Kendall's Tau of bivariate normal












3












$begingroup$


Prove Kendall's tau of a bivariate normal is given by



$$rho_tau (X_1,X_2)=frac{2}{pi}arcsinrho$$



I can derive the bivariate normal as



$$F(x_1,x_2)=frac{1}{2pisqrt{1-rho^2}}e^{-frac{1}{2}x_1Sigma^{-1}x_2}$$



where, for simplicity, I am assuming $X_1$ and $X_2$ have mean $0$. I also am aware that the formula for Kendall's tau involves an integral over some domain involving $X_1$, $X_2$, and $rho$, multiplied by $4$.



So it seems like all the pieces are there; the integral should result in the answer, if only the integral over the exponential term w.r.t. $X_1$ and $X_2$ were $1$. But this is not the case, so I am stuck.






probability integration statistics correlation







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$endgroup$












  • $begingroup$
    See here
    $endgroup$
    – mrtaurho
    Jan 1 at 21:16










  • $begingroup$
    I saw this but, as @Avraham writes, access is not given to the necessary papers for a rigorous proof.
    $endgroup$
    – E Werner
    Jan 1 at 21:29






  • 1




    $begingroup$
    Anyway just stating a task is not appropriate for MSE. Therefore please include some more details with an edit.
    $endgroup$
    – mrtaurho
    Jan 1 at 21:41










  • $begingroup$
    At least write down the expression for Kendall's tau and indicate where you are stuck.
    $endgroup$
    – StubbornAtom
    Jan 2 at 14:35
















3












$begingroup$


Prove Kendall's tau of a bivariate normal is given by



$$rho_tau (X_1,X_2)=frac{2}{pi}arcsinrho$$



I can derive the bivariate normal as



$$F(x_1,x_2)=frac{1}{2pisqrt{1-rho^2}}e^{-frac{1}{2}x_1Sigma^{-1}x_2}$$



where, for simplicity, I am assuming $X_1$ and $X_2$ have mean $0$. I also am aware that the formula for Kendall's tau involves an integral over some domain involving $X_1$, $X_2$, and $rho$, multiplied by $4$.



So it seems like all the pieces are there; the integral should result in the answer, if only the integral over the exponential term w.r.t. $X_1$ and $X_2$ were $1$. But this is not the case, so I am stuck.






probability integration statistics correlation







share|cite|improve this question











$endgroup$












  • $begingroup$
    See here
    $endgroup$
    – mrtaurho
    Jan 1 at 21:16










  • $begingroup$
    I saw this but, as @Avraham writes, access is not given to the necessary papers for a rigorous proof.
    $endgroup$
    – E Werner
    Jan 1 at 21:29






  • 1




    $begingroup$
    Anyway just stating a task is not appropriate for MSE. Therefore please include some more details with an edit.
    $endgroup$
    – mrtaurho
    Jan 1 at 21:41










  • $begingroup$
    At least write down the expression for Kendall's tau and indicate where you are stuck.
    $endgroup$
    – StubbornAtom
    Jan 2 at 14:35














3












3








3





$begingroup$


Prove Kendall's tau of a bivariate normal is given by



$$rho_tau (X_1,X_2)=frac{2}{pi}arcsinrho$$



I can derive the bivariate normal as



$$F(x_1,x_2)=frac{1}{2pisqrt{1-rho^2}}e^{-frac{1}{2}x_1Sigma^{-1}x_2}$$



where, for simplicity, I am assuming $X_1$ and $X_2$ have mean $0$. I also am aware that the formula for Kendall's tau involves an integral over some domain involving $X_1$, $X_2$, and $rho$, multiplied by $4$.



So it seems like all the pieces are there; the integral should result in the answer, if only the integral over the exponential term w.r.t. $X_1$ and $X_2$ were $1$. But this is not the case, so I am stuck.






probability integration statistics correlation







share|cite|improve this question











$endgroup$




Prove Kendall's tau of a bivariate normal is given by



$$rho_tau (X_1,X_2)=frac{2}{pi}arcsinrho$$



I can derive the bivariate normal as



$$F(x_1,x_2)=frac{1}{2pisqrt{1-rho^2}}e^{-frac{1}{2}x_1Sigma^{-1}x_2}$$



where, for simplicity, I am assuming $X_1$ and $X_2$ have mean $0$. I also am aware that the formula for Kendall's tau involves an integral over some domain involving $X_1$, $X_2$, and $rho$, multiplied by $4$.



So it seems like all the pieces are there; the integral should result in the answer, if only the integral over the exponential term w.r.t. $X_1$ and $X_2$ were $1$. But this is not the case, so I am stuck.






probability integration statistics correlation




probability integration statistics correlation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 23:04









mrtaurho

6,09271641




6,09271641










asked Jan 1 at 21:13









E WernerE Werner

605




605












  • $begingroup$
    See here
    $endgroup$
    – mrtaurho
    Jan 1 at 21:16










  • $begingroup$
    I saw this but, as @Avraham writes, access is not given to the necessary papers for a rigorous proof.
    $endgroup$
    – E Werner
    Jan 1 at 21:29






  • 1




    $begingroup$
    Anyway just stating a task is not appropriate for MSE. Therefore please include some more details with an edit.
    $endgroup$
    – mrtaurho
    Jan 1 at 21:41










  • $begingroup$
    At least write down the expression for Kendall's tau and indicate where you are stuck.
    $endgroup$
    – StubbornAtom
    Jan 2 at 14:35


















  • $begingroup$
    See here
    $endgroup$
    – mrtaurho
    Jan 1 at 21:16










  • $begingroup$
    I saw this but, as @Avraham writes, access is not given to the necessary papers for a rigorous proof.
    $endgroup$
    – E Werner
    Jan 1 at 21:29






  • 1




    $begingroup$
    Anyway just stating a task is not appropriate for MSE. Therefore please include some more details with an edit.
    $endgroup$
    – mrtaurho
    Jan 1 at 21:41










  • $begingroup$
    At least write down the expression for Kendall's tau and indicate where you are stuck.
    $endgroup$
    – StubbornAtom
    Jan 2 at 14:35
















$begingroup$
See here
$endgroup$
– mrtaurho
Jan 1 at 21:16




$begingroup$
See here
$endgroup$
– mrtaurho
Jan 1 at 21:16












$begingroup$
I saw this but, as @Avraham writes, access is not given to the necessary papers for a rigorous proof.
$endgroup$
– E Werner
Jan 1 at 21:29




$begingroup$
I saw this but, as @Avraham writes, access is not given to the necessary papers for a rigorous proof.
$endgroup$
– E Werner
Jan 1 at 21:29




1




1




$begingroup$
Anyway just stating a task is not appropriate for MSE. Therefore please include some more details with an edit.
$endgroup$
– mrtaurho
Jan 1 at 21:41




$begingroup$
Anyway just stating a task is not appropriate for MSE. Therefore please include some more details with an edit.
$endgroup$
– mrtaurho
Jan 1 at 21:41












$begingroup$
At least write down the expression for Kendall's tau and indicate where you are stuck.
$endgroup$
– StubbornAtom
Jan 2 at 14:35




$begingroup$
At least write down the expression for Kendall's tau and indicate where you are stuck.
$endgroup$
– StubbornAtom
Jan 2 at 14:35










1 Answer
1






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oldest

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1












$begingroup$

Kendall's $tau$ is defined as
$$
tau(X_1,X_2)=mathsf{P}((X_1-Y_1)(X_2-Y_2)>0)-mathsf{P}((X_1-Y_1)(X_2-Y_2)<0),
$$
where $Yequiv(Y_1,Y_2)$ is an independent copy of $Xequiv(X_1,X_2)$.




Let $Z:=X-Y$ and note that $Zsim N(0,2Sigma)$, where $Sigma=operatorname {Var}(X)$. In addition,
$$
Zoverset{d}{sim}sqrt{2}(Sigma_{11}(V_1cos(varphi)+V_2sin(varphi)),Sigma_{22}V_2),
$$
where $varphiequiv arcsinrho$ and $Vequiv(V_1,V_2)sim N(0,I_2)$. Then, by symmetry,
begin{align}
tau(X_1,X_2)&=2mathsf{P}(Z_1Z_2>0)-1=4mathsf{P}(Z_1>0,Z_2>0)-1 \
&=4mathsf{P}(V_1cos(varphi)+V_2sin(varphi)>0,V_2>0)-1.
end{align}



It is known that $Voverset{d}{=}R(cos(Phi),sin(Phi))$, where $Phisim U[-pi,pi]$ independent of $R=|V|_2$. Therefore,
begin{align}
tau(X_1,X_2)&=4mathsf{P}(cos(Phi)cos(varphi)+sin(Phi)sin(varphi)>0,sin(Phi)>0)-1 \
&=4mathsf{P}(Phiin (varphi-pi/2,varphi+pi/2)cap[0,pi])-1 \
&=frac{2}{pi}(varphi+pi/2)-1=frac{2}{pi}arcsin rho.
end{align}






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    1












    $begingroup$

    Kendall's $tau$ is defined as
    $$
    tau(X_1,X_2)=mathsf{P}((X_1-Y_1)(X_2-Y_2)>0)-mathsf{P}((X_1-Y_1)(X_2-Y_2)<0),
    $$
    where $Yequiv(Y_1,Y_2)$ is an independent copy of $Xequiv(X_1,X_2)$.




    Let $Z:=X-Y$ and note that $Zsim N(0,2Sigma)$, where $Sigma=operatorname {Var}(X)$. In addition,
    $$
    Zoverset{d}{sim}sqrt{2}(Sigma_{11}(V_1cos(varphi)+V_2sin(varphi)),Sigma_{22}V_2),
    $$
    where $varphiequiv arcsinrho$ and $Vequiv(V_1,V_2)sim N(0,I_2)$. Then, by symmetry,
    begin{align}
    tau(X_1,X_2)&=2mathsf{P}(Z_1Z_2>0)-1=4mathsf{P}(Z_1>0,Z_2>0)-1 \
    &=4mathsf{P}(V_1cos(varphi)+V_2sin(varphi)>0,V_2>0)-1.
    end{align}



    It is known that $Voverset{d}{=}R(cos(Phi),sin(Phi))$, where $Phisim U[-pi,pi]$ independent of $R=|V|_2$. Therefore,
    begin{align}
    tau(X_1,X_2)&=4mathsf{P}(cos(Phi)cos(varphi)+sin(Phi)sin(varphi)>0,sin(Phi)>0)-1 \
    &=4mathsf{P}(Phiin (varphi-pi/2,varphi+pi/2)cap[0,pi])-1 \
    &=frac{2}{pi}(varphi+pi/2)-1=frac{2}{pi}arcsin rho.
    end{align}






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Kendall's $tau$ is defined as
      $$
      tau(X_1,X_2)=mathsf{P}((X_1-Y_1)(X_2-Y_2)>0)-mathsf{P}((X_1-Y_1)(X_2-Y_2)<0),
      $$
      where $Yequiv(Y_1,Y_2)$ is an independent copy of $Xequiv(X_1,X_2)$.




      Let $Z:=X-Y$ and note that $Zsim N(0,2Sigma)$, where $Sigma=operatorname {Var}(X)$. In addition,
      $$
      Zoverset{d}{sim}sqrt{2}(Sigma_{11}(V_1cos(varphi)+V_2sin(varphi)),Sigma_{22}V_2),
      $$
      where $varphiequiv arcsinrho$ and $Vequiv(V_1,V_2)sim N(0,I_2)$. Then, by symmetry,
      begin{align}
      tau(X_1,X_2)&=2mathsf{P}(Z_1Z_2>0)-1=4mathsf{P}(Z_1>0,Z_2>0)-1 \
      &=4mathsf{P}(V_1cos(varphi)+V_2sin(varphi)>0,V_2>0)-1.
      end{align}



      It is known that $Voverset{d}{=}R(cos(Phi),sin(Phi))$, where $Phisim U[-pi,pi]$ independent of $R=|V|_2$. Therefore,
      begin{align}
      tau(X_1,X_2)&=4mathsf{P}(cos(Phi)cos(varphi)+sin(Phi)sin(varphi)>0,sin(Phi)>0)-1 \
      &=4mathsf{P}(Phiin (varphi-pi/2,varphi+pi/2)cap[0,pi])-1 \
      &=frac{2}{pi}(varphi+pi/2)-1=frac{2}{pi}arcsin rho.
      end{align}






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Kendall's $tau$ is defined as
        $$
        tau(X_1,X_2)=mathsf{P}((X_1-Y_1)(X_2-Y_2)>0)-mathsf{P}((X_1-Y_1)(X_2-Y_2)<0),
        $$
        where $Yequiv(Y_1,Y_2)$ is an independent copy of $Xequiv(X_1,X_2)$.




        Let $Z:=X-Y$ and note that $Zsim N(0,2Sigma)$, where $Sigma=operatorname {Var}(X)$. In addition,
        $$
        Zoverset{d}{sim}sqrt{2}(Sigma_{11}(V_1cos(varphi)+V_2sin(varphi)),Sigma_{22}V_2),
        $$
        where $varphiequiv arcsinrho$ and $Vequiv(V_1,V_2)sim N(0,I_2)$. Then, by symmetry,
        begin{align}
        tau(X_1,X_2)&=2mathsf{P}(Z_1Z_2>0)-1=4mathsf{P}(Z_1>0,Z_2>0)-1 \
        &=4mathsf{P}(V_1cos(varphi)+V_2sin(varphi)>0,V_2>0)-1.
        end{align}



        It is known that $Voverset{d}{=}R(cos(Phi),sin(Phi))$, where $Phisim U[-pi,pi]$ independent of $R=|V|_2$. Therefore,
        begin{align}
        tau(X_1,X_2)&=4mathsf{P}(cos(Phi)cos(varphi)+sin(Phi)sin(varphi)>0,sin(Phi)>0)-1 \
        &=4mathsf{P}(Phiin (varphi-pi/2,varphi+pi/2)cap[0,pi])-1 \
        &=frac{2}{pi}(varphi+pi/2)-1=frac{2}{pi}arcsin rho.
        end{align}






        share|cite|improve this answer









        $endgroup$



        Kendall's $tau$ is defined as
        $$
        tau(X_1,X_2)=mathsf{P}((X_1-Y_1)(X_2-Y_2)>0)-mathsf{P}((X_1-Y_1)(X_2-Y_2)<0),
        $$
        where $Yequiv(Y_1,Y_2)$ is an independent copy of $Xequiv(X_1,X_2)$.




        Let $Z:=X-Y$ and note that $Zsim N(0,2Sigma)$, where $Sigma=operatorname {Var}(X)$. In addition,
        $$
        Zoverset{d}{sim}sqrt{2}(Sigma_{11}(V_1cos(varphi)+V_2sin(varphi)),Sigma_{22}V_2),
        $$
        where $varphiequiv arcsinrho$ and $Vequiv(V_1,V_2)sim N(0,I_2)$. Then, by symmetry,
        begin{align}
        tau(X_1,X_2)&=2mathsf{P}(Z_1Z_2>0)-1=4mathsf{P}(Z_1>0,Z_2>0)-1 \
        &=4mathsf{P}(V_1cos(varphi)+V_2sin(varphi)>0,V_2>0)-1.
        end{align}



        It is known that $Voverset{d}{=}R(cos(Phi),sin(Phi))$, where $Phisim U[-pi,pi]$ independent of $R=|V|_2$. Therefore,
        begin{align}
        tau(X_1,X_2)&=4mathsf{P}(cos(Phi)cos(varphi)+sin(Phi)sin(varphi)>0,sin(Phi)>0)-1 \
        &=4mathsf{P}(Phiin (varphi-pi/2,varphi+pi/2)cap[0,pi])-1 \
        &=frac{2}{pi}(varphi+pi/2)-1=frac{2}{pi}arcsin rho.
        end{align}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 2 at 2:02









        d.k.o.d.k.o.

        10.5k630




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