Actuarial Notes

Probability Distributions — Reference Sheet

By Actuarial Notes

Every named distribution on the Exam P syllabus, one card each. Tap a card to open its live plot — drag the parameters, switch between PMF/PDF and CDF, and draw Monte-Carlo samples to watch the shape the formula produces. The density, moments, MGF and the notes that matter sit underneath the plot.

Discrete

Binomial Distribution Successes in a fixed number of independent trials, each with the same success probability.

P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}, \quad k = 0, 1, \ldots, n

E[X] = np, \qquad \text{Var}(X) = np(1-p)

M_X(t) = \left(1 - p + pe^t\right)^n

  • Requires independent trials, exactly two outcomes, and a constant p — a question that breaks any of the three is pointing at another distribution
  • \text{Var}(X) < E[X] always, so binomial counts are underdispersed relative to Poisson
  • For "at least one", use the complement 1 - (1-p)^n rather than summing the tail
  • At n = 1 this is the Bernoulli distribution, with mean p and variance p(1-p) — every other count model on this page is built out of it

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Poisson Distribution Events over an interval of time or space, arriving independently at a constant average rate.

P(X = k) = \frac{e^{-\lambda}\lambda^k}{k!}, \quad k = 0, 1, 2, \ldots

E[X] = \text{Var}(X) = \lambda

M_X(t) = \exp\!\left(\lambda(e^t - 1)\right)

  • The mean equals the variance — the fingerprint that identifies it, and the check that tells you a portfolio is overdispersed when the data disagree
  • Rates scale with the interval: counts over t units are \text{Poi}(\lambda t)
  • The standard claim-count model, and the counting side of the Poisson Process
  • Independent Poissons add: \text{Poi}(\lambda_1) + \text{Poi}(\lambda_2) = \text{Poi}(\lambda_1 + \lambda_2)

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Geometric Distribution Trials until the first success, with the trial count not fixed in advance.

P(X = k) = (1-p)^{k-1}p, \quad k = 1, 2, 3, \ldots

E[X] = \frac{1}{p}, \qquad \text{Var}(X) = \frac{1-p}{p^2}

M_X(t) = \frac{pe^t}{1 - (1-p)e^t}

  • Memoryless: P(X > m+n \mid X > m) = P(X > n) — the discrete counterpart of the exponential
  • Work from the survival function P(X > n) = (1-p)^n ("the first n trials all failed") instead of summing the PMF
  • The other parameterization counts failures Y = X - 1, with E[Y] = (1-p)/p and the same variance

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Negative Binomial Distribution Trials until the r-th success. Reduces to the geometric at r = 1.

P(X = k) = \binom{k-1}{r-1}p^r(1-p)^{k-r}, \quad k = r, r+1, \ldots

E[X] = \frac{r}{p}, \qquad \text{Var}(X) = \frac{r(1-p)}{p^2}

M_X(t) = \left(\frac{pe^t}{1 - (1-p)e^t}\right)^{r}

  • \text{Var}(X) > E[X] always, making it the standard overdispersed alternative to the Poisson for a heterogeneous portfolio
  • It is a Poisson whose rate is itself random: a gamma-mixed Poisson is exactly negative binomial
  • Same trials-versus-failures fork as the geometric, with the two means differing by r

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Hypergeometric Distribution Successes in n draws taken without replacement from a finite population.

P(X = k) = \frac{\dbinom{K}{k}\dbinom{N-K}{n-k}}{\dbinom{N}{n}}

E[X] = \frac{nK}{N}, \qquad \text{Var}(X) = \frac{nK(N-K)(N-n)}{N^2(N-1)}

  • The mean matches a binomial with p = K/N; the variance is smaller by the finite population correction \frac{N-n}{N-1}
  • Trials are not independent — drawing one success lowers the chance of the next
  • No elementary MGF; answer from the PMF directly
  • As N grows with K/N fixed it converges to \text{Bin}(n, K/N) — a large population makes replacement irrelevant

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Discrete Uniform Distribution One of n equally likely outcomes — a fair die, a randomly chosen policy.

P(X = k) = \frac{1}{n}, \quad k = 1, 2, \ldots, n

E[X] = \frac{n+1}{2}, \qquad \text{Var}(X) = \frac{n^2 - 1}{12}

  • Both moments follow from the sum formulas \sum k = n(n+1)/2 and \sum k^2 = n(n+1)(2n+1)/6
  • Shifting the support to \{a, \ldots, b\} shifts the mean to (a+b)/2 and leaves the variance a function of the count alone

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Continuous

Continuous Uniform Distribution Equally likely anywhere in a range, with no point favoured over another.

f(x) = \frac{1}{b-a}, \quad a < x < b

E[X] = \frac{a+b}{2}, \qquad \text{Var}(X) = \frac{(b-a)^2}{12}

F(x) = \frac{x-a}{b-a}

  • Conditioning on a sub-interval leaves it uniform on that sub-interval — the property that collapses most deductible questions to geometry
  • F^{-1}(U) for U \sim \text{Unif}(0,1) generates a draw from any distribution (inverse transform)

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Exponential Distribution Waiting time to the next event, or a memoryless loss size. Here \theta is the mean (scale), not a rate.

f(x) = \frac{1}{\theta}e^{-x/\theta}, \quad x > 0

E[X] = \theta, \qquad \text{Var}(X) = \theta^2, \qquad S(x) = e^{-x/\theta}

M_X(t) = \frac{1}{1 - \theta t}, \quad t < 1/\theta

  • Memoryless: given X > d, the excess X - d is again \text{Exp}(\theta). So E[X \mid X > d] = d + \theta and E[(X-d)_+] = \theta e^{-d/\theta} — the two identities most deductible questions reduce to
  • Constant hazard rate 1/\theta: no aging
  • Work from S(x), not F(x) — nearly every exponential question is a tail probability

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Gamma Distribution Waiting time to the \alpha-th event, and the workhorse skewed severity model. Shape \alpha, scale \theta.

f(x) = \frac{x^{\alpha-1}e^{-x/\theta}}{\theta^{\alpha}\Gamma(\alpha)}, \quad x > 0

E[X] = \alpha\theta, \qquad \text{Var}(X) = \alpha\theta^2

M_X(t) = (1 - \theta t)^{-\alpha}, \quad t < 1/\theta

  • Coefficient of variation 1/\sqrt{\alpha} — a larger shape means a tighter, more symmetric curve
  • For integer \alpha the survival function is a Poisson sum: P(X > x) = \sum_{k=0}^{\alpha-1}\frac{e^{-x/\theta}(x/\theta)^k}{k!}
  • Special cases: exponential at \alpha = 1, chi-squared at \alpha = n/2,\ \theta = 2

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Normal Distribution Sums and averages of many independent pieces. The second parameter is the variance.

f(x) = \frac{1}{\sigma\sqrt{2\pi}}\exp\!\left(-\frac{(x-\mu)^2}{2\sigma^2}\right), \quad x \in \mathbb{R}

E[X] = \mu, \qquad \text{Var}(X) = \sigma^2

M_X(t) = \exp\!\left(\mu t + \tfrac{1}{2}\sigma^2 t^2\right)

  • No closed-form CDF: standardize to Z = (X-\mu)/\sigma and read \Phi, using \Phi(-z) = 1 - \Phi(z) for negative arguments
  • Any linear combination of independent normals is exactly normal — no CLT needed. See Probabilities for Linear Combinations
  • Useful percentiles: z_{0.90} = 1.282, z_{0.95} = 1.645, z_{0.975} = 1.960, z_{0.99} = 2.326
  • Approximating a discrete count needs the continuity correction P(X \le k) \approx \Phi\!\left(\frac{k + 0.5 - \mu}{\sigma}\right)

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Lognormal Distribution A positive, right-skewed quantity whose logarithm is normal. The parameters live on the log scale.

f(x) = \frac{1}{x\sigma\sqrt{2\pi}}\exp\!\left(-\frac{(\ln x - \mu)^2}{2\sigma^2}\right), \quad x > 0

E[X] = e^{\mu + \sigma^2/2}, \qquad \text{Var}(X) = e^{2\mu + \sigma^2}\left(e^{\sigma^2} - 1\right)

F(x) = \Phi\!\left(\frac{\ln x - \mu}{\sigma}\right)

  • Every probability question becomes a normal question after taking logs
  • The median is e^{\mu}, strictly below the mean — the gap is the right skew
  • The MGF does not exist for t > 0; moments come from E[X^k] = e^{k\mu + k^2\sigma^2/2} instead

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Beta Distribution Rates, proportions and probabilities — anything confined to (0, 1).

f(x) = \frac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha, \beta)}, \quad 0 < x < 1

B(\alpha,\beta) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}

E[X] = \frac{\alpha}{\alpha+\beta}, \qquad \text{Var}(X) = \frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}

  • Skewed right when \alpha < \beta, left when \alpha > \beta, symmetric when \alpha = \beta; uniform at \alpha = \beta = 1
  • For integer parameters the density is a plain polynomial and integrates directly — no special functions needed
  • The k-th order statistic of n i.i.d. uniforms is exactly \text{Beta}(k,\, n-k+1)

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Reference

Which distribution is this question describing? Most distribution questions are identification problems, not calculation problems. Read for what is being counted or measured and whether the number of trials is fixed.

Counting things

  • Successes in a fixed number of independent trials → binomial (n, p)
  • Successes in a fixed number of draws without replacement → hypergeometric (N, K, n)
  • Trials until the first success → geometric (p)
  • Trials until the r-th success → negative binomial (r, p)
  • Events over an interval of time or space, no trial count at all → Poisson (\lambda)
  • One of n equally likely outcomes → discrete uniform

Measuring things

  • Equally likely anywhere in a range → uniform (a, b)
  • Waiting time to the next event, or a memoryless loss → exponential (\theta)
  • Waiting time to the r-th event, or a skewed severity → gamma (\alpha, \theta)
  • Sums and averages of many independent pieces → normal (\mu, \sigma^2)
  • A positive, right-skewed quantity whose log is normal → lognormal (\mu, \sigma^2)
  • A rate or proportion in (0,1) → beta (\alpha, \beta)

Three fast discriminators

  • Variance-to-mean ratio. Below 1 → binomial. Exactly 1 → Poisson. Above 1 → negative binomial. This single check settles most count-model questions.
  • Replacement. With replacement is binomial; without is hypergeometric. When the population is large relative to the sample the two nearly agree and the binomial is the intended shortcut.
  • Fixed trials? "In 10 policies…" is binomial or hypergeometric. "Until the third claim…" is geometric or negative binomial. "Per month…" is Poisson.

Background: Discrete Univariate Distributions and Continuous Univariate Distributions.

Moments and MGFs at a glance Everything above in one lookup, for when you only need the number.

Discrete

Distribution Support Mean Variance MGF
Bernoulli (p) \{0,1\} p p(1-p) 1-p+pe^t
Binomial (n,p) 0,\ldots,n np np(1-p) (1-p+pe^t)^n
Poisson (\lambda) 0,1,2,\ldots \lambda \lambda e^{\lambda(e^t-1)}
Geometric (p), trials 1,2,\ldots 1/p (1-p)/p^2 pe^t/[1-(1-p)e^t]
Neg. binomial (r,p), trials r,r+1,\ldots r/p r(1-p)/p^2 \left(pe^t/[1-(1-p)e^t]\right)^r
Hypergeometric (N,K,n) \max(0,n{+}K{-}N),\ldots,\min(n,K) nK/N \dfrac{nK(N-K)(N-n)}{N^2(N-1)} not elementary
Discrete uniform (n) 1,\ldots,n (n+1)/2 (n^2-1)/12 \dfrac{e^t(1-e^{nt})}{n(1-e^t)}

Continuous

Distribution Support Mean Variance MGF
Uniform (a,b) (a,b) (a+b)/2 (b-a)^2/12 \dfrac{e^{tb}-e^{ta}}{t(b-a)}
Exponential (\theta) (0,\infty) \theta \theta^2 (1-\theta t)^{-1}
Gamma (\alpha,\theta) (0,\infty) \alpha\theta \alpha\theta^2 (1-\theta t)^{-\alpha}
Chi-squared (k) (0,\infty) k 2k (1-2t)^{-k/2}
Normal (\mu,\sigma^2) \mathbb{R} \mu \sigma^2 e^{\mu t + \sigma^2t^2/2}
Lognormal (\mu,\sigma^2) (0,\infty) e^{\mu+\sigma^2/2} e^{2\mu+\sigma^2}(e^{\sigma^2}-1) does not exist
Beta (\alpha,\beta) (0,1) \dfrac{\alpha}{\alpha+\beta} \dfrac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)} not elementary

The exponential and gamma MGFs require t < 1/\theta; the geometric and negative binomial require (1-p)e^t < 1.

How the distributions relate Half of these cards are the same distribution wearing a different parameter set.

Special cases

  • \text{Bin}(1, p) = \text{Bernoulli}(p)
  • \text{NegBin}(1, p) = \text{Geom}(p)
  • \text{Gamma}(1, \theta) = \text{Exp}(\theta)
  • \text{Gamma}(k/2, 2) = \chi^2_k
  • \text{Beta}(1, 1) = \text{Unif}(0, 1)

Sums of independent variables — the closure properties that turn a convolution into a parameter addition:

Sum Result Condition
\text{Bin}(n_1,p) + \text{Bin}(n_2,p) \text{Bin}(n_1+n_2,\, p) same p
\text{Poi}(\lambda_1) + \text{Poi}(\lambda_2) \text{Poi}(\lambda_1+\lambda_2) —
r i.i.d. \text{Geom}(p) \text{NegBin}(r,\, p) same p
\text{NegBin}(r_1,p) + \text{NegBin}(r_2,p) \text{NegBin}(r_1+r_2,\, p) same p
n i.i.d. \text{Exp}(\theta) \text{Gamma}(n,\, \theta) same scale
\text{Gamma}(\alpha_1,\theta) + \text{Gamma}(\alpha_2,\theta) \text{Gamma}(\alpha_1+\alpha_2,\, \theta) same scale
\sum c_i N(\mu_i, \sigma_i^2) N\!\left(\sum c_i\mu_i,\ \sum c_i^2\sigma_i^2\right) exact, any c_i
k i.i.d. N(0,1) squared \chi^2_k —

Note what is not there: hypergeometrics do not add, lognormals do not add (their products are lognormal), and binomials with different p do not add.

Limits and approximations

  • \text{Bin}(n,p) \to \text{Poi}(\lambda) as n \to \infty, p \to 0 with np = \lambda fixed
  • \text{Hypergeometric}(N,K,n) \to \text{Bin}(n, K/N) as N \to \infty
  • \text{Bin}(n,p) \to normal for large n with np and n(1-p) both large, with the continuity correction
  • \text{Poi}(\lambda) \to normal for large \lambda; \text{Gamma}(\alpha,\theta) \to normal for large \alpha
  • Anything summed enough times → normal, by the Central Limit Theorem

Mixtures and processes

  • A Poisson whose rate is gamma-distributed is exactly negative binomial — the standard justification for using it on heterogeneous claim counts
  • In a Poisson process of rate \lambda: counts over an interval of length t are \text{Poi}(\lambda t), the wait to the next event is \text{Exp}(1/\lambda), and the wait to the r-th is \text{Gamma}(r,\, 1/\lambda). Counts and waiting times are two views of one process

Transformations — see Transformations of Random Variables

  • X \sim N(\mu,\sigma^2) \implies e^X \sim \text{Lognormal}(\mu,\sigma^2), and conversely \ln X \sim N
  • U \sim \text{Unif}(0,1) \implies F^{-1}(U) has CDF F; for the exponential this is -\theta\ln U
  • \min(X_1,\ldots,X_n) of independent exponentials is exponential with the rates added
  • cX for X \sim \text{Gamma}(\alpha,\theta) is \text{Gamma}(\alpha, c\theta) — scale families rescale, shapes do not move

Parameterization traps These cost more marks than the algebra does. Check the convention before substituting.

  • Exponential and gamma: scale or rate? These cards use the scale \theta, so E[X] = \theta and E[X] = \alpha\theta. Texts using the rate write \lambda = 1/\theta and get 1/\lambda and \alpha/\beta. Every mean and variance inverts.
  • Geometric and negative binomial: trials or failures? Counting trials gives E[X] = 1/p and r/p, with support starting at 1 and r. Counting failures shifts both down by 1 and r. The variance is identical either way, so a mismatch shows up only in the mean — and in every answer after it.
  • Normal: variance or standard deviation? N(\mu, \sigma^2) names the variance. N(100, 25) has standard deviation 5, not 25.
  • Lognormal: whose mean? \mu and \sigma describe \ln X, not X. The mean of X is e^{\mu+\sigma^2/2}, always above the median e^{\mu}.
  • Hypergeometric: which letter is which? N is the population, K the successes in it, n the sample. Swapping K and n leaves the mean nK/N unchanged, which is exactly why the error survives long enough to wreck the variance.
  • Variance of a difference. \text{Var}(X - Y) = \text{Var}(X) + \text{Var}(Y) for independent X, Y. Variances add for a difference too — a difference is more variable than either component, never less.

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