Understanding the Problem: Correlation Between Linear Combinations

The question asks for the correlation coefficient between two new variates, $(x + y)$ and $(x - y)$, given that the original variates, $x$ and $y$, are uncorrelated. We are also given their standard deviations, $\sigma_x$ and $\sigma_y$. To solve this, we need to use the definition of the correlation coefficient and the properties of variance and covariance for uncorrelated variates.

Key Concepts for Uncorrelated Variates

When two variates $x$ and $y$ are uncorrelated, their covariance is zero. Mathematically, this means:

$\text{Cov}(x, y) = 0$

The standard deviation ($\sigma$) of a variate is the square root of its variance ($\sigma^2$ or Var). So, $\text{Var}(x) = \sigma_x^2$ and $\text{Var}(y) = \sigma_y^2$.

We also need the formulas for the variance and covariance of linear combinations of variates. For any variates $A$ and $B$, and constants $a, b, c, d$:

• $\text{Var}(aA + bB) = a^2 \text{Var}(A) + b^2 \text{Var}(B) + 2ab \text{Cov}(A, B)$
• $\text{Cov}(aA + bB, cA + dB) = ac \text{Var}(A) + bd \text{Var}(B) + (ad + bc) \text{Cov}(A, B)$

Since $x$ and $y$ are uncorrelated, $\text{Cov}(x, y) = 0$. The formulas simplify for uncorrelated variates:

• $\text{Var}(ax + by) = a^2 \text{Var}(x) + b^2 \text{Var}(y) = a^2 \sigma_x^2 + b^2 \sigma_y^2$
• $\text{Cov}(ax + by, cx + dy) = ac \text{Var}(x) + bd \text{Var}(y) = ac \sigma_x^2 + bd \sigma_y^2$

Calculating the Correlation Coefficient

The correlation coefficient between two variates $U$ and $V$ is defined as:

$\rho_{UV} = \frac{\text{Cov}(U, V)}{\sigma_U \sigma_V}$

Let $U = x + y$ and $V = x - y$. We need to calculate $\text{Cov}(U, V)$, $\sigma_U$, and $\sigma_V$.

Step 1: Calculate Covariance Between (x + y) and (x – y)

Using the simplified covariance formula for uncorrelated variates with $a=1, b=1, c=1, d=-1$:

$\text{Cov}(x + y, x - y) = (1)(1) \text{Var}(x) + (1)(-1) \text{Var}(y)$

$\text{Cov}(x + y, x - y) = \text{Var}(x) - \text{Var}(y)$

Since $\text{Var}(x) = \sigma_x^2$ and $\text{Var}(y) = \sigma_y^2$, we have:

$\text{Cov}(x + y, x - y) = \sigma_x^2 - \sigma_y^2$

Step 2: Calculate Variance and Standard Deviation of (x + y)

Using the simplified variance formula for uncorrelated variates with $a=1, b=1$:

$\text{Var}(x + y) = (1)^2 \text{Var}(x) + (1)^2 \text{Var}(y)$

$\text{Var}(x + y) = \text{Var}(x) + \text{Var}(y) = \sigma_x^2 + \sigma_y^2$

The standard deviation $\sigma_{x+y}$ is the square root of the variance:

$\sigma_{x+y} = \sqrt{\sigma_x^2 + \sigma_y^2}$

Step 3: Calculate Variance and Standard Deviation of (x – y)

Using the simplified variance formula for uncorrelated variates with $a=1, b=-1$:

$\text{Var}(x - y) = (1)^2 \text{Var}(x) + (-1)^2 \text{Var}(y)$

$\text{Var}(x - y) = \text{Var}(x) + \text{Var}(y) = \sigma_x^2 + \sigma_y^2$

The standard deviation $\sigma_{x-y}$ is the square root of the variance:

$\sigma_{x-y} = \sqrt{\sigma_x^2 + \sigma_y^2}$

Step 4: Calculate the Correlation Coefficient between (x + y) and (x – y)

Now we use the correlation coefficient formula:

$\rho_{(x+y), (x-y)} = \frac{\text{Cov}(x + y, x - y)}{\sigma_{x+y} \sigma_{x-y}}$

Substitute the results from Steps 1, 2, and 3:

$\rho_{(x+y), (x-y)} = \frac{\sigma_x^2 - \sigma_y^2}{\sqrt{\sigma_x^2 + \sigma_y^2} \cdot \sqrt{\sigma_x^2 + \sigma_y^2}}$

$\rho_{(x+y), (x-y)} = \frac{\sigma_x^2 - \sigma_y^2}{\sigma_x^2 + \sigma_y^2}$

Summary of Calculations

The calculated correlation coefficient between $(x + y)$ and $(x - y)$ is $\frac{\sigma_x^2 - \sigma_y^2}{\sigma_x^2 + \sigma_y^2}$.

Checking the Options

Let's compare our result with the given options:

1. $\frac{{{{\rm{\sigma }}_{\rm{x}}}{{\rm{\sigma }}_{\rm{y}}}}}{{{\rm{\sigma }}_{\rm{x}}^2 + {\rm{\sigma }}_{\rm{y}}^2}}$
2. $\frac{{{{\rm{\sigma }}_{\rm{x}}} + {{\rm{\sigma }}_{\rm{y}}}}}{{2{{\rm{\sigma }}_{\rm{x}}}{{\rm{\sigma }}_{\rm{y}}}}}$
3. $\frac{{{\rm{\sigma }}_{\rm{x}}^2 - {\rm{\sigma }}_{\rm{y}}^2}}{{{\rm{\sigma }}_{\rm{x}}^2 + {\rm{\sigma }}_{\rm{y}}^2}}$
4. $\frac{{{{\rm{\sigma }}_{\rm{y}}} - {{\rm{\sigma }}_{\rm{x}}}}}{{{{\rm{\sigma }}_{\rm{x}}}{{\rm{\sigma }}_{\rm{y}}}}}$

Our result matches option 3.

Revision Table: Key Formulas

Additional Information: Properties and Interpretation

The correlation coefficient measures the linear relationship between two variates. It ranges from -1 to +1.

• A value of +1 indicates a perfect positive linear relationship.
• A value of -1 indicates a perfect negative linear relationship.
• A value of 0 indicates no linear relationship (the variates are uncorrelated).

In this problem, even though $x$ and $y$ are uncorrelated, the linear combinations $(x+y)$ and $(x-y)$ are generally correlated, unless $\sigma_x^2 = \sigma_y^2$.

• If $\sigma_x^2 = \sigma_y^2$, the correlation coefficient becomes $\frac{\sigma_x^2 - \sigma_x^2}{\sigma_x^2 + \sigma_x^2} = \frac{0}{2\sigma_x^2} = 0$. This means if $x$ and $y$ have the same variance (and are uncorrelated), then $x+y$ and $x-y$ are also uncorrelated.
• If $\sigma_x^2 > \sigma_y^2$, the numerator $\sigma_x^2 - \sigma_y^2$ is positive, and the correlation coefficient is positive.
• If $\sigma_x^2 < \sigma_y^2$, the numerator $\sigma_x^2 - \sigma_y^2$ is negative, and the correlation coefficient is negative.

This makes intuitive sense: if $x$ varies more than $y$, then increasing $x$ has a larger effect on both $(x+y)$ and $(x-y)$, tending to make them move in the same direction, resulting in a positive correlation. If $y$ varies more than $x$, increasing $y$ increases $(x+y)$ but decreases $(x-y)$, tending to make them move in opposite directions, resulting in a negative correlation.