FRM Part I · FRM Exam Part I

Common Univariate Random Variables: formula sheet

Full chapter guide

Key formulas

Discrete uniform PMF: P(X = x) = 1 ÷ n, for each of n equally likely values; For consecutive integers a to b, n = b − a + 1.
Discrete uniform mean: E(X) = (a + b) ÷ 2; Valid for consecutive integers from a to b. It is the midpoint.
Discrete uniform variance: Var(X) = (n² − 1) ÷ 12; For consecutive integers, n = b − a + 1. Standard deviation is the square root.
Bernoulli PMF: P(X = 1) = p; P(X = 0) = 1 − p; Also written P(X = x) = p^x × (1 − p)^(1 − x) for x = 0 or 1.
Bernoulli mean: E(X) = p; The expected value equals the success probability.
Bernoulli variance: Var(X) = p(1 − p); Maximum value is 0.25 at p = 0.5.
Binomial link: Y = X₁ + X₂ + … + Xₙ; E(Y) = np; Var(Y) = np(1 − p); Holds for n independent Bernoulli trials with the same p.
Binomial probability: P(X = k) = C(n, k) × p^k × (1 − p)^(n − k); C(n, k) = n! ÷ [k! × (n − k)!]. k runs from 0 to n. Trials independent with constant p.
Binomial mean: E(X) = np; Expected number of successes.
Binomial variance: Var(X) = np(1 − p); Standard deviation is √[np(1 − p)].
Poisson probability: P(X = k) = e^(−λ) × λ^k ÷ k!; k = 0, 1, 2, ... λ is the average count in the interval.
Poisson mean and variance: E(X) = λ and Var(X) = λ; Both equal λ. If the interval scales, λ scales in proportion.
Poisson approximation to binomial: λ = np; Works well when n is large and p is small. It is an approximation, not exact.
At least one event: P(X ≥ 1) = 1 − P(X = 0); Binomial: 1 − (1 − p)^n. Poisson: 1 − e^(−λ).
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.
Standardization (z-score): z = (X − μ) ÷ σ; Z ~ N(0, 1). Then P(X ≤ x) = N(z). For a left-tail probability below the mean, z is negative and N(−z) = 1 − N(z).
Confidence interval for a normal variable: μ ± z × σ; Two-sided critical values: 90% → 1.645, 95% → 1.96, 99% → 2.576.
One-tail critical values (VaR): 95% → 1.645; 97.5% → 1.96; 99% → 2.326; 99.9% → 3.09; Use these for loss at a given confidence level. They are rounded values; exam questions normally give or accept them.
Parametric VaR (normal returns): VaR = (z × σ − μ) × portfolio value; Often μ is taken as 0 for short horizons. Scale σ by √T for T days if returns are independent.
Linear combination of normals: aX + bY has mean aμx + bμy and variance a²σx² + b²σy² + 2abρσxσy; Holds for jointly normal X and Y. The result is normal.
Lognormal definition: If ln X ~ N(μ, σ²), then X = exp(μ + σZ); μ and σ are parameters of the log, not of X.
Lognormal mean, median and variance: E[X] = exp(μ + σ²/2); Median = exp(μ); Var(X) = (exp(σ²) − 1) × exp(2μ + σ²); The mean exceeds the median whenever σ > 0.
Stock price at time T (geometric Brownian motion): ln(S_T ÷ S_0) ~ N((μ − σ²/2)T, σ²T); E[S_T] = S_0 × exp(μT); μ is the expected annual return and σ the annual volatility. The median price is S_0 × exp((μ − σ²/2)T).
CLT distribution of the sample mean: X̄ ≈ N(μ, σ² ÷ n); Holds for large n with i.i.d. observations and finite variance. It is an approximation.
Standard error of the sample mean: SE = σ ÷ √n; Use when the population σ is known.
Estimated standard error: SE = s ÷ √n; Use when σ is unknown and replaced by the sample standard deviation s.
Standardized sample mean: z = (X̄ − μ) ÷ (σ ÷ √n); Look up z in the standard normal table.
Distribution of a sum: ΣXᵢ ≈ N(nμ, nσ²); The sum has standard deviation σ√n, not σ ÷ √n.
Confidence interval for the mean: X̄ ± z × SE; z = 1.645 for 90%, 1.96 for 95%, 2.576 for 99% (two-sided).
Chi-squared definition: χ²(k) = Z₁² + Z₂² + … + Z_k², with Z_i independent standard normal; Mean = k, variance = 2k. Values are always ≥ 0 and right-skewed.
Variance test statistic: χ² = (n − 1) s² ÷ σ₀², with n − 1 degrees of freedom; Assumes the data are normally distributed. σ₀² is the variance under the null hypothesis.
Student's t definition: t(k) = Z ÷ √(χ²(k) ÷ k); Z and the chi-squared variable are independent. Mean 0 for k > 1.
t variance: Var(t) = k ÷ (k − 2), for k > 2; Always greater than 1, which is why tails are fatter than the normal. It tends to 1 as k grows.
t statistic for a mean: t = (x̄ − μ₀) ÷ (s ÷ √n), with n − 1 degrees of freedom; Use when the population variance is unknown and estimated by s.
F definition: F(d₁, d₂) = [χ²(d₁) ÷ d₁] ÷ [χ²(d₂) ÷ d₂]; The two chi-squared variables are independent. d₁ is numerator df, d₂ is denominator df.
F test for equal variances: F = s₁² ÷ s₂², with n₁ − 1 and n₂ − 1 degrees of freedom; Convention: put the larger sample variance on top, so F ≥ 1. Assumes both samples are normal and independent.
Link between t and F: [t(k)]² = F(1, k); The square of a t variable with k df is an F variable with 1 and k df.
Useful t critical values (two-tailed, 5%): df 10: 2.228 | df 15: 2.131 | df 30: 2.042 | very large df: 1.960; Learn the pattern: values fall toward the normal value of 1.96 as df rises.
Mixture density: f(x) = w1·f1(x) + w2·f2(x), with w1 + w2 = 1; Weights are regime probabilities and must sum to 1. This mixes densities, not random variables.
Mixture mean: μ = w1·μ1 + w2·μ2; The weighted average of component means.
Mixture second moment: E[X²] = w1·(σ1² + μ1²) + w2·(σ2² + μ2²); Use this to get the variance. Do not average the variances alone when means differ.
Mixture variance: σ² = E[X²] − μ²; With equal means this equals w1·σ1² + w2·σ2².
Kurtosis for a zero-mean mixture of normals: Kurtosis = 3·(w1·σ1⁴ + w2·σ2⁴) ÷ (w1·σ1² + w2·σ2²)²; Valid only when both component means are zero. Each normal has fourth moment 3σ⁴. Result is at least 3.
Excess kurtosis: Excess kurtosis = Kurtosis − 3; A normal has kurtosis 3 and excess kurtosis 0.
Skewness: Skewness = E[(X − μ)³] ÷ σ³; Zero when the component means are equal, because the mixture is then symmetric. Different means can make it non-zero.

Quick revision

Bernoulli: P(1) = p, mean p, variance p(1 − p).
Binomial: mean np, variance np(1 − p); n independent trials with the same p.
Poisson: mean and variance both equal λ; models counts of events in a fixed interval.
Discrete uniform on n equally likely values has probability 1 ÷ n for each.
Continuous uniform on [a, b]: mean (a + b) ÷ 2, variance (b − a)² ÷ 12.
Normal: fully described by mean and variance; skewness 0 and kurtosis 3.
Standardise with z = (x − μ) ÷ σ before using normal tables.
If log returns are normal, the price is lognormal and cannot be negative.
CLT: the sample mean of independent, identically distributed observations with finite variance is approximately normal for large n, with standard error σ ÷ √n.
Student's t has heavier tails than the normal and approaches it as degrees of freedom grow.
Chi-squared is a sum of squared standard normals and is used for variance tests; F is a ratio of two scaled chi-squared variables.
Mixtures of normals with different variances give fat tails and excess kurtosis.

Common mistakes

Using (b − a)² ÷ 12 or n²÷12 for discrete uniform variance Fix: For the discrete case use (n² − 1) ÷ 12, where n = b − a + 1.
Taking n = b − a instead of b − a + 1 Fix: Values 0 to 9 give n = 10, not 9. Always count the values.
Using binomial when trials are not independent or p varies Fix: State the assumptions: fixed n, constant p, independence. If they fail, the binomial does not apply.
Forgetting the combination term C(n, k) Fix: Always multiply by C(n, k), the number of ways to choose which k trials succeed.
Dividing the variance by 12 before squaring the width, or using (b − a) ÷ 12. 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. Fix: Density is not probability. For a continuous variable, P(X = x) = 0. Multiply the density by interval width.
Using 1.96 for 95% one-tailed VaR Fix: VaR is one-sided. At 95% the loss tail is 5%, so use 1.645. Use 1.96 only when 2.5% sits in each tail.
Treating the lognormal parameters as the mean and variance of the price Fix: They describe ln X. The mean of X is exp(μ + σ²/2) and the median is exp(μ). Write 'log-mean' and 'log-variance' next to them in your working.
Dividing σ by n instead of √n. Fix: Variance of the mean is σ² ÷ n. Take the square root to get SE = σ ÷ √n.
Using the variance as σ in the SE formula. Fix: Take √σ² first. Check the units given in the question.

Exam tips

Expect direct formula questions: give the mean or variance from stated parameters. Memorise the formulas cold.
Watch the wording on the range of a uniform variable. Check if it is 0 to n or 1 to n, since that changes n and the mean.
Bernoulli questions often appear inside default or VaR exceedance contexts. Spot the 0/1 indicator and use p.
If options include a Bernoulli variance above 0.25, drop them at once.
Use the calculator only for square roots. The rest is quick mental arithmetic.
Read the setup first. A fixed number of obligors points to binomial. 'Average number per period' points to Poisson.
Questions on 'at least one' almost always reward the complement. Do not add terms.
Remember that Poisson mean equals variance. Exam options often test this directly.