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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*}

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,
$sqrt{3}K = 1$.

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