Finding a distribution with a given correlation












1












$begingroup$


Below is a problem that I made up and my attempt at a solution to it. I am hoping that somebody here can help me finish it. I believe there is a unique
answer to the problem.

Thanks,

Bob



Problem:

Let $X$ and $Y$ be uniformly distributed independent variables on the interval $(-1,1)$. Let $K$ be
a real number. Let $Z = Y + KX$ such that the correlation of $X$ and $Z$ is $frac{1}{2}$. Find $K$.

Answer:

If we select $K = 0$ then we get a correlation of $0$. If we select $K$ to be a very large number
then the correlation will be close to $1$.
begin{align*}
rho &= frac{1}{2} \
u_x &= 0 \
u_y &= 0 \
u_z &= u_y + K(u_y) = 0 + K(0) = 0 \
sigma_x^2 &= frac{(1 - -1)^2}{12} = frac{4}{12} \
sigma_x^2 &= frac{1}{3} \
sigma_x &= frac{1}{sqrt{3}} \
sigma_y &= frac{1}{sqrt{3}} \
sigma^2_z &= sigma^2_y + K^2 sigma_x^2 + K(0) \
sigma^2_z &= frac{1}{3} + K^2 left( frac{1}{3} right) \
rho &= frac{sigma_x sigma_z}{sigma_{xz}} \
sigma_{xz} &= frac{sigma_x sigma_z}{rho} =
frac{ left( frac{1}{sqrt{3}} right) left( frac{1}{3} + K^2 left( frac{1}{3} right) right) }{frac{1}{2} }\
sigma_{xz} &= left( frac{2}{sqrt{3}} right) left( frac{1}{3} + K^2 left( frac{1}{3} right) right) \
sigma_{xz} &= left( frac{2}{3 sqrt{3}} right) left( K^2 + 1 right) \
end{align*}










share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Below is a problem that I made up and my attempt at a solution to it. I am hoping that somebody here can help me finish it. I believe there is a unique
    answer to the problem.

    Thanks,

    Bob



    Problem:

    Let $X$ and $Y$ be uniformly distributed independent variables on the interval $(-1,1)$. Let $K$ be
    a real number. Let $Z = Y + KX$ such that the correlation of $X$ and $Z$ is $frac{1}{2}$. Find $K$.

    Answer:

    If we select $K = 0$ then we get a correlation of $0$. If we select $K$ to be a very large number
    then the correlation will be close to $1$.
    begin{align*}
    rho &= frac{1}{2} \
    u_x &= 0 \
    u_y &= 0 \
    u_z &= u_y + K(u_y) = 0 + K(0) = 0 \
    sigma_x^2 &= frac{(1 - -1)^2}{12} = frac{4}{12} \
    sigma_x^2 &= frac{1}{3} \
    sigma_x &= frac{1}{sqrt{3}} \
    sigma_y &= frac{1}{sqrt{3}} \
    sigma^2_z &= sigma^2_y + K^2 sigma_x^2 + K(0) \
    sigma^2_z &= frac{1}{3} + K^2 left( frac{1}{3} right) \
    rho &= frac{sigma_x sigma_z}{sigma_{xz}} \
    sigma_{xz} &= frac{sigma_x sigma_z}{rho} =
    frac{ left( frac{1}{sqrt{3}} right) left( frac{1}{3} + K^2 left( frac{1}{3} right) right) }{frac{1}{2} }\
    sigma_{xz} &= left( frac{2}{sqrt{3}} right) left( frac{1}{3} + K^2 left( frac{1}{3} right) right) \
    sigma_{xz} &= left( frac{2}{3 sqrt{3}} right) left( K^2 + 1 right) \
    end{align*}










    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Below is a problem that I made up and my attempt at a solution to it. I am hoping that somebody here can help me finish it. I believe there is a unique
      answer to the problem.

      Thanks,

      Bob



      Problem:

      Let $X$ and $Y$ be uniformly distributed independent variables on the interval $(-1,1)$. Let $K$ be
      a real number. Let $Z = Y + KX$ such that the correlation of $X$ and $Z$ is $frac{1}{2}$. Find $K$.

      Answer:

      If we select $K = 0$ then we get a correlation of $0$. If we select $K$ to be a very large number
      then the correlation will be close to $1$.
      begin{align*}
      rho &= frac{1}{2} \
      u_x &= 0 \
      u_y &= 0 \
      u_z &= u_y + K(u_y) = 0 + K(0) = 0 \
      sigma_x^2 &= frac{(1 - -1)^2}{12} = frac{4}{12} \
      sigma_x^2 &= frac{1}{3} \
      sigma_x &= frac{1}{sqrt{3}} \
      sigma_y &= frac{1}{sqrt{3}} \
      sigma^2_z &= sigma^2_y + K^2 sigma_x^2 + K(0) \
      sigma^2_z &= frac{1}{3} + K^2 left( frac{1}{3} right) \
      rho &= frac{sigma_x sigma_z}{sigma_{xz}} \
      sigma_{xz} &= frac{sigma_x sigma_z}{rho} =
      frac{ left( frac{1}{sqrt{3}} right) left( frac{1}{3} + K^2 left( frac{1}{3} right) right) }{frac{1}{2} }\
      sigma_{xz} &= left( frac{2}{sqrt{3}} right) left( frac{1}{3} + K^2 left( frac{1}{3} right) right) \
      sigma_{xz} &= left( frac{2}{3 sqrt{3}} right) left( K^2 + 1 right) \
      end{align*}










      share|cite|improve this question









      $endgroup$




      Below is a problem that I made up and my attempt at a solution to it. I am hoping that somebody here can help me finish it. I believe there is a unique
      answer to the problem.

      Thanks,

      Bob



      Problem:

      Let $X$ and $Y$ be uniformly distributed independent variables on the interval $(-1,1)$. Let $K$ be
      a real number. Let $Z = Y + KX$ such that the correlation of $X$ and $Z$ is $frac{1}{2}$. Find $K$.

      Answer:

      If we select $K = 0$ then we get a correlation of $0$. If we select $K$ to be a very large number
      then the correlation will be close to $1$.
      begin{align*}
      rho &= frac{1}{2} \
      u_x &= 0 \
      u_y &= 0 \
      u_z &= u_y + K(u_y) = 0 + K(0) = 0 \
      sigma_x^2 &= frac{(1 - -1)^2}{12} = frac{4}{12} \
      sigma_x^2 &= frac{1}{3} \
      sigma_x &= frac{1}{sqrt{3}} \
      sigma_y &= frac{1}{sqrt{3}} \
      sigma^2_z &= sigma^2_y + K^2 sigma_x^2 + K(0) \
      sigma^2_z &= frac{1}{3} + K^2 left( frac{1}{3} right) \
      rho &= frac{sigma_x sigma_z}{sigma_{xz}} \
      sigma_{xz} &= frac{sigma_x sigma_z}{rho} =
      frac{ left( frac{1}{sqrt{3}} right) left( frac{1}{3} + K^2 left( frac{1}{3} right) right) }{frac{1}{2} }\
      sigma_{xz} &= left( frac{2}{sqrt{3}} right) left( frac{1}{3} + K^2 left( frac{1}{3} right) right) \
      sigma_{xz} &= left( frac{2}{3 sqrt{3}} right) left( K^2 + 1 right) \
      end{align*}







      probability statistics






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      asked Dec 13 '18 at 23:20









      BobBob

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          $begingroup$

          As you show, Pearson's $r$ should be a function of $k$. We will find that function and solve for
          $r(k) = frac{1}{2}$.



          $r(A,B)$ may be calculated from the covariance $cov(A,B)$ and the standard deviations $sigma_A$ and $sigma_B$ as $frac{cov(A,B)} { sigma_A sigma_B}$.



          The covariance itself may be calculated as $E(AB) - E(A)E(B)$.



          With $A$ and $B$ independent and each symmetric around $0$, the second term is $0$.



          So with $Z = KX + Y$, we have $cov(X,Z) = E[X(kX+Y)] = E(KX^2) + E(XY)$.



          Again, with $X$ and $Y$ independent and each symmetric around $0$, the second term is $0$.



          $$ cov (X,Z) = E(KX^2) = KE(X^2) $$



          The second moment $E(X^2)$ of a continuous uniform distribution $(a^2 + ab + b^2)/3$ where $a$ and $b$
          are the bounds of the distribution. In this case, $(-1)^2 + (-1)*1 + 1^2 = 1 -1 + 1 = 1$.



          $$ cov(X,Z) = K frac{1}{3} = frac{K}{3}$$





          The variance $V(X)$ of a uniform distribution is $(b-a)^2/12$.
          begin{align*}
          V(X) &= frac{[1-(-1)]^2}{ 12 } = frac{2^2}{12} = frac{1}{3} \
          V(Z) &= V(KX+Y) = K^2*V(X)*V(Y) = K^2V(X) + V(X) = (K^2+1)V(X) \
          V(Z) &= frac{(K^2+1)}{3} \
          V(X)V(Z) &= frac{K^2+1}{9} \
          sigma(X)sigma(Z) &= sqrt{ frac{K^2+1}{9} } = frac{ sqrt{K^2+1} }{ 3 }\
          r(X,Z) &= frac{ frac{K}{3}}{ frac{sqrt{K^2+1}}{3}} = frac{K}{ sqrt{K^2+1)} } \
          end{align*}





          So for $r = frac{1}{2}$, $frac{K}{sqrt{K^2+1}} = frac{1}{2}$.



          $$ 2K = sqrt{K^2+1} $$



          We will square both sides, which will give two solutions for $K$, only one of which will be relevant.



          begin{align*}
          4K^2 &= K^2 + 1 \
          3K^2 - 1 &= 0 \
          end{align*}



          using $a^2 - b^2 = (a+b)(a-b)$, we see $(sqrt{3}K +1)(sqrt{3}K - 1) = 0$.



          $$ sqrt{3}K = 1 text{ OR } -1$$



          Since we can see in the original problem K must be positive to yield a positive correlation,
          $sqrt{3}K = 1$.



          $$ k = frac{1} {sqrt{3}} $$






          share|cite|improve this answer











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            $begingroup$

            As you show, Pearson's $r$ should be a function of $k$. We will find that function and solve for
            $r(k) = frac{1}{2}$.



            $r(A,B)$ may be calculated from the covariance $cov(A,B)$ and the standard deviations $sigma_A$ and $sigma_B$ as $frac{cov(A,B)} { sigma_A sigma_B}$.



            The covariance itself may be calculated as $E(AB) - E(A)E(B)$.



            With $A$ and $B$ independent and each symmetric around $0$, the second term is $0$.



            So with $Z = KX + Y$, we have $cov(X,Z) = E[X(kX+Y)] = E(KX^2) + E(XY)$.



            Again, with $X$ and $Y$ independent and each symmetric around $0$, the second term is $0$.



            $$ cov (X,Z) = E(KX^2) = KE(X^2) $$



            The second moment $E(X^2)$ of a continuous uniform distribution $(a^2 + ab + b^2)/3$ where $a$ and $b$
            are the bounds of the distribution. In this case, $(-1)^2 + (-1)*1 + 1^2 = 1 -1 + 1 = 1$.



            $$ cov(X,Z) = K frac{1}{3} = frac{K}{3}$$





            The variance $V(X)$ of a uniform distribution is $(b-a)^2/12$.
            begin{align*}
            V(X) &= frac{[1-(-1)]^2}{ 12 } = frac{2^2}{12} = frac{1}{3} \
            V(Z) &= V(KX+Y) = K^2*V(X)*V(Y) = K^2V(X) + V(X) = (K^2+1)V(X) \
            V(Z) &= frac{(K^2+1)}{3} \
            V(X)V(Z) &= frac{K^2+1}{9} \
            sigma(X)sigma(Z) &= sqrt{ frac{K^2+1}{9} } = frac{ sqrt{K^2+1} }{ 3 }\
            r(X,Z) &= frac{ frac{K}{3}}{ frac{sqrt{K^2+1}}{3}} = frac{K}{ sqrt{K^2+1)} } \
            end{align*}





            So for $r = frac{1}{2}$, $frac{K}{sqrt{K^2+1}} = frac{1}{2}$.



            $$ 2K = sqrt{K^2+1} $$



            We will square both sides, which will give two solutions for $K$, only one of which will be relevant.



            begin{align*}
            4K^2 &= K^2 + 1 \
            3K^2 - 1 &= 0 \
            end{align*}



            using $a^2 - b^2 = (a+b)(a-b)$, we see $(sqrt{3}K +1)(sqrt{3}K - 1) = 0$.



            $$ sqrt{3}K = 1 text{ OR } -1$$



            Since we can see in the original problem K must be positive to yield a positive correlation,
            $sqrt{3}K = 1$.



            $$ k = frac{1} {sqrt{3}} $$






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              As you show, Pearson's $r$ should be a function of $k$. We will find that function and solve for
              $r(k) = frac{1}{2}$.



              $r(A,B)$ may be calculated from the covariance $cov(A,B)$ and the standard deviations $sigma_A$ and $sigma_B$ as $frac{cov(A,B)} { sigma_A sigma_B}$.



              The covariance itself may be calculated as $E(AB) - E(A)E(B)$.



              With $A$ and $B$ independent and each symmetric around $0$, the second term is $0$.



              So with $Z = KX + Y$, we have $cov(X,Z) = E[X(kX+Y)] = E(KX^2) + E(XY)$.



              Again, with $X$ and $Y$ independent and each symmetric around $0$, the second term is $0$.



              $$ cov (X,Z) = E(KX^2) = KE(X^2) $$



              The second moment $E(X^2)$ of a continuous uniform distribution $(a^2 + ab + b^2)/3$ where $a$ and $b$
              are the bounds of the distribution. In this case, $(-1)^2 + (-1)*1 + 1^2 = 1 -1 + 1 = 1$.



              $$ cov(X,Z) = K frac{1}{3} = frac{K}{3}$$





              The variance $V(X)$ of a uniform distribution is $(b-a)^2/12$.
              begin{align*}
              V(X) &= frac{[1-(-1)]^2}{ 12 } = frac{2^2}{12} = frac{1}{3} \
              V(Z) &= V(KX+Y) = K^2*V(X)*V(Y) = K^2V(X) + V(X) = (K^2+1)V(X) \
              V(Z) &= frac{(K^2+1)}{3} \
              V(X)V(Z) &= frac{K^2+1}{9} \
              sigma(X)sigma(Z) &= sqrt{ frac{K^2+1}{9} } = frac{ sqrt{K^2+1} }{ 3 }\
              r(X,Z) &= frac{ frac{K}{3}}{ frac{sqrt{K^2+1}}{3}} = frac{K}{ sqrt{K^2+1)} } \
              end{align*}





              So for $r = frac{1}{2}$, $frac{K}{sqrt{K^2+1}} = frac{1}{2}$.



              $$ 2K = sqrt{K^2+1} $$



              We will square both sides, which will give two solutions for $K$, only one of which will be relevant.



              begin{align*}
              4K^2 &= K^2 + 1 \
              3K^2 - 1 &= 0 \
              end{align*}



              using $a^2 - b^2 = (a+b)(a-b)$, we see $(sqrt{3}K +1)(sqrt{3}K - 1) = 0$.



              $$ sqrt{3}K = 1 text{ OR } -1$$



              Since we can see in the original problem K must be positive to yield a positive correlation,
              $sqrt{3}K = 1$.



              $$ k = frac{1} {sqrt{3}} $$






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                As you show, Pearson's $r$ should be a function of $k$. We will find that function and solve for
                $r(k) = frac{1}{2}$.



                $r(A,B)$ may be calculated from the covariance $cov(A,B)$ and the standard deviations $sigma_A$ and $sigma_B$ as $frac{cov(A,B)} { sigma_A sigma_B}$.



                The covariance itself may be calculated as $E(AB) - E(A)E(B)$.



                With $A$ and $B$ independent and each symmetric around $0$, the second term is $0$.



                So with $Z = KX + Y$, we have $cov(X,Z) = E[X(kX+Y)] = E(KX^2) + E(XY)$.



                Again, with $X$ and $Y$ independent and each symmetric around $0$, the second term is $0$.



                $$ cov (X,Z) = E(KX^2) = KE(X^2) $$



                The second moment $E(X^2)$ of a continuous uniform distribution $(a^2 + ab + b^2)/3$ where $a$ and $b$
                are the bounds of the distribution. In this case, $(-1)^2 + (-1)*1 + 1^2 = 1 -1 + 1 = 1$.



                $$ cov(X,Z) = K frac{1}{3} = frac{K}{3}$$





                The variance $V(X)$ of a uniform distribution is $(b-a)^2/12$.
                begin{align*}
                V(X) &= frac{[1-(-1)]^2}{ 12 } = frac{2^2}{12} = frac{1}{3} \
                V(Z) &= V(KX+Y) = K^2*V(X)*V(Y) = K^2V(X) + V(X) = (K^2+1)V(X) \
                V(Z) &= frac{(K^2+1)}{3} \
                V(X)V(Z) &= frac{K^2+1}{9} \
                sigma(X)sigma(Z) &= sqrt{ frac{K^2+1}{9} } = frac{ sqrt{K^2+1} }{ 3 }\
                r(X,Z) &= frac{ frac{K}{3}}{ frac{sqrt{K^2+1}}{3}} = frac{K}{ sqrt{K^2+1)} } \
                end{align*}





                So for $r = frac{1}{2}$, $frac{K}{sqrt{K^2+1}} = frac{1}{2}$.



                $$ 2K = sqrt{K^2+1} $$



                We will square both sides, which will give two solutions for $K$, only one of which will be relevant.



                begin{align*}
                4K^2 &= K^2 + 1 \
                3K^2 - 1 &= 0 \
                end{align*}



                using $a^2 - b^2 = (a+b)(a-b)$, we see $(sqrt{3}K +1)(sqrt{3}K - 1) = 0$.



                $$ sqrt{3}K = 1 text{ OR } -1$$



                Since we can see in the original problem K must be positive to yield a positive correlation,
                $sqrt{3}K = 1$.



                $$ k = frac{1} {sqrt{3}} $$






                share|cite|improve this answer











                $endgroup$



                As you show, Pearson's $r$ should be a function of $k$. We will find that function and solve for
                $r(k) = frac{1}{2}$.



                $r(A,B)$ may be calculated from the covariance $cov(A,B)$ and the standard deviations $sigma_A$ and $sigma_B$ as $frac{cov(A,B)} { sigma_A sigma_B}$.



                The covariance itself may be calculated as $E(AB) - E(A)E(B)$.



                With $A$ and $B$ independent and each symmetric around $0$, the second term is $0$.



                So with $Z = KX + Y$, we have $cov(X,Z) = E[X(kX+Y)] = E(KX^2) + E(XY)$.



                Again, with $X$ and $Y$ independent and each symmetric around $0$, the second term is $0$.



                $$ cov (X,Z) = E(KX^2) = KE(X^2) $$



                The second moment $E(X^2)$ of a continuous uniform distribution $(a^2 + ab + b^2)/3$ where $a$ and $b$
                are the bounds of the distribution. In this case, $(-1)^2 + (-1)*1 + 1^2 = 1 -1 + 1 = 1$.



                $$ cov(X,Z) = K frac{1}{3} = frac{K}{3}$$





                The variance $V(X)$ of a uniform distribution is $(b-a)^2/12$.
                begin{align*}
                V(X) &= frac{[1-(-1)]^2}{ 12 } = frac{2^2}{12} = frac{1}{3} \
                V(Z) &= V(KX+Y) = K^2*V(X)*V(Y) = K^2V(X) + V(X) = (K^2+1)V(X) \
                V(Z) &= frac{(K^2+1)}{3} \
                V(X)V(Z) &= frac{K^2+1}{9} \
                sigma(X)sigma(Z) &= sqrt{ frac{K^2+1}{9} } = frac{ sqrt{K^2+1} }{ 3 }\
                r(X,Z) &= frac{ frac{K}{3}}{ frac{sqrt{K^2+1}}{3}} = frac{K}{ sqrt{K^2+1)} } \
                end{align*}





                So for $r = frac{1}{2}$, $frac{K}{sqrt{K^2+1}} = frac{1}{2}$.



                $$ 2K = sqrt{K^2+1} $$



                We will square both sides, which will give two solutions for $K$, only one of which will be relevant.



                begin{align*}
                4K^2 &= K^2 + 1 \
                3K^2 - 1 &= 0 \
                end{align*}



                using $a^2 - b^2 = (a+b)(a-b)$, we see $(sqrt{3}K +1)(sqrt{3}K - 1) = 0$.



                $$ sqrt{3}K = 1 text{ OR } -1$$



                Since we can see in the original problem K must be positive to yield a positive correlation,
                $sqrt{3}K = 1$.



                $$ k = frac{1} {sqrt{3}} $$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 19 '18 at 2:52

























                answered Dec 18 '18 at 23:21









                Dvd AvinsDvd Avins

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