The basic concept
The binomial distribution counts successes in a fixed number of independent trials, each with the same probability of success. “Success” is a mathematical label. In credit modelling, it can mean a default, even though default is economically undesirable.
Write X ~ Binomial(n,p), where n is a non-negative integer and 0 ≤ p ≤ 1. The possible counts are 0, 1, …, n.
Assumptions before calculation
| Assumption | Question for a credit portfolio |
|---|---|
| Fixed number of trials | How many loans are in the portfolio? |
| Two outcomes per trial | Does each loan default within the stated year or survive? |
| Common success probability | Do all loans really share the same default probability? |
| Independent trials | Could a common shock create dependence? |
Heterogeneous default probabilities require a different count model. Dependent defaults can create a much heavier joint loss tail than an independent binomial model.
Probability mass, mean and variance
C(n,k) counts ways to choose the k successful trials. The probability mass function (PMF) gives the probability at an integer k. The cumulative distribution function (CDF) adds these masses up to k.
Worked example: ten independent loans
Assume ten loans have independent annual default probabilities of 10%. Let X count defaults.
| Question | Calculation | Result |
|---|---|---|
| No defaults | 0.9¹⁰ | 34.87% |
| Exactly one default | 10 × 0.1 × 0.9⁹ | 38.74% |
| At least one default | 1 − 0.9¹⁰ | 65.13% |
| At most two defaults | P(0) + P(1) + P(2) | 92.98% |
| Expected defaults | 10 × 0.1 | 1.00 |
| Standard deviation | √(10 × 0.1 × 0.9) | 0.9487 |
Expected defaults of one do not mean exactly one default is certain. In fact, its probability is below 40%.
Connecting default counts to losses
If each default causes a fixed ₹6 lakh loss, portfolio loss L = ₹6 lakh × X. Expected loss is ₹6 lakh × 1 = ₹6 lakh. This translation assumes equal fixed loss severity and no recoveries beyond those already embedded in ₹6 lakh.
When loss severity varies, a default count alone does not determine portfolio loss. Exposure size, recovery and dependence must also enter the model.
Excel illustration
| Task | Formula |
|---|---|
| Exactly one default | =BINOM.DIST(1,10,0.1,FALSE) |
| At most two defaults | =BINOM.DIST(2,10,0.1,TRUE) |
| At least one default | =1-BINOM.DIST(0,10,0.1,TRUE) |
| At least three defaults | =1-BINOM.DIST(2,10,0.1,TRUE) |
For “at least three”, subtract the probability of two or fewer, rather than three or fewer.
Check your understanding
If the number of defaults is one on average, is the default probability 100%?
No. The expected count is np. With n = 10 and p = 0.1, the expected count is one while each individual trial’s success probability is 10%.
When is a normal approximation reasonable?
A common teaching rule is that both np and n(1−p) should be sufficiently large, often at least 5 or 10. The required accuracy and tail of interest matter. Use exact binomial calculations when available.
Interactive distribution laboratory
Change the parameters and threshold, then switch between cumulative probability and density or mass. Values are numerical approximations for learning.
Horizontal axis: the variable’s value. Vertical axis: density, probability mass, or cumulative probability as labelled. Shading highlights values at or below the threshold. The plot shows a finite viewing window.
Sources and further reading
- National Institute of Standards and Technology: Gallery of Distributions
- Distribution definition and properties
- Microsoft: Statistical functions reference
All numerical finance and project examples are hypothetical teaching illustrations. Probability assumptions require validation before practical use.