Question:medium

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

Show 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.
Updated On: Feb 18, 2026
  • \( \text{Var}(X)+\text{Var}(Y) \)
  • \( \text{Var}(X)-\text{Var}(Y) \)
  • \( \text{Var}(X)+\text{Var}(Y) - 2\text{Cov}(X,Y) \)
  • \( \text{Var}(X)-\text{Var}(Y) - 2\text{Cov}(X,Y) \)
Show Solution

The Correct Option is C

Solution and Explanation

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

Step 2: Apply the definition to \( \text{Var}(X-Y) \).
\( E[X-Y] = E[X] - E[Y] \). Define \( \mu_X = E[X] \) 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: Identify 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 yields the final formula:\[ \text{Var}(X-Y) = \text{Var}(X) + \text{Var}(Y) - 2\text{Cov}(X,Y) \]

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