Question

Let \( X \) be the random variable taking values \( 1, 2, \dots, n \) for a fixed positive integer \( n \). If \( P(X = k) = \frac{1}{n} \) for \( 1 \leq k \leq n \), then the variance of \( X \) is:

• A. \( \frac{n^2 - 1}{12} \)
• B. \( \frac{n^2 + 1}{12} \)
• C. \( \frac{n^2 - 1}{6} \)
• D. \( \frac{(n+1)(n+2)}{6} \)

Hint:

For uniform discrete distributions, use known formulas: \( E(X) = \frac{n+1}{2} \), \( E(X^2) = \frac{(n+1)(2n+1)}{6} \), and \( \text{Var}(X) = E(X^2) - [E(X)]^2 \).

Answer 1

The Correct Option is A

Step 1: Given that \( X \) takes values \( 1, 2, \dots, n \) with equal probability, this is a discrete uniform distribution. \[ P(X = k) = \frac{1}{n}
\text{for } k = 1, 2, \dots, n \] Step 2: The mean of a discrete uniform distribution is: \[ \mu = E(X) = \frac{1 + 2 + \dots + n}{n} = \frac{n(n+1)/2}{n} = \frac{n+1}{2} \] Step 3: Compute \( E(X^2) \): \[ E(X^2) = \frac{1^2 + 2^2 + \dots + n^2}{n} = \frac{n(n+1)(2n+1)}{6n} = \frac{(n+1)(2n+1)}{6} \] Step 4: Use the formula for variance: \[ \text{Var}(X) = E(X^2) - [E(X)]^2 \] \[ = \frac{(n+1)(2n+1)}{6} - \left( \frac{n+1}{2} \right)^2 = \frac{(n+1)}{6} \left( 2n+1 - \frac{3(n+1)}{2} \right) = \frac{n^2 - 1}{12} \]