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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*}

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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, $\sqrt{3}K = 1$.

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

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