root{x_{2} over theta}
    ,{dd x_{2} over theta}}
    \[5mm] & =
    {pi over 4}
    underbrace{pars{int_{0}^{infty}x^{1/2}expo{-x}dd x}^{2}}
    _{ds{= Gamma^{2}pars{3/2} = {pi/4}}} theta =
    bbx{{pi^{2} over 16},theta}
    \[1cm]
    mathbb{E}bracks{L^{2}} & equiv
    int_{0}^{infty}{expo{-x_{1}/theta} over theta}
    int_{0}^{infty}{expo{-x_{2}/theta} over theta}
    pars{{pi over 4}root{x_{1}x_{2}}}^{2}dd x_{1},dd x_{2}
    \[5mm] & =
    {pi^{2} over 16}pars{thetaint_{0}^{infty}
    expo{-x_{1}/theta},{x_{1} over theta}
    ,{dd x_{1} over theta}}
    pars{thetaint_{0}^{infty}
    expo{-x_{2}/theta},{x_{1} over theta}
    ,{dd x_{2} over theta}}
    \[5mm] & =
    bbx{{pi^{2} over 16},theta^{2}}
    \[1cm]
    mbox{Var}pars{L} & =
    bbx{{pi^{2} over 16}pars{1 - {pi^{2} over 16}}theta^{2}}
    end{align}






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
      newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
      newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
      newcommand{dd}{mathrm{d}}
      newcommand{ds}[1]{displaystyle{#1}}
      newcommand{expo}[1]{,mathrm{e}^{#1},}
      newcommand{ic}{mathrm{i}}
      newcommand{mc}[1]{mathcal{#1}}
      newcommand{mrm}[1]{mathrm{#1}}
      newcommand{pars}[1]{left(,{#1},right)}
      newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
      newcommand{root}[2]{,sqrt[#1]{,{#2},},}
      newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
      newcommand{verts}[1]{leftvert,{#1},rightvert}$
      begin{align}
      mathbb{E}bracks{L} & equiv
      int_{0}^{infty}{expo{-x_{1}/theta} over theta}
      int_{0}^{infty}{expo{-x_{2}/theta} over theta}
      pars{{pi over 4}root{x_{1}x_{2}}}dd x_{1},dd x_{2}
      \[5mm] & =
      {pi over 4}pars{root{theta}
      int_{0}^{infty}expo{-x_{1}/theta}
      ,root{x_{1} over theta},{dd x_{1} over theta}}
      \[2mm] & phantom{===,}
      pars{root{theta}
      int_{0}^{infty}expo{-x_{2}/theta},root{x_{2} over theta}
      ,{dd x_{2} over theta}}
      \[5mm] & =
      {pi over 4}
      underbrace{pars{int_{0}^{infty}x^{1/2}expo{-x}dd x}^{2}}
      _{ds{= Gamma^{2}pars{3/2} = {pi/4}}} theta =
      bbx{{pi^{2} over 16},theta}
      \[1cm]
      mathbb{E}bracks{L^{2}} & equiv
      int_{0}^{infty}{expo{-x_{1}/theta} over theta}
      int_{0}^{infty}{expo{-x_{2}/theta} over theta}
      pars{{pi over 4}root{x_{1}x_{2}}}^{2}dd x_{1},dd x_{2}
      \[5mm] & =
      {pi^{2} over 16}pars{thetaint_{0}^{infty}
      expo{-x_{1}/theta},{x_{1} over theta}
      ,{dd x_{1} over theta}}
      pars{thetaint_{0}^{infty}
      expo{-x_{2}/theta},{x_{1} over theta}
      ,{dd x_{2} over theta}}
      \[5mm] & =
      bbx{{pi^{2} over 16},theta^{2}}
      \[1cm]
      mbox{Var}pars{L} & =
      bbx{{pi^{2} over 16}pars{1 - {pi^{2} over 16}}theta^{2}}
      end{align}






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
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        newcommand{expo}[1]{,mathrm{e}^{#1},}
        newcommand{ic}{mathrm{i}}
        newcommand{mc}[1]{mathcal{#1}}
        newcommand{mrm}[1]{mathrm{#1}}
        newcommand{pars}[1]{left(,{#1},right)}
        newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
        newcommand{root}[2]{,sqrt[#1]{,{#2},},}
        newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
        newcommand{verts}[1]{leftvert,{#1},rightvert}$
        begin{align}
        mathbb{E}bracks{L} & equiv
        int_{0}^{infty}{expo{-x_{1}/theta} over theta}
        int_{0}^{infty}{expo{-x_{2}/theta} over theta}
        pars{{pi over 4}root{x_{1}x_{2}}}dd x_{1},dd x_{2}
        \[5mm] & =
        {pi over 4}pars{root{theta}
        int_{0}^{infty}expo{-x_{1}/theta}
        ,root{x_{1} over theta},{dd x_{1} over theta}}
        \[2mm] & phantom{===,}
        pars{root{theta}
        int_{0}^{infty}expo{-x_{2}/theta},root{x_{2} over theta}
        ,{dd x_{2} over theta}}
        \[5mm] & =
        {pi over 4}
        underbrace{pars{int_{0}^{infty}x^{1/2}expo{-x}dd x}^{2}}
        _{ds{= Gamma^{2}pars{3/2} = {pi/4}}} theta =
        bbx{{pi^{2} over 16},theta}
        \[1cm]
        mathbb{E}bracks{L^{2}} & equiv
        int_{0}^{infty}{expo{-x_{1}/theta} over theta}
        int_{0}^{infty}{expo{-x_{2}/theta} over theta}
        pars{{pi over 4}root{x_{1}x_{2}}}^{2}dd x_{1},dd x_{2}
        \[5mm] & =
        {pi^{2} over 16}pars{thetaint_{0}^{infty}
        expo{-x_{1}/theta},{x_{1} over theta}
        ,{dd x_{1} over theta}}
        pars{thetaint_{0}^{infty}
        expo{-x_{2}/theta},{x_{1} over theta}
        ,{dd x_{2} over theta}}
        \[5mm] & =
        bbx{{pi^{2} over 16},theta^{2}}
        \[1cm]
        mbox{Var}pars{L} & =
        bbx{{pi^{2} over 16}pars{1 - {pi^{2} over 16}}theta^{2}}
        end{align}






        share|cite|improve this answer









        $endgroup$



        $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
        newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
        newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
        newcommand{dd}{mathrm{d}}
        newcommand{ds}[1]{displaystyle{#1}}
        newcommand{expo}[1]{,mathrm{e}^{#1},}
        newcommand{ic}{mathrm{i}}
        newcommand{mc}[1]{mathcal{#1}}
        newcommand{mrm}[1]{mathrm{#1}}
        newcommand{pars}[1]{left(,{#1},right)}
        newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
        newcommand{root}[2]{,sqrt[#1]{,{#2},},}
        newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
        newcommand{verts}[1]{leftvert,{#1},rightvert}$
        begin{align}
        mathbb{E}bracks{L} & equiv
        int_{0}^{infty}{expo{-x_{1}/theta} over theta}
        int_{0}^{infty}{expo{-x_{2}/theta} over theta}
        pars{{pi over 4}root{x_{1}x_{2}}}dd x_{1},dd x_{2}
        \[5mm] & =
        {pi over 4}pars{root{theta}
        int_{0}^{infty}expo{-x_{1}/theta}
        ,root{x_{1} over theta},{dd x_{1} over theta}}
        \[2mm] & phantom{===,}
        pars{root{theta}
        int_{0}^{infty}expo{-x_{2}/theta},root{x_{2} over theta}
        ,{dd x_{2} over theta}}
        \[5mm] & =
        {pi over 4}
        underbrace{pars{int_{0}^{infty}x^{1/2}expo{-x}dd x}^{2}}
        _{ds{= Gamma^{2}pars{3/2} = {pi/4}}} theta =
        bbx{{pi^{2} over 16},theta}
        \[1cm]
        mathbb{E}bracks{L^{2}} & equiv
        int_{0}^{infty}{expo{-x_{1}/theta} over theta}
        int_{0}^{infty}{expo{-x_{2}/theta} over theta}
        pars{{pi over 4}root{x_{1}x_{2}}}^{2}dd x_{1},dd x_{2}
        \[5mm] & =
        {pi^{2} over 16}pars{thetaint_{0}^{infty}
        expo{-x_{1}/theta},{x_{1} over theta}
        ,{dd x_{1} over theta}}
        pars{thetaint_{0}^{infty}
        expo{-x_{2}/theta},{x_{1} over theta}
        ,{dd x_{2} over theta}}
        \[5mm] & =
        bbx{{pi^{2} over 16},theta^{2}}
        \[1cm]
        mbox{Var}pars{L} & =
        bbx{{pi^{2} over 16}pars{1 - {pi^{2} over 16}}theta^{2}}
        end{align}







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        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 27 '18 at 16:39









        Felix MarinFelix Marin

        67.5k7107141




        67.5k7107141






























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