Questions: Discrete Random Variables

Question 1

Which of the following could be a valid PMF for a random variable taking values {1, 2, 3}?

• A. p(1) = 0.5, p(2) = 0.4, p(3) = 0.2
• B. p(1) = 0.1, p(2) = -0.1, p(3) = 1.0
• C. p(1) = 0.3, p(2) = 0.5, p(3) = 0.2
• D. p(1) = 0.4, p(2) = 0.4, p(3) = 0.4

Answer: (C)

Option C is the only valid PMF: all values are non-negative and they sum to exactly 1 (0.3 + 0.5 + 0.2 = 1.0). Option A fails because the probabilities sum to 1.1. Option B fails because p(2) = -0.1 is negative — probabilities can never be negative. Option D fails because the probabilities sum to 1.2. Both conditions — non-negativity AND summing to 1 — are required; either alone is not sufficient.

Question 2

A discrete random variable X has PMF: p(0) = 0.1, p(1) = 0.4, p(2) = 0.3, p(3) = 0.2. What is P(1 ≤ X < 3)?

• A. 0.4 — only p(1), since X=1 is the left endpoint
• B. 0.7 — summing p(1) + p(2) = 0.4 + 0.3
• C. 0.9 — summing p(1) + p(2) + p(3) = 0.4 + 0.3 + 0.2
• D. 0.6 — summing p(2) + p(3) = 0.3 + 0.2

Answer: (B)

The event 1 ≤ X < 3 means X can equal 1 or 2 (since X is discrete and integer-valued, 'X < 3' means X ≤ 2). For discrete random variables, event probabilities are found by summing the PMF over all values in the event: p(1) + p(2) = 0.4 + 0.3 = 0.7. Option C is the classic error of treating the strict inequality as inclusive — X = 3 is excluded because 3 < 3 is false.

Question 3

A discrete random variable can mainly take on a finite number of distinct values.

• T. True
• F. False

Answer: False

False. A discrete random variable can take on countably infinitely many values. The standard example is the number of coin flips until the first heads: the possible values are {1, 2, 3, 4, …} with no upper bound, yet the values are still discrete (isolated, listable). 'Discrete' means the values form a countable set — not that the set must be finite.

Question 4

For a discrete random variable, the probability that X falls in a range [a, b] is found by summing p(x) over all values x satisfying a ≤ x ≤ b.

• T. True
• F. False

Answer: True

True. This is exactly how event probabilities work for discrete distributions. The PMF assigns a specific probability to each individual value, and any event is a union of those individual outcomes. Summing p(x) over the relevant values uses the additivity of probability for disjoint events. This is fundamentally different from continuous distributions, where you integrate a probability density function — for discrete variables, integration is replaced by summation.

Question 5

Explain what the two required properties of a probability mass function are, and why each property is necessary.

Model answer: The two required properties are: (1) non-negativity — p(x) ≥ 0 for every value x — because probabilities cannot be negative; and (2) normalization — the sum of p(x) over all possible values equals exactly 1 — because the random variable must take on some value with certainty (total probability = 1). Non-negativity ensures p(x) is interpretable as a probability at each point. Normalization ensures the probabilities form a coherent model of a complete experiment where exactly one outcome occurs.

Together these two properties are necessary and sufficient: any non-negative function on a countable set that sums to 1 is a valid PMF and defines a legitimate discrete random variable. If either property fails — say, a value has probability -0.1, or all values sum to 0.8 — the function cannot represent a genuine probability distribution.