Finding a distribution with a given correlation to a Normal Distribution












0












$begingroup$


Below is my attempt to solve a problem which I believe has a solution. My work shows there is no solution. I believe I did something wrong. It could be related to the algebra.



Thanks,

Bob



Problem:

Let $X$ and $Y$ independent normally distributed random variables with a mean of $0$ and a standard deviation
of $1$. Let $K$ be a positive 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 &= 1 \
sigma_y &= 1 \
sigma_z^2 &= sigma_y^2 + K^2 sigma_x^2 + K(sigma_{xy}) \
sigma_{xy} &= 0 \
sigma_z^2 &= 1^2 + K^2 1^2 + K(0) = K^2 + 1 \
end{align*}
begin{align*}
sigma_{xz} &= E(XZ) - E(X)E(Z) \
E(X) &= 0 \
sigma_{xz} &= E(XZ) - (0)E(Z) = E(XZ) \
E(XZ) &= E( X(Y+KX)) = E(XY + KX^2) = E(XY) + KE(X^2) \
E(X^2) &= 1 \
end{align*}
Now we need to find $E(XY)$.
begin{align*}
E(XY) &= E(X) E(Y) = 0(0) \
E(XY) &= 0 \
E(XZ) &= 0 + K(1) = K \
sigma_{xz} &= K \
rho_{xz} &= frac{sigma_x sigma_z }{sigma_{xz}} \
frac{1}{2} &= frac{ sqrt{K^2+1} }{K} \
frac{K}{2} &= sqrt{K^2+1} \
frac{K^2}{4} &= K^2 + 1 \
frac{-3K^2}{4} &= 1 \
end{align*}
This equation has no real roots.






probability







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Your correlation formula is inverse. Correlation is equal to covariance divided by product of standard deviations. Fix that and you get $K=1/sqrt{3}$ and $K=-1/sqrt{3}$
    $endgroup$
    – Sauhard Sharma
    Dec 23 '18 at 18:43


















0












$begingroup$


Below is my attempt to solve a problem which I believe has a solution. My work shows there is no solution. I believe I did something wrong. It could be related to the algebra.



Thanks,

Bob



Problem:

Let $X$ and $Y$ independent normally distributed random variables with a mean of $0$ and a standard deviation
of $1$. Let $K$ be a positive 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 &= 1 \
sigma_y &= 1 \
sigma_z^2 &= sigma_y^2 + K^2 sigma_x^2 + K(sigma_{xy}) \
sigma_{xy} &= 0 \
sigma_z^2 &= 1^2 + K^2 1^2 + K(0) = K^2 + 1 \
end{align*}
begin{align*}
sigma_{xz} &= E(XZ) - E(X)E(Z) \
E(X) &= 0 \
sigma_{xz} &= E(XZ) - (0)E(Z) = E(XZ) \
E(XZ) &= E( X(Y+KX)) = E(XY + KX^2) = E(XY) + KE(X^2) \
E(X^2) &= 1 \
end{align*}
Now we need to find $E(XY)$.
begin{align*}
E(XY) &= E(X) E(Y) = 0(0) \
E(XY) &= 0 \
E(XZ) &= 0 + K(1) = K \
sigma_{xz} &= K \
rho_{xz} &= frac{sigma_x sigma_z }{sigma_{xz}} \
frac{1}{2} &= frac{ sqrt{K^2+1} }{K} \
frac{K}{2} &= sqrt{K^2+1} \
frac{K^2}{4} &= K^2 + 1 \
frac{-3K^2}{4} &= 1 \
end{align*}
This equation has no real roots.






probability







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Your correlation formula is inverse. Correlation is equal to covariance divided by product of standard deviations. Fix that and you get $K=1/sqrt{3}$ and $K=-1/sqrt{3}$
    $endgroup$
    – Sauhard Sharma
    Dec 23 '18 at 18:43
















0












0








0





$begingroup$


Below is my attempt to solve a problem which I believe has a solution. My work shows there is no solution. I believe I did something wrong. It could be related to the algebra.



Thanks,

Bob



Problem:

Let $X$ and $Y$ independent normally distributed random variables with a mean of $0$ and a standard deviation
of $1$. Let $K$ be a positive 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 &= 1 \
sigma_y &= 1 \
sigma_z^2 &= sigma_y^2 + K^2 sigma_x^2 + K(sigma_{xy}) \
sigma_{xy} &= 0 \
sigma_z^2 &= 1^2 + K^2 1^2 + K(0) = K^2 + 1 \
end{align*}
begin{align*}
sigma_{xz} &= E(XZ) - E(X)E(Z) \
E(X) &= 0 \
sigma_{xz} &= E(XZ) - (0)E(Z) = E(XZ) \
E(XZ) &= E( X(Y+KX)) = E(XY + KX^2) = E(XY) + KE(X^2) \
E(X^2) &= 1 \
end{align*}
Now we need to find $E(XY)$.
begin{align*}
E(XY) &= E(X) E(Y) = 0(0) \
E(XY) &= 0 \
E(XZ) &= 0 + K(1) = K \
sigma_{xz} &= K \
rho_{xz} &= frac{sigma_x sigma_z }{sigma_{xz}} \
frac{1}{2} &= frac{ sqrt{K^2+1} }{K} \
frac{K}{2} &= sqrt{K^2+1} \
frac{K^2}{4} &= K^2 + 1 \
frac{-3K^2}{4} &= 1 \
end{align*}
This equation has no real roots.






probability







share|cite|improve this question









$endgroup$




Below is my attempt to solve a problem which I believe has a solution. My work shows there is no solution. I believe I did something wrong. It could be related to the algebra.



Thanks,

Bob



Problem:

Let $X$ and $Y$ independent normally distributed random variables with a mean of $0$ and a standard deviation
of $1$. Let $K$ be a positive 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 &= 1 \
sigma_y &= 1 \
sigma_z^2 &= sigma_y^2 + K^2 sigma_x^2 + K(sigma_{xy}) \
sigma_{xy} &= 0 \
sigma_z^2 &= 1^2 + K^2 1^2 + K(0) = K^2 + 1 \
end{align*}
begin{align*}
sigma_{xz} &= E(XZ) - E(X)E(Z) \
E(X) &= 0 \
sigma_{xz} &= E(XZ) - (0)E(Z) = E(XZ) \
E(XZ) &= E( X(Y+KX)) = E(XY + KX^2) = E(XY) + KE(X^2) \
E(X^2) &= 1 \
end{align*}
Now we need to find $E(XY)$.
begin{align*}
E(XY) &= E(X) E(Y) = 0(0) \
E(XY) &= 0 \
E(XZ) &= 0 + K(1) = K \
sigma_{xz} &= K \
rho_{xz} &= frac{sigma_x sigma_z }{sigma_{xz}} \
frac{1}{2} &= frac{ sqrt{K^2+1} }{K} \
frac{K}{2} &= sqrt{K^2+1} \
frac{K^2}{4} &= K^2 + 1 \
frac{-3K^2}{4} &= 1 \
end{align*}
This equation has no real roots.






probability




probability






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asked Dec 23 '18 at 18:19









BobBob

928515




928515








  • 1




    $begingroup$
    Your correlation formula is inverse. Correlation is equal to covariance divided by product of standard deviations. Fix that and you get $K=1/sqrt{3}$ and $K=-1/sqrt{3}$
    $endgroup$
    – Sauhard Sharma
    Dec 23 '18 at 18:43
















  • 1




    $begingroup$
    Your correlation formula is inverse. Correlation is equal to covariance divided by product of standard deviations. Fix that and you get $K=1/sqrt{3}$ and $K=-1/sqrt{3}$
    $endgroup$
    – Sauhard Sharma
    Dec 23 '18 at 18:43










1




1




$begingroup$
Your correlation formula is inverse. Correlation is equal to covariance divided by product of standard deviations. Fix that and you get $K=1/sqrt{3}$ and $K=-1/sqrt{3}$
$endgroup$
– Sauhard Sharma
Dec 23 '18 at 18:43






$begingroup$
Your correlation formula is inverse. Correlation is equal to covariance divided by product of standard deviations. Fix that and you get $K=1/sqrt{3}$ and $K=-1/sqrt{3}$
$endgroup$
– Sauhard Sharma
Dec 23 '18 at 18:43












2 Answers
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active

oldest

votes


















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$begingroup$

The formula for the correlation is incorrect (you mixed the numerator with the denominator). The corrected one is below (as already stated in the comments).



$$rho_{xz}=frac{sigma_{xz}}{sigma_xsigma_z}$$



It can help to check your assumptions in cases like this, for example your claim that 'if $K$ is very large the correlation approaches $1$' does not hold with $frac{sqrt{K^2+1}}{K}$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    $Cov(X,Z)=E(XZ)=E(X(Y+kX))=kE(X^2)+E(X)E(Y)=k,Var(Z)=1+k^2.$ So we want $1/2=cor(X,Z)=frac{cov(X,Z)}{sqrt{var(X)var(Z)}}=k/sqrt{1+k^2}.$ So, $k=1/sqrt 3$ should work.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

      votes









      0












      $begingroup$

      The formula for the correlation is incorrect (you mixed the numerator with the denominator). The corrected one is below (as already stated in the comments).



      $$rho_{xz}=frac{sigma_{xz}}{sigma_xsigma_z}$$



      It can help to check your assumptions in cases like this, for example your claim that 'if $K$ is very large the correlation approaches $1$' does not hold with $frac{sqrt{K^2+1}}{K}$.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        The formula for the correlation is incorrect (you mixed the numerator with the denominator). The corrected one is below (as already stated in the comments).



        $$rho_{xz}=frac{sigma_{xz}}{sigma_xsigma_z}$$



        It can help to check your assumptions in cases like this, for example your claim that 'if $K$ is very large the correlation approaches $1$' does not hold with $frac{sqrt{K^2+1}}{K}$.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          The formula for the correlation is incorrect (you mixed the numerator with the denominator). The corrected one is below (as already stated in the comments).



          $$rho_{xz}=frac{sigma_{xz}}{sigma_xsigma_z}$$



          It can help to check your assumptions in cases like this, for example your claim that 'if $K$ is very large the correlation approaches $1$' does not hold with $frac{sqrt{K^2+1}}{K}$.






          share|cite|improve this answer









          $endgroup$



          The formula for the correlation is incorrect (you mixed the numerator with the denominator). The corrected one is below (as already stated in the comments).



          $$rho_{xz}=frac{sigma_{xz}}{sigma_xsigma_z}$$



          It can help to check your assumptions in cases like this, for example your claim that 'if $K$ is very large the correlation approaches $1$' does not hold with $frac{sqrt{K^2+1}}{K}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 23 '18 at 18:52









          Cows quackCows quack

          1184




          1184























              0












              $begingroup$

              $Cov(X,Z)=E(XZ)=E(X(Y+kX))=kE(X^2)+E(X)E(Y)=k,Var(Z)=1+k^2.$ So we want $1/2=cor(X,Z)=frac{cov(X,Z)}{sqrt{var(X)var(Z)}}=k/sqrt{1+k^2}.$ So, $k=1/sqrt 3$ should work.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                $Cov(X,Z)=E(XZ)=E(X(Y+kX))=kE(X^2)+E(X)E(Y)=k,Var(Z)=1+k^2.$ So we want $1/2=cor(X,Z)=frac{cov(X,Z)}{sqrt{var(X)var(Z)}}=k/sqrt{1+k^2}.$ So, $k=1/sqrt 3$ should work.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $Cov(X,Z)=E(XZ)=E(X(Y+kX))=kE(X^2)+E(X)E(Y)=k,Var(Z)=1+k^2.$ So we want $1/2=cor(X,Z)=frac{cov(X,Z)}{sqrt{var(X)var(Z)}}=k/sqrt{1+k^2}.$ So, $k=1/sqrt 3$ should work.






                  share|cite|improve this answer









                  $endgroup$



                  $Cov(X,Z)=E(XZ)=E(X(Y+kX))=kE(X^2)+E(X)E(Y)=k,Var(Z)=1+k^2.$ So we want $1/2=cor(X,Z)=frac{cov(X,Z)}{sqrt{var(X)var(Z)}}=k/sqrt{1+k^2}.$ So, $k=1/sqrt 3$ should work.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 23 '18 at 18:55









                  John_WickJohn_Wick

                  1,616111




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