What is a discrete distribution?
A discrete probability distribution describes a random variable that can take on only countable values (for example: 0, 1, 2, … or a short list of categories such as {Heads, Tails}). Each possible outcome has an associated probability. The distribution lists those outcomes and their probabilities so you can answer questions like “What is the probability of exactly k occurrences?”
Key definitions
– Random variable: a quantity whose value is determined by a random process.
– Discrete random variable: a random variable with countable possible values.
– Probability mass function (pmf): the function p(x) that gives the probability that a discrete random variable equals x.
– Continuous distribution: a distribution for a variable that can take any value in an interval (uncountably many values); probabilities are described with a density function, not a pmf.
Why discrete distributions matter
Discrete models are used whenever outcomes are naturally counted or are categorical. Common finance examples include counts of trades per day, number of defaults in a portfolio, success/failure outcomes for an investment idea, and modelling up/down moves in a binomial option-pricing tree. Discrete models are also commonly used inside Monte Carlo simulations when the simulated variable takes integer or categorical values.
Common types of discrete distributions
– Bernoulli: single trial with two outcomes (often coded 1 for “success” and 0 for “failure”).
– Binomial: number of successes in n independent Bernoulli trials with success probability p.
– Multinomial: counts across more than two categories over n trials (generalization of binomial).
– Poisson: counts of events occurring in a fixed interval when events occur independently and with a constant average rate (can model rare counts like number of trades).
Two basic mathematical requirements for any discrete probability distribution
1. For every outcome x: 0 ≤ p(x) ≤ 1.
2. The probabilities sum to 1: sum over all possible x of p(x) = 1.
How to recognize a discrete distribution
Ask these questions:
– Can you list all possible outcomes (finite or countably infinite)?
– Are probabilities assigned to individual outcomes rather than intervals?
If yes to both, the distribution is discrete. Examples: die rolls, flip outcomes, count of trades per day.
How to build a simple discrete probability model (step‑by‑step checklist)
1. Define the random variable and list all possible outcomes.
2. Choose an appropriate model (Bernoulli, binomial, Poisson, multinomial, or empirical probabilities).
3. Assign or estimate the probabilities for each outcome (from theory or data).
4. Verify the two requirements: probabilities are between 0 and 1, and they sum to 1.
5. Compute desired probabilities (point probabilities, cumulative probabilities, expected value, variance).
6. Validate the model against observed data and revise if necessary.
Short worked examples
A) Binomial example — exact probability
Problem: Five independent trials, each with success probability p = 0.6. What is the probability of exactly k = 3 successes?
Formula: P(X = k) = C(n, k) p^k (1 − p)^(n−k), where C(n, k) = n! / (k!(n−k)!)
Calculation:
C(5,3) = 10
p^3 = 0.6^3 = 0.216
(1 − p)^2 = 0.4^2 = 0.16
P(X = 3) = 10 × 0.216 × 0.16 = 0.3456
Interpretation: There is about a 34.6% chance of exactly 3 successes in 5 trials.
B) Poisson example — probability of zero events
Problem: Average number of events (λ) = 0.5 per day. What is the probability of zero events in a day?
Formula: P(X = k) = e^(−λ) λ^k / k!
Calculation for k = 0: P(0) = e^(−0.5) × 0.5^0 / 0! = e^(−0.5) ≈ 0.6065
Interpretation: About a 60.7% chance of zero events that day.
Notes and practical points
– Choose the model that matches the data-generating process: use binomial for fixed number of independent trials, Poisson for counts over continuous intervals with low rates, and multinomial when more than two categories are possible per trial.
– In finance, discrete distributions often appear for decision outcomes, count data (trades, defaults), and discrete-state option models (binomial trees).
– Monte Carlo simulations can aggregate many discrete outcomes; the underlying variables determine whether the simulation output is discrete or continuous.
Discrete vs continuous — quick contrast
– Discrete: outcomes you can list; probabilities attach to exact values. Example: roll of a die.
– Continuous:
– Continuous: outcomes form an interval (uncountably many values); probabilities attach to ranges rather than exact points. Example: intraday return expressed as a real number. For continuous variables you use probability density functions (PDFs) and integrals instead of probability mass functions (PMFs) and sums.
Probability mass function (PMF) — definition and checklist
– Definition: For a discrete random variable X, the PMF p(x) = P(X = x) gives the probability that X takes the exact value x. The PMF satisfies p(x) ≥ 0 for all x and Σx p(x) = 1.
– Use this checklist to work with a PMF:
1. Identify the support (set of possible x values).
2. Write the PMF formula for that support.
3. Compute normalization (optional check that probabilities sum to 1).
4. Compute mean E[X] = Σx x p(x) and variance Var[X] = Σx (x − E[X])^2 p(x
5. (Optional) Compute the cumulative distribution function (CDF). The CDF F(x) = P(X ≤ x) = Σ_{t ≤ x} p(t). Use it when you need probabilities for ranges (≤ or > sample mean)? (yes → consider negative binomial)
Quick worked examples
1) Binomial (fixed trials, independent, same p)
Problem: Probability of exactly 3 successes in n = 10 trials with p = 0.2.
Formula: P(X = k) = C(n,k) p^k (1−p)^{n−k}
Compute:
C(10,3) = 120
P(X = 3) = 120 × 0.2^3 × 0.8^7
0.2^3 = 0.008, 0.8^7 ≈ 0.2097152
P(X = 3) ≈ 120 × 0.008 × 0.2097152 ≈ 0.20133
2) Poisson as an approximation to binomial (rare events)
Problem: n = 1,000 trials, p = 0.002 (rare). Use Poisson with λ = n p = 2.
P(X = 0) = e^{−λ} = e^{−2} ≈ 0.1353
Interpretation: about a 13.5% chance of zero events.
3) Hypergeometric (without replacement)
Problem: Population N = 50, K = 10 “success” items, draw n = 5; probability of exactly 2 successes.
Formula: P(X = k) = [C(K,k) C(N−K, n−k)] / C(N,n)
Compute:
C(10,2) = 45; C(40,3) = 9,880; C(50,5) = 2,118,760
P(X = 2) = (45 × 9,880) / 2,118,760 ≈ 444,600 / 2,118,760 ≈ 0.2098
Parameter estimation (quick)
– Binomial: p̂ = x / n where x = observed successes, n = trials. Example: 30 successes out of 100 → p̂ = 0.30.
– Poisson: λ̂ = sample mean of counts. Example: average events per
period (e.g., per hour). Example: if you observe counts of events in 10 hours as {3, 5, 2, 4, 6, 1, 4, 3, 2, 4}, the sample mean = (3+5+2+4+6+1+4+3+2+4)/10 = 34/10 = 3.4, so the Poisson MLE λ̂ = 3.4.
– Hypergeometric: If you need to estimate the number of successes K in a finite population N from draws without replacement, the method of moments or maximum likelihood can be used, but closed-form unbiased estimators are less simple than for binomial/Poisson. A practical approach is to use sample proportion multiplied by population size: K̂ ≈ (x/n)·N, where x is observed successes in sample size n. This is approximate