PROBABILITY MASS FUNCTION

Tags: #◇Computer Science #◇STT101A #◇Conceptual #◇Probability Mass Function

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◇Overview

The Probability Mass Function (PMF) of a discrete random variable $X$, denoted $p_X(x)$ or $f(x) = P(X = x)$, assigns a probability mass to each possible discrete value $x$ in the support $R_X$.

◇Key Definitions

• PMF Definition: $p_X(x) = P(X = x) = P(\{\omega \in S : X(\omega) = x\})$ for $x \in \mathbb{R}$.
• Support ($R_X$): The countable set of values with strictly positive probability: $R_X = \{x \in \mathbb{R} : p_X(x) > 0\}$.
• Two Defining Axioms of a Valid PMF:
  1. Non-negativity: $p_X(x) \ge 0$ for all $x \in \mathbb{R}$.
  2. Normalization (Sums to 1): $\sum_{x \in R_X} p_X(x) = 1$.
• Relation to CDF $F_X(x)$: $F_X(x) = \sum_{t \le x} p_X(t)$, and for ordered support $x_1 < x_2 < \cdots$, $p_X(x_i) = F_X(x_i) - F_X(x_{i-1})$.

◇Worked Example

Problem: Verify that $p(x) = \frac{x^2}{30}$ for $x \in \{1, 2, 3, 4\}$ is a valid PMF, and compute $P(X \ge 3)$.

Solution:

1. Non-negative: $p(x) > 0$ for each $x \in \{1, 2, 3, 4\}$.
2. Sum: $\sum_{x=1}^4 \frac{x^2}{30} = \frac{1 + 4 + 9 + 16}{30} = \frac{30}{30} = 1$. (Valid PMF!)
3. $P(X \ge 3) = p(3) + p(4) = \frac{9}{30} + \frac{16}{30} = \frac{25}{30} = \frac{5}{6}$.

◇References & Navigation

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• Related Notes: ◇Normal Distribution, ◇Random Variables