## **Learning Outcomes**

This article explains the three core discrete distributions used throughout FRM Quantitative Analysis—Bernoulli, binomial, and Poisson—and how to compute probabilities and moments (mean and variance) for each. After reading, you will be able to:

- Identify when Bernoulli, binomial, and Poisson models are appropriate in risk problems.
- Compute probabilities using each distribution’s probability mass function (PMF) and interpret cumulative distribution function (CDF) statements.
- Compute and interpret expected value and variance for each distribution, including common risk management applications (defaults, VaR exceedances, and event counts).

## **GARP FRM Part 1 Syllabus**

For the GARP FRM Part 1 exam, you are expected to understand discrete distributions in the broader context of probability and random variables, with a focus on the following syllabus points:

- Distinguish key properties and identify common occurrences of the Bernoulli, binomial, and Poisson distributions.
- Describe and distinguish a PMF from a CDF, and compute expectations for discrete random variables.
- Apply distribution parameters correctly (e.g., $p$, $n$, $\lambda$) and interpret mean/variance results in finance and risk contexts.

## **Test Your Knowledge**

_Attempt these questions before reading this article. If you find some difficult or cannot remember the answers, remember to look more closely at that area during your revision._

1. A risk manager models whether a counterparty defaults over the next year (default vs no default). The most appropriate distribution for the default indicator is:
   a) Normal  
   b) Bernoulli  
   c) Poisson  
   d) Lognormal

2. A portfolio contains 25 similar loans, each with a 2% annual probability of default, independent across loans. What is the expected number of defaults in one year?

3. Trade arrivals follow a Poisson distribution with an average of 3 trades per minute. What is $P(X=0)$ trades in the next minute?

4. True/False (brief explanation required): In a Poisson distribution, the variance equals the mean.

## **What Do You Know Already?**

- Can you explain the difference between a PMF and a CDF for a discrete random variable?
- If $X \sim \text{Binomial}(n,p)$, do you remember the formulas for $E[X]$ and $\text{Var}(X)$ and what assumptions justify them?
- If events occur at an average rate of 6 per hour, can you correctly set $\lambda$ for a 10-minute interval?

Use these prompts to gauge your starting point; then focus your revision where your answers were least confident.

## **Introduction**

Discrete distributions appear whenever risk is naturally expressed as a count or a yes/no indicator. Credit risk problems often start with a default event (default vs no default) for a single name, which is naturally represented as a Bernoulli random variable. Portfolios then introduce counts (how many defaults out of $n$ obligors), which leads to the binomial distribution under simplifying assumptions. Operational risk and market microstructure problems often focus on event counts per time interval (fraud attempts per day, system outages per month, trades per minute), where the Poisson distribution is a standard baseline model.

For the FRM Part 1 exam, you must be able to (i) recognize which distribution fits the story, (ii) write down the correct PMF, (iii) compute probabilities and CDF statements, and (iv) compute and interpret the mean and variance. Many exam errors come from using the wrong parameter (especially the Poisson $\lambda$ after rescaling time) or from failing to define “success” correctly in binomial settings.

> **Key Term: Discrete random variable**  
> A random variable that can take only a countable set of values (e.g., 0,1,2,...). Examples include a default indicator (0/1) and the number of defaults in a portfolio (0 through n).

## **Discrete random variables, PMFs, CDFs, and moments**

A discrete distribution assigns probabilities to individual outcomes. Two functions summarize this:

1. The probability mass function (PMF): gives the probability of each specific value.
2. The cumulative distribution function (CDF): gives the probability the variable is less than or equal to a threshold.

> **Key Term: Probability mass function (PMF)**  
> For a discrete random variable X, the PMF is:
>
> $$
> f(x)=P(X=x).
> $$
>
> A valid PMF has:
>
> $$
> f(x)\ge 0 \quad \forall x
> $$
>
> and:
>
> $$
> \sum_x f(x)=1
> $$
>
> over the support of X.
>
> **Key Term: Cumulative distribution function (CDF)**  
> The CDF is:
>
> $$
> F(x)=P(X\le x).
> $$
>
> For discrete X, F(x) is a step function formed by summing PMF values up to x.

### Expectations (mean) and variance for discrete random variables

For a discrete random variable with outcomes $\{x_1,\dots,x_k\}$:
$$E[X] = \sum_{i=1}^k x_i P(X=x_i).$$

Two expectation properties used repeatedly in FRM calculations are:

- If $c$ is a constant, $E[cX]=cE[X]$.
- For random variables $X$ and $Y$, $E[X+Y]=E[X]+E[Y]$.

Variance is:
$$\text{Var}(X) = E[X^2] - (E[X])^2,$$
where
$$E[X^2] = \sum_{i=1}^k x_i^2 P(X=x_i).$$

### Worked Example 1.1: From PMF to CDF, then compute the mean

A discrete random variable $X$ takes values $\{0,1,2\}$ with:

- $P(X=0)=0.20$
- $P(X=1)=0.50$
- $P(X=2)=0.30$

Compute (i) $F(1)$ and (ii) $E[X]$.

> **Answer:**  
> Step 1 (CDF).
>
> $$
> F(1)=P(X\le 1)=P(X=0)+P(X=1)=0.20+0.50=0.70.
> $$
>
> Step 2 (expected value).  
> $$E[X]=0(0.20)+1(0.50)+2(0.30)=0+0.50+0.60=1.10.$$

## **Bernoulli distribution**

The Bernoulli distribution models a single trial with two outcomes, coded as 1 and 0. In risk management, “1” is often an event such as default, fraud detection, or a VaR exceedance; “0” means it did not occur.

> **Key Term: Bernoulli random variable (Bernoulli trial)**  
> A binary random variable Y that takes values 1 (success/event) with probability p and 0 (failure/no event) with probability 1-p. Notation:
>
> $$
> Y \sim \text{Bernoulli}(p).
> $$

### PMF and CDF

PMF:
$$P(Y=y)=p^y(1-p)^{1-y},\quad y\in\{0,1\}.$$
So:

- $P(Y=1)=p$
- $P(Y=0)=1-p$

CDF (step function):

$$
F_Y(y)=
\begin{cases}
0 & y<0\\
1-p & 0\le y<1\\
1 & y\ge 1
\end{cases}
$$

### Mean and variance

Because $Y$ is 1 with probability $p$ and 0 with probability $1-p$:
$$E[Y]=p,$$
$$\text{Var}(Y)=p(1-p).$$

Interpretation:

- If $p$ is close to 0 or 1, outcomes are mostly the same, so variance is small.
- Variance is largest at $p=0.5$, where uncertainty between 0 and 1 is greatest.

### Risk management applications

Common mappings:

- Credit risk (single obligor): $Y=1$ if the obligor defaults over the horizon; $p$ is the probability of default (PD).
- Market risk backtesting: $Y=1$ if the portfolio loss exceeds VaR on a given day; $p$ should be close to the VaR tail probability (e.g., 1% for 99% VaR) if the model is calibrated and assumptions hold.
- Fraud/AML flagging: $Y=1$ if a transaction is fraudulent (or flagged), $p$ is the relevant event rate.

### Worked Example 2.1: Expected loss with a Bernoulli default indicator

A single loan has exposure at default (EAD) of \$5,000,000. Default over one year is modeled as $Y\sim \text{Bernoulli}(p=0.02)$. If default occurs, loss given default (LGD) is 60% of EAD; otherwise the loss is zero. Compute the expected loss (EL).

> **Answer:**  
> Step 1 (define the loss random variable).  
> Loss:
>
> $$
> L = Y \times (\text{LGD}\times \text{EAD}) = Y \times (0.60 \times 5{,}000{,}000).
> $$
>
> Step 2 (use linearity of expectation).  
> $$E[L]=E[Y]\times (0.60\times 5{,}000{,}000)=p\times 3{,}000{,}000.$$
>
> Step 3 (compute).  
> $$E[L]=0.02\times 3{,}000{,}000=60{,}000.$$
>
> The expected one-year credit loss is \$60,000 under this simple model.

## **Binomial distribution**

The binomial distribution models the number of “successes” in $n$ independent Bernoulli trials, each with the same success probability $p$.

Typical FRM interpretations:

- Number of defaults among $n$ similar obligors (under strong simplifying assumptions).
- Number of VaR exceedances over $n$ days (under model correctness and independence).
- Number of “up days” in $n$ trading days for a stock, if daily direction is simplified to up/down.

> **Key Term: Binomial distribution**  
> If X counts the number of successes in n independent Bernoulli(p) trials, then:
>
> $$
> X\sim \text{Binomial}(n,p).
> $$

### Assumptions (what “binomial” really means)

Binomial modeling requires:

1. Fixed number of trials, $n$.
2. Each trial has two outcomes (success/failure).
3. Constant probability of success $p$ in each trial.
4. Independence across trials.

In finance, assumptions (3) and (4) are often questionable (defaults and VaR exceedances tend to cluster). The binomial is still widely used as a starting point and as a benchmark.

### PMF

For $x\in\{0,1,\dots,n\}$:
$$P(X=x)=\binom{n}{x}p^x(1-p)^{n-x},$$
where
$$\binom{n}{x}=\frac{n!}{x!(n-x)!}.$$

Interpretation of terms:

- $\binom{n}{x}$ counts how many distinct sequences of $x$ successes can occur in $n$ trials.
- $p^x(1-p)^{n-x}$ is the probability of any one specific sequence with $x$ successes and $n-x$ failures.

### Mean and variance for the binomial distribution

For $X\sim \text{Binomial}(n,p)$:
$$E[X]=np,$$
$$\text{Var}(X)=np(1-p).$$

A useful way to remember these is through the “sum of Bernoullis” representation. If $X=\sum_{i=1}^n Y_i$ and each $Y_i$ is Bernoulli$(p)$ and independent, then:

- $E[X]=\sum E[Y_i]=np$
- $\text{Var}(X)=\sum \text{Var}(Y_i)=n p(1-p)$ (independence makes the covariance terms equal to zero)

### Worked Example 3.1: Probability of multiple defaults in a small portfolio

A portfolio has $n=10$ loans. Each loan has a one-year PD of $p=0.03$, and defaults are assumed independent. Let $X$ be the number of defaults over one year, $X\sim \text{Binomial}(10,0.03)$.

Compute:

1. $E[X]$ and $\text{Var}(X)$
2. $P(X=0)$
3. $P(X\ge 2)$

> **Answer:**
>
> 1. Mean and variance  
>    $$E[X]=np=10(0.03)=0.30.$$
>    $$\text{Var}(X)=np(1-p)=10(0.03)(0.97)=0.291.$$
> 2. Probability of zero defaults  
>    $$P(X=0)=\binom{10}{0}(0.03)^0(0.97)^{10}=1\times 1\times 0.97^{10}.$$
>    Numerically, 0.97^10 ≈ 0.7374, so:
>    $$P(X=0)\approx 0.7374.$$
> 3. Probability of two or more defaults  
>    Use the complement:
>    $$P(X\ge 2)=1-P(X=0)-P(X=1).$$
>    We already have P(X=0) ≈ 0.7374.
>
> Compute P(X=1):
> $$P(X=1)=\binom{10}{1}(0.03)^1(0.97)^9 = 10(0.03)(0.97^9).$$
> Since 0.97^9 ≈ 0.7602:
> $$P(X=1)\approx 10(0.03)(0.7602)=0.2281.$$
>
> Therefore:
> $$P(X\ge 2)\approx 1-0.7374-0.2281=0.0345.$$
>
> Under these assumptions, there is about a 3.45% chance of two or more defaults in one year.

### **Exam Warning**

> A common binomial mistake is failing to define “success” clearly (e.g., “default” vs “no default,” “loss exceeds VaR” vs “loss does not exceed VaR”). The value of p must match your definition of success. Another frequent error is assuming independence when the question states (or strongly suggests) dependence or clustering.

### Binomial approximations you should recognize

You may see approximations used to simplify calculations:

- Normal approximation (large $n$ with $np$ and $n(1-p)$ not small):  
  $$X \approx N(np, np(1-p)).$$  
  This is often introduced later with the normal distribution, but you should know why $np$ and $np(1-p)$ appear: they are the binomial mean and variance.

- Poisson approximation (rare event, large $n$, small $p$ with $\lambda=np$):  
  $$\text{Binomial}(n,p) \approx \text{Poisson}(\lambda=np).$$  
  This is widely used for default counts when PDs are small and portfolios are large.

## **Poisson distribution**

The Poisson distribution models the number of events occurring in a fixed interval when events arrive at a stable average rate and (in the standard story) arrivals are independent within the interval.

Risk contexts:

- Operational risk: number of fraud events per day, number of system incidents per month.
- Credit risk (time-based): number of defaults in a year for a large pool (as an approximation).
- Market microstructure: number of trades per minute or order arrivals per second.

> **Key Term: Poisson rate parameter (λ)**  
> In:
>
> $$
> X\sim\text{Poisson}(\lambda),
> $$
>
> λ is the expected number of events in the interval being modeled. If the event rate is r per unit time, then for an interval of length t (in the same units):
>
> $$
> \lambda=rt.
> $$

### PMF for the Poisson distribution

For $x\in\{0,1,2,\dots\}$:
$$P(X=x)=\frac{\lambda^x e^{-\lambda}}{x!}.$$

### Mean and variance for the Poisson distribution

A defining property used constantly in exam problems:
$$E[X]=\lambda,\qquad \text{Var}(X)=\lambda.$$

So the standard deviation is:
$$\sigma_X=\sqrt{\lambda}.$$

### Time scaling and additivity

Two practical properties:

1. Time scaling (rate $\rightarrow$ interval mean):  
   If the average rate is $r$ events per hour, then in $t$ hours the count is Poisson with $\lambda=rt$.

2. Additivity (independent intervals or independent sources):  
   If $X_1\sim\text{Poisson}(\lambda_1)$ and $X_2\sim\text{Poisson}(\lambda_2)$ are independent, then:
   $$X_1+X_2 \sim \text{Poisson}(\lambda_1+\lambda_2).$$
   This is useful for combining event counts across desks, business lines, or time blocks (when independence is a stated assumption).

### Worked Example 4.1: Call arrivals (classic Poisson calculation)

On average, 911 emergency switchboards receive 0.1 calls per second. Assume calls follow a Poisson distribution.

1. What is the expected number of calls in one minute?
2. What is the probability of exactly 5 calls in one minute?

> **Answer:**  
> Step 1 (convert to the interval and compute λ).  
> Rate is 0.1 calls/second. One minute is 60 seconds, so:
> $$\lambda = 0.1 \times 60 = 6.$$
> Therefore, X ~ Poisson(6) for one minute.
>
> Step 2 (expected value).  
> $$E[X]=\lambda=6.$$
>
> Step 3 (probability of exactly 5).  
> Use the Poisson PMF with x=5 and λ=6:
> $$P(X=5)=\frac{6^5 e^{-6}}{5!}.$$
> Compute components:
>
> - 6^5 = 7776
> - 5! = 120
> - e^{-6} ≈ 0.00247875
>
> So:
> $$P(X=5)\approx \frac{7776}{120}\times 0.00247875 = 64.8\times 0.00247875 \approx 0.1607.$$
>
> The probability of exactly 5 calls in one minute is about 16.1%.

### Link to default timing (Poisson and exponential)

In many basic “arrival process” models, if the time between events is exponentially distributed, then the number of events up to time $t$ follows a Poisson distribution. In credit risk language, a constant hazard rate over time leads to exponential waiting times to default, and default counts over a fixed horizon can be modeled (or approximated) with a Poisson distribution. For FRM Part 1, the operational focus is usually on correctly setting $\lambda$ and computing probabilities.

## **Answers to Test Your Knowledge**

1. b) Bernoulli. The default indicator is binary (0/1).
2. $E[X]=np=25(0.02)=0.5$ defaults.
3. For $\lambda=3$, $P(X=0)=e^{-3}\approx 0.0498$.
4. True. For $X\sim\text{Poisson}(\lambda)$, $E[X]=\lambda$ and $\text{Var}(X)=\lambda$.

## **Summary**

Bernoulli, binomial, and Poisson distributions are the main discrete models used in FRM quantitative questions involving binary events and event counts. The Bernoulli distribution models a single event indicator and is summarized by one parameter, $p$. The binomial distribution generalizes Bernoulli to $n$ independent trials with constant success probability $p$, giving mean $np$ and variance $np(1-p)$. The Poisson distribution models counts over an interval with mean (and variance) equal to $\lambda$, the expected count in that interval.

From a risk management standpoint, these distributions appear in credit default modeling (single-name and portfolio counts), VaR backtesting (exceedance counts), and operational risk (incident and loss-event frequencies). Exam performance often depends less on algebra and more on translating the story into the correct distribution, defining “success” correctly, and scaling parameters consistently (especially Poisson rates over time).

## **Key Point Checklist**

_This article has covered the following key knowledge points:_

- A PMF gives $P(X=x)$ for discrete outcomes; a CDF gives $F(x)=P(X\le x)$.
- Bernoulli$(p)$ models a binary indicator with $E[Y]=p$ and $\text{Var}(Y)=p(1-p)$.
- Binomial$(n,p)$ models the number of successes in $n$ independent Bernoulli trials with $E[X]=np$ and $\text{Var}(X)=np(1-p)$.
- Poisson$(\lambda)$ models event counts per interval with $P(X=x)=\lambda^x e^{-\lambda}/x!$ and $E[X]=\text{Var}(X)=\lambda$.
- Common exam errors include misdefining success, misreading independence assumptions, and using the wrong Poisson $\lambda$ after time rescaling.

## **Key Terms and Concepts**

- Discrete random variable
- Probability mass function (PMF)
- Cumulative distribution function (CDF)
- Bernoulli random variable (Bernoulli trial)
- Binomial distribution
- Poisson distribution
- Poisson rate parameter ($\lambda$)
- Expected value (mean)
- Variance


Probability, random variables, and distributions - Discrete distributions (Bernoulli, Binomial, Poisson)