3.2 Probability mass functions (PMFs)

Probability mass functions (PMFs) are key tools for understanding discrete random variables. They give us the probability of a variable taking on specific values, helping us calculate important probabilities and make predictions about discrete events.

PMFs have crucial properties like non-negativity and summing to one. These properties ensure PMFs accurately represent probability distributions. We can use PMFs to find expected values, variances, and other important measures for discrete random variables.

Probability Mass Functions

Definition and Notation

• A probability mass function (PMF) is a function that gives the probability of a discrete random variable taking on a specific value
• The PMF is denoted as $P(X = x)$, where $X$ is the random variable and $x$ is a specific value that $X$ can take
• For a discrete random variable $X$, the PMF is defined for all possible values of $X$
• The PMF is also known as the probability function or the probability distribution of a discrete random variable (probability function, probability distribution)

Properties of PMFs

• Non-negativity: A PMF must be non-negative for all possible values of the random variable $X$, i.e., $P(X = x) ≥ 0$ for all $x$
  • This property ensures that probabilities are always between 0 and 1, inclusive
• Summing to one: The sum of the PMF values for all possible values of $X$ must equal one, i.e., $Σ P(X = x) = 1$ over all $x$
  • This property ensures that the total probability of all possible outcomes is 1
• The PMF of a discrete random variable $X$ uniquely determines its probability distribution, as it specifies the probability of each possible value of $X$

Calculating Probabilities with PMFs

Evaluating PMFs

• To calculate the probability of a discrete random variable $X$ taking on a specific value $x$, evaluate the PMF at that value: $P(X = x)$
  • Example: If $X$ represents the number of heads in two coin flips and $P(X = 1) = 0.5$, then the probability of getting exactly one head in two coin flips is 0.5
• The probability of $X$ taking on a value within a range can be calculated by summing the PMF values for all $x$ in that range: $P(a ≤ X ≤ b) = Σ P(X = x)$ for $a ≤ x ≤ b$
  • Example: If $X$ represents the number of defective items in a batch of 5, and $P(X = 0) = 0.7$, $P(X = 1) = 0.2$, and $P(X = 2) = 0.1$, then the probability of having at most one defective item is $P(0 ≤ X ≤ 1) = P(X = 0) + P(X = 1) = 0.7 + 0.2 = 0.9$

Cumulative Distribution Function (CDF)

• The cumulative distribution function (CDF) of a discrete random variable $X$ can be calculated using the PMF: $F(x) = P(X ≤ x) = Σ P(X = k)$ for all $k ≤ x$
  • The CDF gives the probability that the random variable $X$ takes on a value less than or equal to $x$
  • Example: Using the previous example, the CDF for $X ≤ 1$ is $F(1) = P(X ≤ 1) = P(X = 0) + P(X = 1) = 0.7 + 0.2 = 0.9$

Validating PMFs

Non-negativity and Summing to One

• A valid PMF must satisfy two conditions: non-negativity and summing to one
  • Non-negativity: The PMF must be non-negative for all possible values of the random variable $X$, i.e., $P(X = x) ≥ 0$ for all $x$
  • Summing to one: The sum of the PMF values for all possible values of $X$ must equal one, i.e., $Σ P(X = x) = 1$ over all $x$
• If a given function satisfies both conditions, it is a valid PMF; otherwise, it is not a valid PMF
  • Example: Let $P(X = 0) = 0.3$, $P(X = 1) = 0.5$, and $P(X = 2) = 0.2$. This is a valid PMF because all values are non-negative and they sum to 1
  • Example: Let $P(X = 0) = 0.4$, $P(X = 1) = 0.7$, and $P(X = 2) = -0.1$. This is not a valid PMF because $P(X = 2)$ is negative and the values do not sum to 1

Properties of PMFs

Expected Value and Variance

• The expected value (mean) of a discrete random variable $X$ can be calculated using its PMF: $E(X) = Σ x * P(X = x)$ over all $x$
  • Example: If $X$ represents the number of heads in two coin flips, with $P(X = 0) = 0.25$, $P(X = 1) = 0.5$, and $P(X = 2) = 0.25$, then $E(X) = 0 * 0.25 + 1 * 0.5 + 2 * 0.25 = 1$
• The variance of a discrete random variable $X$ can be calculated using its PMF: $Var(X) = E(X^2) - [E(X)]^2$, where $E(X^2) = Σ x^2 * P(X = x)$ over all $x$
  • Example: Using the previous example, $E(X^2) = 0^2 * 0.25 + 1^2 * 0.5 + 2^2 * 0.25 = 0 + 0.5 + 1 = 1.5$, and $Var(X) = 1.5 - 1^2 = 0.5$