Question

Let \(X\) be a discrete random variable, \(\mu\) be its mean, and \(P(X=x)\) be the probability function. The variance of \(X\) is

• A. \(\displaystyle \sum x^2P(X=x)-\sum x(P(X=x))^2\)
• B. \(\displaystyle \sum (x-\mu)P(X=x)\)
• C. \(\displaystyle \sum x^2P(X=x)-\mu^2\)
• D. \(\displaystyle \sum x^2P(X=x)-\left(\sum x\right)^2\)

Hint:

Remember the two most important formulas for a discrete random variable: \[ E(X)=\sum xP(X=x), \] \[ \boxed{\operatorname{Var}(X)=E(X^2)-[E(X)]^2.} \] These formulas are frequently used in probability and statistics.

Answer 1

The Correct Option is C

Concept: For a discrete random variable \(X\), \[ \mu=E(X)=\sum xP(X=x), \] and the variance is defined as \[ \operatorname{Var}(X)=E[(X-\mu)^2]. \] Expanding this expression gives the standard formula \[ \operatorname{Var}(X)=E(X^2)-[E(X)]^2. \]

Step 1: Write the expression for the second moment. The second moment about the origin is \[ E(X^2)=\sum x^2P(X=x). \]

Step 2: Use the variance formula. Since \[ \operatorname{Var}(X)=E(X^2)-[E(X)]^2, \] and \[ E(X)=\mu, \] we obtain \[ \operatorname{Var}(X) = \sum x^2P(X=x)-\mu^2. \] Hence, \[ \boxed{\operatorname{Var}(X)=\sum x^2P(X=x)-\mu^2.} \] Therefore, the correct option is \[ \boxed{(C)\;\sum x^2P(X=x)-\mu^2.} \]