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




    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















up vote
1
down vote

favorite












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













up vote
1
down vote

favorite









up vote
1
down vote

favorite











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










1 Answer
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$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
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newcommand{expo}[1]{,mathrm{e}^{#1},}
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newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
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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    up vote
    1
    down vote



    accepted










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

























      up vote
      1
      down vote



      accepted










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























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
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        newcommand{mrm}[1]{mathrm{#1}}
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        newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
        newcommand{root}[2]{,sqrt[#1]{,{#2},},}
        newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
        newcommand{verts}[1]{leftvert,{#1},rightvert}$
        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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        share|cite|improve this answer



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        answered Nov 27 at 16:39









        Felix Marin

        66.7k7107139




        66.7k7107139






























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            Berounka

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            Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc...
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            Sphinx de Gizeh

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            Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
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            Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

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            3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
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