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Explain the discrete uniform and the binomial distribution

CFA® Level I · Quantitative Methods · Statistical Distributions for Financial Asset Prices and Returns · QM.6.a

The idea

A discrete uniform distribution gives every outcome the same probability. A binomial distribution counts the number of successes in a fixed number of independent trials, each with the same probability of success p. Its mean is n times p and its variance is n times p times (1 - p). The independence and constant-probability assumptions are what usually fail in finance: consecutive price moves are not independent trials, and a default probability is not constant across a cycle. The binomial is nevertheless the foundation of the binomial option pricing model, where the two outcomes are an up move and a down move.

Formulas

Binomial probability

P(x) = nCx × p^(x) × (1 − p)^(n − x)

nCx
the number of ways x successes can fall among n trials
p
the probability of success on one trial, the same for every trial
n
the number of trials
x
the number of successes being asked about

Trials that are independent and identical, with two outcomes each. The mean is n × p and the variance n × p × (1 − p), which is worth carrying separately because most questions ask for one of those rather than a single probability.

A practice question

From the question bank, with the reason each wrong option is wrong.

What does a binomial random variable count?

  1. The number of successes in a fixed number of independent trials — correct
  2. The time until the first successWhy not: That is the geometric distribution. The binomial fixes the number of trials and counts successes; the geometric fixes the success and counts trials.
  3. The average outcome across all trialsWhy not: An average is a summary of the results, not the variable itself. The binomial variable is the count, and n x p is its expected value.

The binomial counts successes across n trials, each independent and each with the same probability p. Its two parameters are n and p.

A harder one

A binomial variable has 20 trials and a success probability of 0.25. What is its expected number of successes?

  1. 0.25Why not: The probability of success on a single trial. Across 20 trials the expected count is 20 times that.
  2. 5 — correct
  3. 3.75Why not: n x p x (1 − p) = 20 x 0.25 x 0.75 — the variance, not the mean. The two formulas differ by the extra (1 − p) and are easily swapped.

E(X) = n × p = 20 × 0.25 = 5. The figure 3.75 is the variance, n × p × (1 − p), which answers a different question.

The hardest tier

Which binomial assumption is most obviously violated by modelling consecutive daily stock returns as trials?

  1. That there is a fixed number of trialsWhy not: The number of trading days in a period is fixed and known, so this assumption holds comfortably.
  2. That the probability of success is between zero and oneWhy not: Any probability satisfies this by definition. It is not an assumption a model can fail.
  3. That the trials are independent and the probability constant — correct

Volatility clusters and probabilities drift with conditions, so daily moves are neither independent nor identically distributed. The model can still be a useful approximation, but the assumption should be stated rather than assumed.

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