Discrete distributions: uniform, binomial and Poisson

What is a discrete uniform distribution?

A discrete uniform distribution has n equally spaced outcomes from a to b, each with probability 1/n. Throwing a fair die is the classic case: six outcomes, each 1/6.

$$\mu = \dfrac{a+b}{2}, \quad \sigma^2 = \dfrac{(b-a+1)^2 - 1}{12}$$

The variance formula assumes the outcomes are whole numbers one unit apart.

What is a binomial distribution?

A Bernoulli trial has two outcomes: success with probability p and failure with probability 1 − p. Count the successes in n independent Bernoulli trials and you have a binomial distribution .

$$\mu = np, \quad \sigma^2 = np(1-p)$$

$$P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}$$

The binomial coefficient C(n, x) = n! ÷ [x!(n − x)!] counts the different orders in which x successes can fall among n trials.

Shape follows p. The skewness is (1 − 2p) ÷ √[np(1 − p)]. At p = 0.5 it is zero, so the distribution is symmetric. Below 0.5 it is positive (the long tail runs right); above 0.5 it is negative (the tail runs left). Here p = 0.60 gives −0.2 ÷ 2.19 = −0.091, a slight left skew. Read the direction from the formula rather than memorizing it.

When n is large and both np and n(1 − p) are above 5, the normal approximation to the binomial is reasonable: X is approximately normal with mean np and variance np(1 − p). In the signal example np = 12 and n(1 − p) = 8, so it qualifies.

$$X \approx N\big(np,\; np(1-p)\big) \text{ if } np > 5 \text{ and } n(1-p) > 5$$

How does a binomial tree describe a stock price?

Each period the price is multiplied by u (up) or d (down). After n periods with i ups and j downs the price is S₀ × uⁱ × dʲ, and the number of ups is binomial. A binomial stock-price tree lays these prices out node by node.

$$S = S_0\,u^{i}\,d^{j}$$

The same tree prices options in Derivatives (the binomial model), and the probability trees lesson later in this module uses it for conditional forecasts.

What is a Poisson distribution?

If events arrive independently at a constant average rate λ per unit of time, the number of events in an interval of length t follows a Poisson distribution . For one unit of time the mean and the variance both equal λ. Finance uses it for defaults in a loan book, arrivals of orders and the timing of cash flows in and out.

$$P(k) = \dfrac{(\lambda t)^k e^{-\lambda t}}{k!}$$

$$\mu = \sigma^2 = \lambda \quad \text{(one unit of time)}$$

The Poisson is also what the binomial becomes with many trials and a tiny success probability, keeping np = λ. For 1,000 loans, each with a default probability of 0.002, the binomial chance of no default is 0.998¹⁰⁰⁰ = 0.1351. The Poisson with λ = 2 gives e^(−2) = 0.1353.

Discrete distributions in financecountable outcomes with a known mean and variance

• Discrete uniform Memorize every whole-number outcome equally likely
• Bernoulli trial Use it one trial: success with probability p, failure 1 − p
  • Binomial Memorize successes in n independent trials; mean np
    • Poissonn large, p small, np = λ; events at a constant rate

How does default intensity give a default probability?

A reduced-form credit model treats default as an outside random event arriving at rate λ, the default intensity (also called the hazard rate ). A structural model, by contrast, says default happens when a firm’s assets fall below its liabilities.

With a reduced-form model, the chance of surviving to time t is e^(−λt). So the chance of default by t is one minus that, and the expected time to the first default is 1/λ.

$$P(\text{default by } t) = 1 - e^{-\lambda t}, \quad E[t_d] = \dfrac{1}{\lambda}$$

Use the Binomial tab to find the p that makes the distribution symmetric and to switch on the normal curve. On the Poisson tab, raise λ and watch the skewness fall; the credit view plots survival and default against time, with the λt trap drawn as a dashed line.

How do you do this in a spreadsheet?

Know the idea

Spreadsheets give the mass and cumulative values directly.

Task	Excel and Google Sheets
Binomial P(X = x) / P(X ≤ x)	BINOM.DIST(x, n, p, FALSE) / BINOM.DIST(x, n, p, TRUE)
Poisson P(X = x) / P(X ≤ x)	POISSON.DIST(x, λ, FALSE) / POISSON.DIST(x, λ, TRUE)

FALSE gives the probability of exactly x (the mass function), TRUE the cumulative probability.

Go deeper Why the Poisson is the binomial with many trials and a tiny p

Take n trials with success probability p = λ/n. The binomial mean is np = λ and the variance is np(1 − p), which approaches λ as n grows and p shrinks. That is why the Poisson has a single parameter that sets both moments. The 1,000-loan check above shows the two answers (0.1351 and 0.1353) already agree to three decimals.

Go deeper Why the time between events is exponential

If events are Poisson with rate λ, the waiting time to the next event has the exponential distribution, and the chance of no event up to t is e^(−λt). That is exactly the survival probability in the default section. It also makes the process “memoryless”: a borrower that has survived five years has the same hazard rate going forward as a new one.

Go deeper Kurtosis of the discrete distributions (for reference)

Excess kurtosis (kurtosis minus 3) is 1/λ for the Poisson and [1 − 6p(1 − p)] ÷ [np(1 − p)] for the binomial. For the discrete uniform with n outcomes it is −6(n² + 1) ÷ [5(n² − 1)], which is negative: the distribution is flatter than the normal. You do not need to memorize these figures.

Go deeper Counts and yes/no outcomes in regression

A Poisson regression models counts, such as the number of defaults in a quarter. A logistic regression models a yes/no outcome, such as default or no default. The next lesson meets the logistic distribution that sits behind the second.