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












1














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










  • @jesterll edited.
    – OvermanZarathustra
    Nov 27 at 13:20






  • 3




    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 $E(L^2)-E(L)^2$.
    – yurnero
    Nov 27 at 13:50








  • 1




    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?
    – StubbornAtom
    Nov 27 at 16:14
















1














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




















  • 1




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










  • @jesterll edited.
    – OvermanZarathustra
    Nov 27 at 13:20






  • 3




    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 $E(L^2)-E(L)^2$.
    – yurnero
    Nov 27 at 13:50








  • 1




    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?
    – StubbornAtom
    Nov 27 at 16:14














1












1








1







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















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






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edited Nov 27 at 13:20

























asked Nov 27 at 10:41









OvermanZarathustra

156




156








  • 1




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










  • @jesterll edited.
    – OvermanZarathustra
    Nov 27 at 13:20






  • 3




    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 $E(L^2)-E(L)^2$.
    – yurnero
    Nov 27 at 13:50








  • 1




    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?
    – StubbornAtom
    Nov 27 at 16:14














  • 1




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










  • @jesterll edited.
    – OvermanZarathustra
    Nov 27 at 13:20






  • 3




    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 $E(L^2)-E(L)^2$.
    – yurnero
    Nov 27 at 13:50








  • 1




    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?
    – StubbornAtom
    Nov 27 at 16:14








1




1




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




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












@jesterll edited.
– OvermanZarathustra
Nov 27 at 13:20




@jesterll edited.
– OvermanZarathustra
Nov 27 at 13:20




3




3




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 $E(L^2)-E(L)^2$.
– yurnero
Nov 27 at 13:50






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 $E(L^2)-E(L)^2$.
– yurnero
Nov 27 at 13:50






1




1




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?
– StubbornAtom
Nov 27 at 16:14




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?
– StubbornAtom
Nov 27 at 16:14










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newcommand{mrm}[1]{mathrm{#1}}
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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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    $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
    newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
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    newcommand{ds}[1]{displaystyle{#1}}
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    newcommand{root}[2]{,sqrt[#1]{,{#2},},}
    newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
    newcommand{verts}[1]{leftvert,{#1},rightvert}$
    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


























      1














      $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
      newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
      newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
      newcommand{dd}{mathrm{d}}
      newcommand{ds}[1]{displaystyle{#1}}
      newcommand{expo}[1]{,mathrm{e}^{#1},}
      newcommand{ic}{mathrm{i}}
      newcommand{mc}[1]{mathcal{#1}}
      newcommand{mrm}[1]{mathrm{#1}}
      newcommand{pars}[1]{left(,{#1},right)}
      newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
      newcommand{root}[2]{,sqrt[#1]{,{#2},},}
      newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
      newcommand{verts}[1]{leftvert,{#1},rightvert}$
      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
























        1












        1








        1






        $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
        newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
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        newcommand{mrm}[1]{mathrm{#1}}
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        newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
        newcommand{root}[2]{,sqrt[#1]{,{#2},},}
        newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
        newcommand{verts}[1]{leftvert,{#1},rightvert}$
        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












        $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
        newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
        newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
        newcommand{dd}{mathrm{d}}
        newcommand{ds}[1]{displaystyle{#1}}
        newcommand{expo}[1]{,mathrm{e}^{#1},}
        newcommand{ic}{mathrm{i}}
        newcommand{mc}[1]{mathcal{#1}}
        newcommand{mrm}[1]{mathrm{#1}}
        newcommand{pars}[1]{left(,{#1},right)}
        newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
        newcommand{root}[2]{,sqrt[#1]{,{#2},},}
        newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
        newcommand{verts}[1]{leftvert,{#1},rightvert}$
        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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        answered Nov 27 at 16:39









        Felix Marin

        67.1k7107141




        67.1k7107141






























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            up vote 4 down vote favorite 1 Some time ago, I set up skydrive for my wife on her mac. But now she wants to access the files online, I cannot remember what Microsoft account I setup and linked her skydive app to and nor can I see anything in the skydrive app settings to tell me either? How can I see the name of the Microsoft account the skydive app is currently linked to? I would like to know this for skydrive app for Mac and PC onedrive share | improve this question asked Feb 25 '13 at 4:52 Gary Barrett 260 2 5 12
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            Grease: Live!

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            A tradução deste artigo está abaixo da qualidade média aceitável. É possível que tenha sido feita por um tradutor automático ou por alguém que não conhece bem o português ou a língua original do texto. Caso queira colaborar com a Wikipédia, tente encontrar a página original e melhore este verbete conforme o guia de tradução. Grease: Live! Grease: Live!  (PRT/BRA)  Estados Unidos 2016 •   cor •   130 min. min   Direção Thomas Kail Alex Rudzinski Produção Marc Platt Adam Siegel Roteiro Allan Carr Robert Cary Warren Casey Jim Jacobs Jonathan Tolins Bronte Woodard Baseado em Grease (musical) Grease (filme) Elenco Julianne Hough Aaron Tveit Vanessa Hudgens Carlos Pena Jr. Keke Palmer Carly Rae Jepsen Kether Donohue Jordan Fisher David Del Rio Andrew Call Gênero Musical Música Tom Kitt Figurino William Ivey Long Companhia(s) produtora(s) Warner Bros. Studios Distribuição Fox Lançamento 31 de Janei
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            When does type information flow backwards in C++?

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            up vote 62 down vote favorite 19 I just watched Stephan T. Lavavej talk at CppCon 2018 on "Class Template Argument Deduction", where at some point he incidentally says: In C++ type information almost never flows backwards ... I had to say "almost" because there's one or two cases, possibly more but very few . Despite trying to figure out which cases he might be referring to, I couldn't come up with anything. Hence the question: In which cases the C++17 standard mandates that type information propagate backwards? c++ language-lawyer c++17 share | improve this question edited 7 hours ago scohe0
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