# Finding a distribution with a given correlation

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*}

## 1 Answer

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

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