Probability Distributions in Risk Context
Choose a distribution from the random variable and its support: Bernoulli for one binary event, binomial for independent repeated binary trials, and a continuous distribution for a continuously varying loss or return. Tail assumptions matter as much as the mean.
The method at a glance
Illustrative cohort: 200 borrowers, 40 flagged, 12 defaults.
| Group | Default | No default |
|---|---|---|
| Flagged | 8 | 32 |
| Unflagged | 4 | 156 |
- Default given flagged
- 8 / 40 = 20%
- Flagged given default
- 8 / 12 = 66.67%
| Concept | How it works | Watch for |
|---|---|---|
| Bernoulli / binomial | A binomial count needs a fixed number of independent trials with common probability. | Ignoring correlated defaults. |
| Normal / lognormal | A normal variable can be negative; a lognormal variable is positive. | Using a lognormal model directly for returns that can be negative. |
| Conditional probability | P(A given B) = P(A and B)/P(B), when P(B) > 0. | Reversing the conditioning event. |
01Bernoulli / binomial
How it works: A binomial count needs a fixed number of independent trials with common probability.
Watch for: Ignoring correlated defaults.
02Normal / lognormal
How it works: A normal variable can be negative; a lognormal variable is positive.
Watch for: Using a lognormal model directly for returns that can be negative.
03Conditional probability
How it works: P(A given B) = P(A and B)/P(B), when P(B) > 0.
Watch for: Reversing the conditioning event.
Original worked example
Default given a risk flag
Among 200 borrowers, 40 are flagged. Eight flagged borrowers and four unflagged borrowers default.
- 1
P(default given flagged) = 8/40 = 20%.
- 2
P(flagged given default) = 8/12 = 66.67%.
- 3
The unconditional default rate is 12/200 = 6%.
20%, 66.67% and 6% answer different questions. State the conditioning group before dividing.
Illustrative assumptions and calculations for learning. Not observed market data or an official exam item.
Apply the method and check its limits
Before you start: Fractions, conditioning groups and discrete versus continuous outcomes.
Use the conditioning group as the denominator
P(A | B) = P(A ∩ B) / P(B), for P(B) > 0
In the borrower example, 8 of the 40 flagged borrowers default, so P(default | flagged) = 20%. Of all 12 defaults, 8 were flagged, so P(flagged | default) = 66.67%. These answer different questions. The first describes default risk within a flagged group; the second describes how many defaults the flag captured.
Check what a distribution permits
A Bernoulli outcome is 0 or 1. A binomial count adds a fixed number of independent Bernoulli trials with a common success probability. A normal variable can take any real value; a lognormal variable is strictly positive. Choosing a convenient shape does not establish independence or adequate tail fit. Shared economic shocks can invalidate a binomial model for borrower defaults.
A tail fit can fail even when the average looks right
Two distributions can share a mean and variance but assign very different probabilities to severe losses. Skewness captures asymmetry; heavy tails increase the chance of extreme observations relative to a thin-tailed benchmark. A central-limit argument about sample means does not automatically justify a normal model for individual losses.
Work through the topic
Choose a distribution with suitable support, calculate conditional probabilities from a clear population, and test how dependence changes the risk conclusion.
Match the distribution to the random variable
A Bernoulli variable represents one event with two outcomes, often coded zero and one. A binomial variable counts successes in a fixed number of independent trials with a common probability. A normal variable can take any real value, while a lognormal variable is strictly positive. Start with the quantity being modeled rather than selecting the distribution whose formula is easiest to remember.
For ten independent loans with identical 2% default probabilities, a binomial count has expected defaults 10 × 0.02 = 0.2. That does not mean one-fifth of a loan will default in the realized portfolio. It is an average over the model’s possible outcomes. The probability of no defaults is 0.98¹⁰, approximately 81.71%; the probability of at least one is approximately 18.29%. These answer different questions from the expected count.
If loans share a common borrower industry or economic driver, independence may be unsuitable. Heterogeneous probabilities also break the simple common-probability binomial specification. Expected counts can still be added from individual event probabilities, but joint-event probabilities need a dependence model. Recognizing that distinction is more useful than forcing every portfolio default question into one familiar distribution.
| Variable | Possible starting model | Condition to challenge |
|---|---|---|
| One default indicator | Bernoulli | What event and horizon define default? |
| Defaults among similar independent loans | Binomial count | Are trials independent with a common probability? |
| A variable with positive and negative values | Normal as a possible approximation | Are symmetry and tail behavior credible? |
| A strictly positive modeled level | Lognormal as a possible approximation | Is the level positive, and are log changes modeled appropriately? |
Distribution choice begins with support and assumptions
- One default indicator
- Possible starting model: Bernoulli
- Condition to challenge: What event and horizon define default?
- Defaults among similar independent loans
- Possible starting model: Binomial count
- Condition to challenge: Are trials independent with a common probability?
- A variable with positive and negative values
- Possible starting model: Normal as a possible approximation
- Condition to challenge: Are symmetry and tail behavior credible?
- A strictly positive modeled level
- Possible starting model: Lognormal as a possible approximation
- Condition to challenge: Is the level positive, and are log changes modeled appropriately?
Use a contingency table to keep conditional probabilities straight
In the opening example there are 200 borrowers. Forty are flagged and eight of those default. Among the 160 unflagged borrowers, four default. The default probability given a flag is 8/40 = 20%. The probability of a flag given default is 8/12 = 66.67%. Reversing the conditioning event changes the denominator, even though the numerator refers to the same eight borrowers.
The unconditional default rate is 12/200 = 6%. It can also be recovered by weighting the conditional rates: 20% × 20% + 2.5% × 80% = 6%. Here the first 20% is the default rate among flagged borrowers and the second is the fraction flagged. Labeling those terms prevents two numerically identical percentages from being treated as the same concept.
A flag can be informative without being a certainty. Thirty-two flagged borrowers do not default, and four defaults occur outside the flagged group. A decision using the flag must consider the costs of missed defaults and false alarms, as well as how the sample was selected. This table describes the supplied sample; it does not establish that the same rates will hold in a different population or future period.
| Group | Default / no default | Default rate within group |
|---|---|---|
| Flagged: 40 borrowers | 8 / 32 | 8 ÷ 40 = 20% |
| Unflagged: 160 borrowers | 4 / 156 | 4 ÷ 160 = 2.5% |
| All: 200 borrowers | 12 / 188 | 12 ÷ 200 = 6% |
Original 200-borrower contingency table
- Flagged: 40 borrowers
- Default / no default: 8 / 32
- Default rate within group: 8 ÷ 40 = 20%
- Unflagged: 160 borrowers
- Default / no default: 4 / 156
- Default rate within group: 4 ÷ 160 = 2.5%
- All: 200 borrowers
- Default / no default: 12 / 188
- Default rate within group: 12 ÷ 200 = 6%
Test the tail and dependence assumptions separately
Two models can have the same mean and variance yet assign different probabilities to extreme losses. Matching those summary statistics does not establish a good tail fit. A normal approximation is especially easy to misuse when an instrument’s payoff is nonlinear or when the modeled quantity has a hard boundary. Inspect the support and the loss mechanism before drawing a smooth bell curve through the data.
Consider two loans each defaulting with probability 4%. Under independence, the probability both default is 0.04² = 0.16%, and the probability neither defaults is 0.96² = 92.16%. Under perfect common default, both default with probability 4% and neither defaults with probability 96%. Expected default count is 0.08 in either model, but the probability of the two-default outcome is dramatically different.
The example isolates dependence by keeping each loan’s individual probability unchanged. It shows why portfolio tail risk cannot be inferred from marginal default rates alone. It does not imply that independence and perfect dependence are the only possibilities. For a real application, estimate or stress plausible joint behavior and explain how limited tail observations affect confidence in the result.
Is zero correlation the same as independence?
Not in general. Zero correlation rules out a particular linear relationship, while independence restricts the full joint distribution. Under specific jointly normal assumptions the relationship is stronger, but those assumptions must be stated. Do not use a near-zero sample correlation as automatic proof that joint tail events are independent.
Can a probability density exceed one?
Yes. For a continuous variable, probability is the area over an interval, not the density height at one point. A uniform variable on an interval of width 0.5 has density 2 within that interval, while total probability remains one. Always distinguish a density from a discrete event probability.
Try a changed assumption
What is the default rate among the 160 unflagged borrowers if four default?
Show answer and reasoning
4/160 = 2.5%. Compare this with 8/40 = 20% for the flagged group. The full population rate is 12/200 = 6%; using 200 as the conditional denominator would answer the wrong question.
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Does zero correlation mean independence?
Not in general. Independence implies zero covariance when moments exist; the reverse requires additional assumptions, such as a jointly normal model.
Sources and scope
Links checked 2026-10-10. GARP establishes exam scope; the technical references support the methods identified below. These are selected study notes and original examples, not complete curriculum coverage or official exam questions.
- GARP FRM study materials and 2026 learning objectives
Exam-domain scope and the official route to the annual learning objectives. This page does not certify coverage of every reading.
- MIT OpenCourseWare: quantitative lecture index
Further reading across quantitative methods. Specific technical references are linked beside the explanations below.
- MIT: conditional probability and independence
Conditional probability and the distinction between conditioning on different events.
- MIT: Probability Theory, lecture 3
Random variables, independence, expectation, variance and normal/lognormal distributions.
Prepared by MiloPrep editorial team. MiloPrep is independent of GARP. References do not imply endorsement or independent review of these notes.