Show the main properties and uses of the Bernoulli, binomial, Poisson, uniform, normal, lognormal, Chi-squared, Student's t, F, exponential and Beta distributions.
The idea
Each named distribution has a job. Bernoulli is one trial with success probability p. Binomial counts successes in n independent trials, with mean np and variance np(1 − p). Poisson counts events arriving at rate λ, with mean and variance both equal to λ. The waiting time between events is exponential with mean 1/λ. Uniform spreads probability evenly over an interval. Normal is symmetric and set by its mean and σ. Lognormal means the log is normal, so values are positive and right-skewed, the standard model for prices. Its mean exp(μ + σ²/2) exceeds its median exp(μ). Chi-squared with k degrees of freedom is the sum of Z² over k independent standard normal variables Z. Student's t is a standard normal divided by the root of an independent chi-squared over k, with fatter tails and variance k/(k − 2) for k above 2. F is a ratio of two independent chi-squared variables, each over its degrees of freedom, used to compare variances. Beta lives on 0 to 1, which suits recovery rates. The usual slip is mixing up the lognormal mean and median.
Formulas
Binomial probability of k successes
P(X = k) = C(n, k) × p^(k) × (1 − p)^(n − k)
- C(n, k)
- the number of ways to choose k of the n trials
- n
- the number of independent trials
- k
- the number of successes
- p
- the probability of success in each trial
Trials must be independent with the same p. The mean is np and the variance np(1 − p).
Mean of a lognormal variable
E[S] = e^(μ + σ² ÷ 2)
- μ
- the mean of ln S
- σ
- the standard deviation of ln S
The median is e^(μ), so the mean exceeds the median. Mixing up the two is the usual slip.
Probability of at least one Poisson event
P(N ≥ 1) = 1 − e^(−λ × t)
- λ
- the arrival rate per unit of time
- t
- the length of the period
Events arrive as a Poisson process, so the waiting time is exponential. Make sure λ and t use the same time unit.
A practice question
Which distribution is most naturally used to model a quantity confined to the interval from 0 to 1, such as an uncertain recovery rate on defaulted debt, because its two shape parameters allow a wide range of skewed or symmetric shapes?
- LognormalWhy not: Lognormal values are positive but unbounded above, so it can put probability on recovery rates above 100 percent.
- ExponentialWhy not: The exponential is a one-parameter waiting-time distribution on zero to infinity with a decreasing density; it is not bounded at 1 and cannot take a hump shape.
- Beta - correct
- NormalWhy not: The normal is symmetric and unbounded on both sides, so it assigns probability to recoveries below 0 and above 100 percent.
The Beta distribution is defined on 0 to 1 and its two shape parameters control mean and skew, so it is the standard choice for recovery rates and other proportions. Normal, lognormal and exponential all put probability outside the unit interval.
A harder one
A portfolio holds 20 loans. Each defaults within one year with probability 5 percent, and defaults are independent. The probability that exactly two loans default is closest to:
- 92.5%Why not: This is the probability of at most two defaults, the cumulative probability, not exactly two.
- 26.4%Why not: This is the probability of at least two defaults (one minus P(0) minus P(1)), not exactly two.
- 0.1%Why not: Omits the binomial coefficient 190 and so uses only 0.05² x 0.95^18, ignoring the number of ways to choose which two loans default.
- 18.9% - correct
Binomial: P(X = 2) = C(20,2) x 0.05² x 0.95^18 = 190 x 0.0025 x 0.3972 = 0.1887, about 18.9 percent. The coefficient counts the ways to choose the two defaulters.
The hardest tier
Defaults of a borrower class arrive as a Poisson process with an intensity of 3 per year. The probability of at least one default in the next six months is closest to:
- 77.7% - correct
- 22.3%Why not: This is exp(-1.5), the probability of no default in six months, not of at least one.
- 95.0%Why not: Uses the full-year intensity of 3 instead of the six-month intensity of 1.5, giving 1 - exp(-3).
- 15.4%Why not: Treats 3 as the mean waiting time rather than the rate, giving 1 - exp(-0.5/3).
Waiting time to the first Poisson event is exponential with rate 3 per year. P(at least one in 0.5 years) = 1 - exp(-3 x 0.5) = 1 - 0.2231 = 0.7769. Equivalently, the Poisson mean over six months is 1.5 and P(N = 0) = exp(-1.5).
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