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














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






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












        $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
        newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
        newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
        newcommand{dd}{mathrm{d}}
        newcommand{ds}[1]{displaystyle{#1}}
        newcommand{expo}[1]{,mathrm{e}^{#1},}
        newcommand{ic}{mathrm{i}}
        newcommand{mc}[1]{mathcal{#1}}
        newcommand{mrm}[1]{mathrm{#1}}
        newcommand{pars}[1]{left(,{#1},right)}
        newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
        newcommand{root}[2]{,sqrt[#1]{,{#2},},}
        newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
        newcommand{verts}[1]{leftvert,{#1},rightvert}$
        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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            Bundesstraße 106

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            Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
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            Verónica Boquete

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            Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
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            Ida-Boy-Ed-Garten

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            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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