Question

For two correlated data series X and Y, which formula for \( \text{Var}(X-Y) \) is correct?

• A. \( \text{Var}(X)+\text{Var}(Y) \)
• B. \( \text{Var}(X)-\text{Var}(Y) \)
• C. \( \text{Var}(X)+\text{Var}(Y) - 2\text{Cov}(X,Y) \)
• D. \( \text{Var}(X)-\text{Var}(Y) - 2\text{Cov}(X,Y) \)

Hint:

A key property of variance is \( \text{Var}(aX \pm bY) = a^2\text{Var}(X) + b^2\text{Var}(Y) \pm 2ab\text{Cov}(X,Y) \). If X and Y are independent, \( \text{Cov}(X,Y)=0 \), and the formula simplifies.

Answer 1

The Correct Option is C

Step 1: Recall the definition of variance.
\( \text{Var}(Z) = E[(Z - E[Z])^2] \). Let \( Z = X-Y \).

Step 2: Apply the definition to \( \text{Var}(X-Y) \).
\( E[X-Y] = E[X] - E[Y] \). Let \( \mu_X = E[X] \) and \( \mu_Y = E[Y] \). \[ \text{Var}(X-Y) = E[((X-Y) - (\mu_X - \mu_Y))^2] \] \[ = E[((X-\mu_X) - (Y-\mu_Y))^2] \]

Step 3: Expand the squared term. \[ = E[(X-\mu_X)^2 - 2(X-\mu_X)(Y-\mu_Y) + (Y-\mu_Y)^2] \]

Step 4: Use the linearity of expectation. \[ = E[(X-\mu_X)^2] - 2E[(X-\mu_X)(Y-\mu_Y)] + E[(Y-\mu_Y)^2] \]

Step 5: Recognize the definitions of variance and covariance. \[ E[(X-\mu_X)^2] = \text{Var}(X) \] \[ E[(Y-\mu_Y)^2] = \text{Var}(Y) \] \[ E[(X-\mu_X)(Y-\mu_Y)] = \text{Cov}(X,Y) \] Substituting these back gives the final formula: \[ \text{Var}(X-Y) = \text{Var}(X) + \text{Var}(Y) - 2\text{Cov}(X,Y) \]