FRM Exam Part I · Common Univariate Random Variables

Continuous Uniform Distribution for FRM Part I

Q: How is the uniform distribution used in FRM Part I?

It is the basic example of a continuous distribution with a simple density and CDF. It also underlies random number generation in Monte Carlo simulation, where U(0, 1) draws are converted to other distributions.

Updated 11 October 2026 · Fact-checked

A continuous uniform random variable on [a, b] has constant density 1 ÷ (b − a), so every equal-length sub-interval has equal probability. Probability over [x1, x2] is (x2 − x1) ÷ (b − a). Mean is (a + b) ÷ 2 and variance is (b − a)² ÷ 12.

Understand Continuous Uniform Distribution

A continuous uniform distribution describes a variable that can take any value between a lower bound a and an upper bound b, with no value region more likely than another. Think of a random time within a window when you know nothing else about when the event happens.

Because the variable is continuous, the probability of any single exact value is zero. Only intervals carry probability. The density is a flat line at height 1 ÷ (b − a). The total area under the line must equal 1, and a rectangle of width (b − a) and height 1 ÷ (b − a) gives exactly 1. That is why the height is what it is.

The probability of landing in a sub-interval is just the rectangle area: width of the sub-interval times the height. So you simply take the fraction of the full range that your interval covers. The CDF rises in a straight line from 0 at a to 1 at b.

The mean sits at the midpoint, (a + b) ÷ 2, by symmetry. The variance is (b − a)² ÷ 12. It depends only on the width of the range, not on where the range sits. The distribution is symmetric, so skewness is 0. Its excess kurtosis is −1.2 (kurtosis 1.8), so it has thinner tails than the normal distribution.

The standard uniform U(0, 1) matters in practice. Random number generators produce it, and Monte Carlo simulation transforms it into other distributions.

Key formulas to remember

Probability density function: f(x) = 1 ÷ (b − a) for a ≤ x ≤ b; f(x) = 0 otherwise; Constant height. Total area under the density equals 1.
Cumulative distribution function: F(x) = 0 for x < a; F(x) = (x − a) ÷ (b − a) for a ≤ x ≤ b; F(x) = 1 for x > b; F(x) = P(X ≤ x). Linear between a and b.
Interval probability: P(x1 ≤ X ≤ x2) = (x2 − x1) ÷ (b − a), for a ≤ x1 ≤ x2 ≤ b; Same as F(x2) − F(x1). Endpoints do not matter because single points have probability 0.
Mean: E(X) = (a + b) ÷ 2; Midpoint of the range. Also the median.
Variance: Var(X) = (b − a)² ÷ 12; Standard deviation = (b − a) ÷ √12, about 0.2887 × (b − a).
Quantile: x_p = a + p × (b − a); Value below which a fraction p of outcomes fall. Inverse of the CDF.
Shape: Skewness = 0; Kurtosis = 1.8 (excess kurtosis = −1.2); Symmetric and light-tailed compared with the normal.

How to solve Continuous Uniform Distribution questions

Use this routine for any question on a continuous uniform variable. Most questions need only the range and simple arithmetic.

1Identify a and b from the wording, such as 'between 2 and 10' or 'uniform on [0, 1]'.
2Compute the width (b − a). Check that your interval of interest lies inside [a, b].
3If the question asks for a probability, clip the interval to [a, b] and compute (upper − lower) ÷ (b − a).
4For 'greater than x', use (b − x) ÷ (b − a). For 'less than x', use (x − a) ÷ (b − a).
5For mean or variance, plug into (a + b) ÷ 2 or (b − a)² ÷ 12. Square the width before dividing by 12.
6For a quantile or percentile, use a + p × (b − a).
7For transformed variables such as Y = cX + d, use E(Y) = c·E(X) + d and Var(Y) = c²·Var(X). Y is still uniform.
8Sanity-check: the answer should lie between 0 and 1 for a probability, and the mean should lie inside [a, b].

Quickest way: Fraction-of-the-range shortcut

When to use it: Use it for almost every probability question. It avoids writing the density or integrating.

Draw a number line from a to b and mark your interval.
Probability = length of your interval ÷ length of the full range.
For the mean, take the midpoint. For the standard deviation, multiply the width by 0.2887.
For a percentile p, move p of the way from a to b.

Common mistakes in Continuous Uniform Distribution

Dividing the variance by 12 before squaring the width, or using (b − a) ÷ 12.
The formula is remembered as 'something over 12' without the square.
Fix: Write (b − a)² ÷ 12. Check units: variance has squared units, so the width must be squared.
Treating the height of the density as a probability, for example saying P(X = 5) = 0.1.
The density 1 ÷ (b − a) looks like a probability.
Fix: Density is not probability. For a continuous variable, P(X = x) = 0. Multiply the density by interval width.
Not clipping the interval to [a, b], such as computing P(X > 12) for U(2, 10) as a positive value.
Students apply the formula without checking the range.
Fix: Outside [a, b] the density is 0. Use only the overlap. P(X > 12) = 0 here.
Using the discrete uniform formulas (n² − 1) ÷ 12 for variance.
Both distributions are called uniform and both have a 12 in the variance.
Fix: For a continuous uniform, variance is (b − a)² ÷ 12. The discrete version applies only to equally spaced integer outcomes.
Confusing the standard deviation with the variance when the question asks for volatility.
Rushing and stopping after the variance step.
Fix: Take the square root. Standard deviation = (b − a) ÷ √12.
Thinking the mean changes the spread, or that shifting the range changes the variance.
Mixing up location and scale.
Fix: Variance depends only on the width (b − a). Shifting both a and b by the same amount moves the mean but not the variance.

Worked examples

Example 1

A loss recovery time X (in days) is uniformly distributed between 10 and 30 days. Find (i) P(15 ≤ X ≤ 24), (ii) the mean, and (iii) the standard deviation.

Show the solution

a = 10, b = 30, so the width is 30 − 10 = 20.
(i) Interval length = 24 − 15 = 9. Probability = 9 ÷ 20 = 0.45.
(ii) Mean = (10 + 30) ÷ 2 = 20 days.
(iii) Variance = 20² ÷ 12 = 400 ÷ 12 = 33.33.
Standard deviation = √33.33 = 5.774 days.

Answer: P = 0.45; mean = 20 days; standard deviation ≈ 5.77 days.

Example 2

A daily P&L adjustment X, in USD millions, is uniform on [−2, 6]. What is the probability that X exceeds 3, and what is the value x such that P(X ≤ x) = 0.90?

Show the solution

a = −2, b = 6, so the width is 6 − (−2) = 8.
P(X > 3) = (b − 3) ÷ (b − a) = (6 − 3) ÷ 8 = 3 ÷ 8 = 0.375.
The 90th percentile uses x = a + p × (b − a) = −2 + 0.90 × 8.
0.90 × 8 = 7.2, so x = −2 + 7.2 = 5.2.
Check with the CDF: (5.2 − (−2)) ÷ 8 = 7.2 ÷ 8 = 0.90.

Answer: P(X > 3) = 0.375; the 90th percentile is USD 5.2 million.

Exam tips

Most questions are one-line arithmetic. Spend time reading a and b carefully, especially when a is negative.
Expect the variance formula to be tested through the standard deviation. Know that (b − a) ÷ √12 is about 0.2887 × (b − a).
Watch for questions that ask for kurtosis or compare tails with a normal. Answer: symmetric, skewness 0, excess kurtosis −1.2.
Check the interval against the range first. Trap options often come from forgetting to clip or from using the density as a probability.
U(0, 1) appears in simulation questions. Know that its mean is 0.5 and variance is 1 ÷ 12 ≈ 0.0833.

Practice questions from Common Univariate Random Variables

Continuous Uniform Distribution in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

Continuous Uniform Distribution: frequently asked questions

What is the mean and variance of a continuous uniform distribution?

For a variable uniform on [a, b], the mean is (a + b) ÷ 2 and the variance is (b − a)² ÷ 12. The standard deviation is the square root of the variance. Only the width of the range affects the variance.

What is the CDF of a uniform distribution?

It is 0 below a, (x − a) ÷ (b − a) between a and b, and 1 above b. It is a straight line rising from 0 to 1. Use it to get interval probabilities as F(x2) − F(x1).

Why is the probability of an exact value zero?

A continuous variable has infinitely many possible values, so each single point has zero area under the density. Only intervals have positive probability. This is why it does not matter whether you use ≤ or < in the interval.

How is the uniform distribution used in FRM Part I?

It is the basic example of a continuous distribution with a simple density and CDF. It also underlies random number generation in Monte Carlo simulation, where U(0, 1) draws are converted to other distributions.