Distributions for Actuaries

Full formulas, parametrisations, interactive plots, relationship maps, worked examples and personal notes.

Select a distribution from the library to explore it.

Probability & Random Variables Basics

Conditional probability, CDFs, expectations, variance, joint distributions and survival functions — drawn from ACTL30001 preliminaries.

1. Conditional Probability

Conditional probability measures how likely an event is given that another event has occurred. It is the foundation of Bayesian reasoning and risk modelling.

Definition

P(A \mid B) = \frac{P(A \cap B)}{P(B)}

Chain rule (when A ⊂ C)

P(A \mid B) = P(A \mid B, C)\,P(C \mid B)

Union formula

P(A \cup B) = P(A) + P(A^c)\,P(B \mid A^c)

2. Total Probability Formula

When $A_1, A_2, \ldots, A_n$ are mutually exclusive and exhaustive (i.e. $A_i \cap A_j = \emptyset$ and $\bigcup_{i=1}^n A_i = \Omega$):

Unconditional probability

P(B) = \sum_{i=1}^{n} P(A_i)\,P(B \mid A_i)

Conditional on C

P(B \mid C) = \sum_{i=1}^{n} P(B \mid A_i, C)\,P(A_i \mid C)

📋 Actuarial Example

$Y \sim \text{Poisson}(\lambda)$ = number of claims. Each claim is large with prob $p$, small with prob $q=1-p$. Let $X$ = number of large claims. Then:
• $X \mid Y=n \sim \text{Bin}(n, p)$
• $E[X] = \lambda p$ (by total expectation)
• $\text{Var}[X] = \lambda p$ (by total variance)
• Marginally: $X \sim \text{Poisson}(\lambda p)$

3. Bayes' Theorem

Bayes' theorem lets you reverse conditional probabilities — updating beliefs given new evidence. Essential for credibility theory.

Bayes' theorem

P(A_i \mid B) = \frac{P(A_i)\,P(B \mid A_i)}{\sum_{j=1}^{n} P(A_j)\,P(B \mid A_j)}

Continuous version

f(\theta \mid x) = \frac{f(x \mid \theta)\,\pi(\theta)}{\int f(x \mid \theta)\,\pi(\theta)\,d\theta} \propto f(x \mid \theta)\,\pi(\theta)

Key insight: The prior $\pi(\theta)$ is updated by the likelihood $f(x|\theta)$ to give the posterior $f(\theta|x)$. This is how credibility theory works in actuarial science.

4. Independence

A and B are independent if

P(A \cap B) = P(A)\,P(B) \iff P(A \mid B) = P(A)

Mutual independence of $X_1, \ldots, X_n$

P(X_1 \in A_1, \ldots, X_n \in A_n) = \prod_{i=1}^{n} P(X_i \in A_i)

1. CDFs, PDFs and Basic Relations

For a continuous random variable $X \geq 0$ with CDF $F(x)$ and PDF $f(x)$:

CDF and PDF

F(x) = P(X \leq x) = \int_0^x f(y)\,dy, \qquad f(x) = F'(x)

Probability over a small interval

P(x < X < x+h) = F(x+h)-F(x) \approx f(x)\,h \quad \text{for small } h

Survival function

\bar{F}(x) = P(X > x) = 1 - F(x) = \int_x^\infty f(y)\,dy

Important: $f(x)$ is a density — a rate, not a probability. $P(X=x)=0$ for any continuous r.v.

2. Expectation for Continuous R.V.

$n$-th moment — two equivalent forms

E[X^n] = \int_0^\infty x^n f(x)\,dx = n\int_0^\infty x^{n-1}[1-F(x)]\,dx

Survival function

\bar{F}(x) = P(X > x) = 1 - F(x) = \int_x^\infty f(y)\,dy

Important: $f(x)$ is a rate, not a probability. $P(X = x) = 0$ for any continuous r.v., and $P(X=x) \neq f(x)$.

3. Expectation for Discrete R.V.

CDF as step function

F(x) = \sum_{y=0}^{x} f(y), \qquad f(x) = F(x) - F(x-1)

$n$-th moment

E[X^n] = \sum_{x=1}^{\infty} x^n f(x) = \sum_{k=0}^{\infty} \left[(k+1)^n - k^n\right][1-F(k)]

4. General Moment Formula (φ-method)

Extremely useful in actuarial science — works for any r.v. $X \geq 0$, whether continuous, discrete or mixed.

General form

E[\varphi(X)] = \varphi(0) + \int_0^\infty \varphi'(x)[1 - F(x)]\,dx

Special case $\varphi(x) = x^n$

E[X^n] = n\int_0^\infty x^{n-1}[1-F(x)]\,dx, \quad n > 0

Limited expected value — $Y = \min(X, a)$

E[Y^n] = n\int_0^a x^{n-1}[1-F(x)]\,dx

📋 Why it matters

For $X \sim \text{Exp}(\lambda)$: $E[X] = \int_0^\infty [1-F(x)]\,dx = \int_0^\infty e^{-\lambda x}\,dx = \frac{1}{\lambda}$. No integration by parts needed!

4b. Truncated / Conditional Expectations

A truncated expectation conditions on the variable lying in a restricted range. This arises directly in policy limits, deductibles, and excess-of-loss pricing.

Conditional expectation given $X < a$

E[X \mid X < a] = \frac{\int_0^a x\,f(x)\,dx}{\int_0^a f(x)\,dx} = \frac{\int_0^a x\,f(x)\,dx}{F(a)}

Conditional expectation given $X > a$ (excess mean)

E[X \mid X > a] = \frac{\int_a^\infty x\,f(x)\,dx}{1-F(a)} = \frac{\int_a^\infty x\,f(x)\,dx}{\bar{F}(a)}

Law of total expectation via truncation

E[X] = E[X \mid X{<}a]\cdot F(a) + E[X \mid X{>}a]\cdot[1-F(a)]

Limited expected value $E[X \wedge a] = E[\min(X,a)]$

E[X \wedge a] = \int_0^a x\,f(x)\,dx + a\,[1-F(a)] = \int_0^a [1-F(x)]\,dx

📋 Insurance Application

Claim $X \sim \text{Exp}(\lambda)$, policy limit $a$. Insurer pays $\min(X,a)$.
$E[X \wedge a] = \int_0^a e^{-\lambda x}\,dx = \frac{1-e^{-\lambda a}}{\lambda}$.
The excess-of-loss layer above $a$ costs $E[X] - E[X \wedge a] = \frac{e^{-\lambda a}}{\lambda}$.

Weighted average of a function (from PDF)

\text{Weighted avg of } f \text{ on } [a,b] = \frac{\int_a^b w(x)f(x)\,dx}{\int_a^b w(x)\,dx}

\text{Simple average: } \frac{\int_a^b f(x)\,dx}{b-a}, \qquad \text{Discrete: } \frac{\sum_{i=1}^n w_i x_i}{\sum_{i=1}^n w_i}

5. Conditional Expectation & Variance

The tower property and variance decomposition are among the most powerful tools in actuarial mathematics.

Conditional expectation — discrete $X$, given $Y = y$

E(X \mid Y = y) = \sum_{x} x \cdot P(X = x \mid Y = y)

Conditional expectation — continuous $X$, given $Y = y$

E(X \mid Y = y) = \int_{-\infty}^{\infty} x \cdot f_{X|Y}(x \mid y)\,dx

Tower property (iterated expectations)

E[X] = E\!\left[E(X \mid Y)\right]

Eve's law (total variance formula)

\text{Var}[X] = E\!\left[\text{Var}(X \mid Y)\right] + \text{Var}\!\left[E(X \mid Y)\right]

When Y is continuous with PDF $g(y)$

E[X] = \int_0^\infty E(X \mid Y=y)\,g(y)\,dy

\text{Var}[X] = \int_0^\infty \text{Var}(X\mid Y=y)\,g(y)\,dy + \int_0^\infty \!\left[E^2(X\mid Y=y) - \left(\int_0^\infty E(X\mid Y=y)g(y)dy\right)^{\!2}\right]g(y)\,dy

When Y is discrete with PMF $g(y)$

E[X] = \sum_{y=0}^{\infty} E(X \mid Y=y)\,g(y)

\text{Var}[X] = \sum_{y=0}^{\infty}\text{Var}(X\mid Y=y)\,g(y) + \sum_{y=0}^{\infty}\!\left[E^2(X\mid Y=y) - \left(E[X]\right)^2\right]g(y)

📋 Worked Example

$Y \sim \text{Poisson}(\lambda)$, $X \mid Y=n \sim \text{Bin}(n,p)$.
$E[X] = E[E[X|Y]] = E[Yp] = \lambda p$.
$\text{Var}[X] = E[\text{Var}(X|Y)] + \text{Var}[E(X|Y)] = E[Yp(1-p)] + \text{Var}[Yp] = \lambda p(1-p) + \lambda p^2 = \lambda p$.

1. Joint Distributions

Joint PDF (continuous)

f_{X,Y}(x,y) \geq 0, \qquad \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f_{X,Y}(x,y)\,dx\,dy = 1

Marginal PDFs

f_X(x) = \int_{-\infty}^{\infty} f_{X,Y}(x,y)\,dy, \qquad f_Y(y) = \int_{-\infty}^{\infty} f_{X,Y}(x,y)\,dx

Conditional PDF

f_{X|Y}(x \mid y) = \frac{f_{X,Y}(x,y)}{f_Y(y)}

Independence condition

X \perp Y \iff f_{X,Y}(x,y) = f_X(x)\,f_Y(y)

2. Order Statistics

Let $X_1, \ldots, X_n$ be i.i.d. with CDF $F$ and PDF $f$. Define $Y = \max(X_i)$ and $Z = \min(X_i)$.

Maximum $Y = \max(X_1, \ldots, X_n)$

G(y) = [F(y)]^n, \qquad g(y) = n[F(y)]^{n-1}f(y)

Minimum $Z = \min(X_1, \ldots, X_n)$

H(z) = 1-[1-F(z)]^n, \qquad h(z) = n[1-F(z)]^{n-1}f(z)

Key special cases

X_i \sim \text{Exp}(\lambda) \implies Z \sim \text{Exp}(n\lambda)

X_i \sim U(0,1) \implies Y \sim \text{Beta}(n,1), \quad Z \sim \text{Beta}(1,n)

3. Competing Exponentials

If $X \sim \text{Exp}(\lambda_1)$ and $Y \sim \text{Exp}(\lambda_2)$ independently:

Race probabilities

P(X < Y) = \frac{\lambda_1}{\lambda_1 + \lambda_2}, \qquad P(X > Y) = \frac{\lambda_2}{\lambda_1 + \lambda_2}

Generalisation (n independent exponentials)

\min(X_1,\ldots,X_n) \sim \text{Exp}\!\left(\sum_{i=1}^n \lambda_i\right)

P\!\left(X_j = \min(X_1,\ldots,X_n)\right) = \frac{\lambda_j}{\sum_{i=1}^n \lambda_i}

📋 Actuarial Interpretation

Three independent risks trigger claims at rates $\lambda_1=1, \lambda_2=2, \lambda_3=3$. The first claim arrives at rate $1+2+3=6$, and Risk 2 causes the first claim with probability $2/6 = 1/3$.

4. Sums of Independent Exponentials

All same rate $\lambda$: Erlang/Gamma

\sum_{i=1}^n X_i \sim \Gamma(n, \lambda) = \text{Erlang}(n, \lambda)

All distinct rates $\lambda_i \neq \lambda_j$: Hyper-exponential

f_S(x) = \sum_{i=1}^n C_{i,n}\,\lambda_i\,e^{-\lambda_i x}, \qquad C_{i,n} = \prod_{j \neq i}\frac{\lambda_j}{\lambda_j - \lambda_i}

$n=2$ case explicitly

f_S(x) = \frac{\lambda_1}{\lambda_1-\lambda_2}\lambda_2 e^{-\lambda_2 x} + \frac{\lambda_2}{\lambda_2-\lambda_1}\lambda_1 e^{-\lambda_1 x}

5. Jensen's Inequality

For convex $g$ ($g'' > 0$): e.g. $g(x)=e^x$, $g(x)=x^2$

E[g(X)] \geq g(E[X])

For concave $g$ ($g'' < 0$): e.g. $g(x)=\ln x$, $g(x)=\sqrt{x}$

E[g(X)] \leq g(E[X])

Memory trick: Convex = cup shape = expectations are pushed up. Concave = cap shape = expectations are pushed down.

1. Hazard Rate (Continuous)

Hazard rate (force of mortality)

\mu_x = \frac{f(x)}{S(x)} = \frac{f(x)}{1-F(x)}, \qquad S(x) = e^{-\int_0^x \mu_y\,dy}

Interpretation: $\mu_x \cdot h \approx P(\text{die in } (x, x+h] \mid \text{alive at } x)$ for small $h$. This is also called the force of mortality $\mu_x$ or hazard function $h(x)$.

2. Hazard Function (Discrete)

For discrete r.v. taking values $0 < x_1 < x_2 < \ldots < x_k$:

Discrete hazard

\lambda_j = P(X = x_j \mid X \geq x_j) = \frac{F(x_j) - F(x_j-)}{1 - F(x_j-)}

Survival from discrete hazards

S(x) = P(X > x) = \prod_{j=1}^{j_x}(1 - \lambda_j), \quad j_x = \max\{j : x_j \leq x\}

3. Proper vs Defective R.V.

Proper: $F(\infty) = 1$ — all mass is finite

P(X < \infty) = 1

Defective: $F(\infty) < 1$ — positive mass at $\infty$

P(\tau < \infty) < 1 \iff P(\tau = \infty) > 0

Actuarial use: Defective distributions model scenarios where an event may never occur — e.g. in ruin theory, a company may never go bankrupt (probability of ruin < 1).

1. Simulating a Discrete R.V.

Generate $u \sim U(0,1)$ and map to the discrete distribution:

Procedure

X = x_j \quad \text{if} \quad \sum_{i=0}^{j-1} p_i \leq u < \sum_{i=0}^{j} p_i

📋 Example: $X \sim \text{Poisson}(1)$

$P(X=0)=e^{-1}\approx 0.368$, $P(X\leq 1)\approx 0.736$, $P(X\leq 2)\approx 0.920$.
If $u=0.251$: $u < 0.368$, so $X=0$.
If $u=0.822$: $0.736 \leq 0.822 < 0.920$, so $X=2$.

2. Inverse Transform Method (Continuous)

Inversion theorem

X = F^{-1}(U), \quad U \sim U(0,1) \implies X \sim F

Proof

P(X \leq x) = P(F^{-1}(U) \leq x) = P(U \leq F(x)) = F(x) \checkmark

📋 Exponential example

$X \sim \text{Exp}(\lambda)$, $F(x) = 1 - e^{-\lambda x}$.
Set $u = 1 - e^{-\lambda x}$, solve: $x = F^{-1}(u) = -\ln(1-u)/\lambda$.
Since $1-U \sim U(0,1)$ also, we can use $x = -\ln(u)/\lambda$.

3. Generalised Inverse Method (Any R.V.)

Generalised inverse

x = F^{-1}(u) := \min\{y : F(y) > u\}

Why generalise? For step functions (discrete CDFs), the regular inverse doesn't exist. The generalised inverse always produces the correct simulated value regardless of distribution type.

1. Linearity of Expectation

Linearity holds for any random variables — no independence required.

Linearity (always true)

$$E[aX + b] = aE[X] + b$$

$$E[X + Y] = E[X] + E[Y]$$

E\!\left[\sum_{i=1}^n X_i\right] = \sum_{i=1}^n E[X_i]

Product rule (independence required)

X \perp Y \implies E[XY] = E[X]\,E[Y]

Key insight: Linearity of expectation always holds. The product rule only holds when $X$ and $Y$ are independent.

2. Variance of Sums

Variance identities

\text{Var}(X) = E[X^2] - (E[X])^2

\text{Var}(aX + b) = a^2\,\text{Var}(X)

Covariance

\text{Cov}(X,Y) = E[XY] - E[X]E[Y] = E[(X-\mu_X)(Y-\mu_Y)]

X \perp Y \implies \text{Cov}(X,Y) = 0 \quad (\text{converse not generally true})

Variance of a sum — general

\text{Var}(X+Y) = \text{Var}(X) + \text{Var}(Y) + 2\,\text{Cov}(X,Y)

\text{Var}\!\left(\sum_{i=1}^n X_i\right) = \sum_{i=1}^n \text{Var}(X_i) + 2\sum_{i < j}\text{Cov}(X_i, X_j)

Independent case

X_i \text{ independent} \implies \text{Var}\!\left(\sum_{i=1}^n X_i\right) = \sum_{i=1}^n \text{Var}(X_i)

X_i \text{ i.i.d.} \implies \text{Var}\!\left(\sum_{i=1}^n X_i\right) = n\,\text{Var}(X_1)

3. MGF of Sums (Independent Variables)

The MGF of a sum of independent variables is the product of their MGFs — the main tool for identifying what distribution a sum follows.

MGF multiplication rule

M_{X+Y}(t) = M_X(t)\cdot M_Y(t), \quad X \perp Y

M_{\sum X_i}(t) = \prod_{i=1}^n M_{X_i}(t), \quad X_i \text{ independent}

Uniqueness: MGF identifies the distribution

M_S(t) = M_Z(t) \text{ for all } t \text{ near } 0 \implies S \overset{d}{=} Z

Key closure results

$X_i \sim \text{Poisson}(\lambda_i)$ indep. $\implies \sum X_i \sim \text{Poisson}(\sum \lambda_i)$
$X_i \sim \text{Exp}(\lambda)$ i.i.d. $\implies \sum_{i=1}^n X_i \sim \text{Gamma}(n,\lambda)$
$X_i \sim N(\mu_i,\sigma_i^2)$ indep. $\implies \sum X_i \sim N(\sum \mu_i, \sum \sigma_i^2)$
$X_i \sim \text{Gamma}(\alpha_i,\beta)$ indep., same $\beta$ $\implies \sum X_i \sim \text{Gamma}(\sum \alpha_i, \beta)$

4. Central Limit Theorem (CLT)

Sums of i.i.d. variables are approximately Normal for large $n$, regardless of the original distribution.

CLT — standardised form

\frac{S_n - n\mu}{\sigma\sqrt{n}} \xrightarrow{d} N(0,1) \quad \text{as } n \to \infty

S_n = \sum_{i=1}^n X_i,\quad \mu=E[X_i],\quad \sigma^2=\text{Var}(X_i)

Normal approximation

S_n \approx N(n\mu,\; n\sigma^2) \quad \text{for large }n

Actuarial application

Portfolio of 1,000 independent claims, each with mean 500 and std dev 200.
$S \approx N(500{,}000,\; 1000 \times 200^2)$, so $\sigma_S = 200\sqrt{1000} \approx 6{,}325$.
95th percentile reserve: $500{,}000 + 1.645 \times 6{,}325 \approx 510{,}400$.

5. Compound Distributions

$S = X_1 + \cdots + X_N$ where $N$ is a random count (e.g. number of claims) and $X_i$ are i.i.d. severities, independent of $N$.

Mean and variance

E[S] = E[N]\cdot E[X]

\text{Var}(S) = E[N]\,\text{Var}(X) + \text{Var}(N)\,(E[X])^2

MGF via PGF

M_S(t) = G_N(M_X(t)), \quad G_N(z) = E[z^N]

Compound Poisson special case ($N \sim \text{Poisson}(\lambda)$)

M_S(t) = \exp(\lambda(M_X(t)-1))

E[S] = \lambda E[X], \quad \text{Var}(S) = \lambda E[X^2]

Example

$N \sim \text{Poisson}(10)$, $X \sim \text{Exp}(1/500)$ (mean 500).
$E[S] = 10 \times 500 = 5{,}000$. Since $E[X^2] = 2 \times 500^2 = 500{,}000$:
$\text{Var}(S) = 10 \times 500{,}000 = 5{,}000{,}000$, $\sigma_S \approx 2{,}236$.

1. Moment Generating Function (MGF)

The MGF $M_X(t)=E[e^{tX}]$ encodes all moments of $X$ in a single function. It uniquely determines the distribution (when it exists in a neighbourhood of $t=0$) and is the key tool for identifying the distribution of a sum.

Definition — connection to the PDF / PMF

M_X(t)=E[e^{tX}]=egin{cases}\displaystyle\sum_x e^{tx}\,p(x)& ext{discrete (sum over PMF)}\[6pt]\displaystyle\int_{-\infty}^\infty e^{tx}f(x)\,dx& ext{continuous (weighted integral of PDF)}\end{cases}

ext{For continuous }X ext{: }M_X(t)=\int e^{tx}f(x)\,dx ext{ is the two-sided Laplace transform of }f ext{ evaluated at }-t

Moment extraction

M_X^{(n)}(0)=E[X^n]\implies E[X]=M_X'(0),\quad E[X^2]=M_X''(0)

ext{Var}(X)=M_X''(0)-[M_X'(0)]^2

ext{Expanding: }M_X(t)=\sum_{n=0}^\infty rac{E[X^n]}{n!}\,t^n=1+E[X]\,t+ rac{E[X^2]}{2}t^2+\cdots

Uniqueness & sums of independent variables

M_X(t)=M_Y(t) ext{ for all }t ext{ near }0\implies X\overset{d}{=}Y\quad ext{(MGF uniquely determines distribution)}

X\perp Y\implies M_{X+Y}(t)=M_X(t)\cdot M_Y(t)

ext{This is why: Poisson}+ ext{Poisson}= ext{Poisson},\; ext{Normal}+ ext{Normal}= ext{Normal},\; ext{Gamma}+ ext{Gamma}= ext{Gamma}

Key MGFs to know

$ ext{Bernoulli}(p)$: $M(t)=1-p+pe^t$
$ ext{Binomial}(n,p)$: $M(t)=(1-p+pe^t)^n$
$ ext{Poisson}(\lambda)$: $M(t)=e^{\lambda(e^t-1)}$
$ ext{Exp}(\lambda)$: $M(t)=\dfrac{\lambda}{\lambda-t}$ for $t{<}\lambda$
$N(\mu,\sigma^2)$: $M(t)=e^{\mu t+\sigma^2 t^2/2}$
$ ext{Gamma}(lpha,eta)$: $M(t)=\left(\dfrac{eta}{eta-t} ight)^lpha$ for $t{<}eta$

2. Probability Generating Function (PGF)

The PGF $G_X(z)=E[z^X]$ is defined for non-negative integer-valued random variables. It encodes the entire PMF as power-series coefficients and is the natural tool for discrete count distributions.

Definition — direct connection to the PMF

G_X(z)=E[z^X]=\sum_{k=0}^\infty P(X=k)\,z^k=p_0+p_1 z+p_2 z^2+\cdots

P(X=k)= rac{G_X^{(k)}(0)}{k!}\quad ext{(}k ext{th derivative at }z=0 ext{, divided by }k! ext{)}

G_X(1)=1\quad ext{(normalisation)},\qquad G_X(0)=P(X=0)

Moments from PGF

E[X]=G_X'(1),\qquad E[X(X-1)]=G_X''(1)

ext{Var}(X)=G_X''(1)+G_X'(1)-[G_X'(1)]^2

Relationship between MGF and PGF

G_X(z)=M_X(\ln z),\qquad M_X(t)=G_X(e^t)

ext{PGF is the MGF with substitution }z=e^t ext{ — they carry the same information for discrete r.v.s}

Sums & compound distributions

X\perp Y\implies G_{X+Y}(z)=G_X(z)\cdot G_Y(z)

S=\sum_{i=1}^N X_i,\;N ext{ random},\;X_i ext{ i.i.d.}\implies G_S(z)=G_N\!\left(G_X(z) ight)\quad ext{(compound PGF)}

Key PGFs to know

$ ext{Bernoulli}(p)$: $G(z)=1-p+pz$
$ ext{Binomial}(n,p)$: $G(z)=(1-p+pz)^n$
$ ext{Poisson}(\lambda)$: $G(z)=e^{\lambda(z-1)}$
$ ext{Geometric}(p)$: $G(z)=\dfrac{pz}{1-(1-p)z}$
$ ext{NegBin}(r,p)$: $G(z)=\left(\dfrac{p}{1-(1-p)z} ight)^r$

MGF vs PGF — when to use which: Use PGF for non-negative integer data (claim counts, number of events) — it directly encodes $P(X=k)$ as coefficients. Use MGF for continuous or mixed distributions — it connects to the PDF via the Laplace transform. Both uniquely determine the distribution when they exist.

Algebraic & Differentiation Basics

Series, summations, differentiation rules, and algebraic tools used throughout actuarial mathematics.

1. Geometric Series

Finite geometric sum

a + ar + ar^2 + \cdots + ar^{n-1} = a\cdot\frac{1-r^n}{1-r}, \quad r \neq 1

Infinite geometric series ($|r| < 1$)

\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}, \qquad \sum_{n=1}^{\infty} r^n = \frac{r}{1-r}

Derived series (differentiate $\sum r^n$)

\sum_{n=0}^{\infty} nr^{n-1} = \frac{1}{(1-r)^2}, \qquad \sum_{n=0}^{\infty} nr^n = \frac{r}{(1-r)^2}

📋 Application: PGF of Geometric

$G(z) = \sum_{k=0}^\infty (1-p)^k p \cdot z^k = p\sum_{k=0}^\infty [(1-p)z]^k = \frac{p}{1-(1-p)z}$ for $|(1-p)z|<1$.

2. Standard Summation Identities

Sum of integers

\sum_{k=1}^{n} k = \frac{n(n+1)}{2}

Sum of squares

\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}

Exponential series

e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}, \quad x \in (-\infty, \infty)

Logarithm series ($|x| < 1$)

-\ln(1-x) = \sum_{n=1}^{\infty} \frac{x^n}{n} = x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots

3. Binomial Theorem

Integer $n$

(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^k b^{n-k}

Negative binomial expansion ($|x| < 1$)

(1-x)^{-r} = \sum_{k=0}^{\infty} \binom{r+k-1}{k} x^k

📋 Application: NB PMF from PGF

$G(z) = \left(\frac{p}{1-(1-p)z}\right)^r = p^r \cdot (1-(1-p)z)^{-r}$. Expand using negative binomial theorem to recover $P(X=k) = \binom{r+k-1}{k}p^r(1-p)^k$.

4. Changing Order of Summation / Integration

Frequently used to simplify double sums and integrals. The key is to identify the same region of integration and re-express it with swapped limits.

Constant bounds — trivially swap

\sum_{x=a}^{b}\sum_{y=c}^{d} f(x,y) = \sum_{y=c}^{d}\sum_{x=a}^{b} f(x,y)

Triangular region — classic example

\sum_{x=0}^{a}\sum_{y=x}^{a} f(x,y) = \sum_{y=0}^{a}\sum_{x=0}^{y} f(x,y)

Continuous version

\int_0^a \int_x^a f(x,y)\,dy\,dx = \int_0^a \int_0^y f(x,y)\,dx\,dy

Strategy: Draw the region of integration. Identify all $(x,y)$ pairs in the region. Then re-express the boundary conditions in terms of the new outer variable.

1. Taylor Expansion

General Taylor series around 0 (Maclaurin)

f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}\,x^n

Key expansions

e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots

\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots, \quad |x| < 1

(1+x)^\alpha = 1 + \alpha x + \frac{\alpha(\alpha-1)}{2!}x^2 + \cdots, \quad |x| < 1

2. Exponential and Logarithm Laws

Exponential rules

e^{a+b} = e^a e^b, \quad (e^a)^b = e^{ab}, \quad e^0 = 1, \quad \frac{d}{dx}e^x = e^x

General exponential $a^x$ (from PDF)

\frac{d}{dx}\left(a^x\right) = a^x \ln a, \quad a > 0,\; a \neq 1

\int a^x\,dx = \frac{a^x}{\ln a} + C, \qquad \text{since } a^x = e^{x \ln a}

\text{Note: } e^x \text{ is the special case } a=e \text{ since } \ln e = 1

Logarithm rules

\ln(ab) = \ln a + \ln b, \quad \ln(a/b) = \ln a - \ln b, \quad \ln(a^n) = n\ln a

\ln e^x = x, \quad e^{\ln x} = x, \quad \frac{d}{dx}\ln x = \frac{1}{x}

3. Gamma and Beta Functions

Gamma function

\Gamma(n) = \int_0^\infty x^{n-1} e^{-x}\,dx, \qquad \Gamma(n) = (n-1)!\text{ for integer }n

\Gamma(n+1) = n\,\Gamma(n), \qquad \Gamma(1/2) = \sqrt{\pi}

Beta function

B(\alpha, \beta) = \int_0^1 x^{\alpha-1}(1-x)^{\beta-1}\,dx = \frac{\Gamma(\alpha)\,\Gamma(\beta)}{\Gamma(\alpha+\beta)}

📋 Why it matters

The Beta function normalises the Gamma and Beta distributions. E.g. $\int_0^\infty x^{\alpha-1}e^{-\beta x}dx = \Gamma(\alpha)/\beta^\alpha$ — this is the normalisation constant for the Gamma PDF.

4. Differentiation Rules

Basic rules

\frac{d}{dx}[af(x)+bg(x)] = af'(x)+bg'(x) \quad \text{(linearity)}

\frac{d}{dx}[f(x)g(x)] = f'(x)g(x)+f(x)g'(x) \quad \text{(product rule)}

\frac{d}{dx}\!\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2} \quad \text{(quotient rule)}

Chain rule

\frac{d}{dx}f(g(x)) = f'(g(x))\cdot g'(x)

\frac{d}{dx}e^{g(x)} = e^{g(x)}g'(x), \quad \frac{d}{dx}[g(x)]^n = n[g(x)]^{n-1}g'(x)

Key derivatives

\frac{d}{dx}x^n = nx^{n-1}, \quad \frac{d}{dx}e^x = e^x, \quad \frac{d}{dx}\ln x = \frac{1}{x}

\frac{d}{dx}a^x = a^x\ln a, \quad \frac{d}{dx}\log_a x = \frac{1}{x\ln a}

\frac{d}{dx}\sin x = \cos x, \quad \frac{d}{dx}\cos x = -\sin x, \quad \frac{d}{dx}\tan x = \sec^2 x

Leibniz rule — differentiation under the integral sign

\frac{d}{d\theta}\int_{a(\theta)}^{b(\theta)} f(x,\theta)\,dx = \int_{a(\theta)}^{b(\theta)} \frac{\partial f}{\partial \theta}\,dx + f(b,\theta)b'(\theta) - f(a,\theta)a'(\theta)

Differentiating the MGF to get moments

M_X(t) = E[e^{tX}] \implies M_X^{(n)}(0) = E[X^n]

E[X] = M_X'(0), \quad E[X^2] = M_X''(0), \quad \text{Var}(X) = M_X''(0) - [M_X'(0)]^2

Moments from MGF — example

$X \sim \text{Exp}(\lambda)$, $M_X(t) = \lambda/(\lambda-t)$.
$M_X'(t) = \lambda/(\lambda-t)^2 \Rightarrow E[X] = 1/\lambda$.
$M_X''(t) = 2\lambda/(\lambda-t)^3 \Rightarrow E[X^2] = 2/\lambda^2$, so $\text{Var}(X)=1/\lambda^2$.

5. L'Hopital's Rule and Useful Limits

L'Hopital's rule

\lim_{x\to a}\frac{f(x)}{g(x)} = \frac{0}{0}\text{ or }\frac{\infty}{\infty} \implies \lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)}

Useful limits in actuarial work

\lim_{x\to\infty} x^n e^{-\lambda x} = 0 \quad (\lambda > 0) \quad \text{exponential dominates polynomial}

\lim_{n\to\infty}\!\left(1+\frac{x}{n}\right)^n = e^x, \quad \lim_{x\to 0}\frac{e^x-1}{x}=1, \quad \lim_{x\to 0}\frac{\ln(1+x)}{x}=1

1. Weighted Averages

Arithmetic weighted average of $x_1, \ldots, x_n$

\bar{x}_w = \frac{\sum_{i=1}^n w_i x_i}{\sum_{i=1}^n w_i}

\text{(Simple average when } w_i = 1/n \text{: } \bar{x} = \frac{1}{n}\sum_{i=1}^n x_i)

Geometric weighted average ($\sum w_i = 1$)

\prod_{i=1}^n x_i^{w_i}

AM–GM inequality

\sum_{i=1}^n w_i x_i \geq \prod_{i=1}^n x_i^{w_i} \qquad (w_i \geq 0,\ \sum w_i = 1)

2. Continuous Weighted Average

Weighted average of function $f$ on $[a,b]$ with weight $w(x) \geq 0$

\frac{\int_a^b w(x)f(x)\,dx}{\int_a^b w(x)\,dx}

Simple average ($w(x)=1$)

\frac{\int_a^b f(x)\,dx}{b-a}

Key Inequalities in Actuarial Maths

Exponential lower bound

e^{-x} > 1 - x, \quad x > 0

Logarithm lower bound

-\ln(1-x) > x, \quad 0 < x < 1

Variance bound

x(1-x) \leq \frac{1}{4}, \quad 0 \leq x \leq 1

Use: The bound $x(1-x) \leq 1/4$ is useful when bounding $\text{Var}(X) = p(1-p) \leq 1/4$ for a Bernoulli$(p)$ variable — guarantees variance is at most $1/4$ regardless of $p$.

1. First-Order Linear ODE

The general form is $f'(t) = a(t)f(t) + b(t)$. Interpreted as: money $f(0)$ in an account earning interest $a(t)$ with continuous deposits at rate $b(t)$.

Solution via integrating factor $I(t) = e^{-\int_0^t a(s)ds}$

f(t) = f(0)\,e^{\int_0^t a(s)\,ds} + \int_0^t e^{\int_s^t a(u)\,du}\,b(s)\,ds

Constant coefficient case $a(t) = a$

f(t) = f(0)\,e^{at} + \int_0^t e^{a(t-s)}\,b(s)\,ds

📋 Actuarial interpretation

First term: the initial deposit $f(0)$ accumulates to $f(0)e^{\int_0^t a(s)ds}$.
Second term: each infinitesimal deposit $b(s)ds$ made at time $s$ accumulates by factor $e^{\int_s^t a(u)du}$ to time $t$. The integral adds all such accumulated deposits.

2. Second-Order ODE with Constant Coefficients

Form: $f''(t) + bf'(t) + cf(t) = 0$, characteristic eq: $r^2 + br + c = 0$

\Delta = b^2 - 4c > 0 \implies f(t) = k_1 e^{r_1 t} + k_2 e^{r_2 t}

\Delta = 0 \implies r = -b/2,\quad f(t) = k_1 e^{rt} + k_2 t e^{rt}

\Delta < 0 \implies r = d_1 \pm id_2,\quad f(t) = k_1 e^{d_1 t}\cos(d_2 t) + k_2 e^{d_1 t}\sin(d_2 t)

3. Pair of ODEs with Constant Coefficients

System: $f'(t) = af(t) + bg(t)$, $g'(t) = cf(t) + dg(t)$

$$f''(t) = (a+d)f'(t) + (bc-ad)f(t)$$

f(0) = y_1,\quad f'(0) = ay_1 + by_2, \qquad g(t) = \frac{1}{b}[f'(t) - af(t)]

Integration Basics

Integration techniques — with focus on actuarial applications involving continuous cash flows, survival functions and moment calculations.

1. Standard Definite and Indefinite Integrals

Power rule

\int x^n\,dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1; \qquad \int_a^b x^n\,dx = \frac{b^{n+1}-a^{n+1}}{n+1}

Exponential

\int e^{ax}\,dx = \frac{e^{ax}}{a} + C; \qquad \int_0^\infty e^{-ax}\,dx = \frac{1}{a}, \quad a > 0

Logarithm

\int \frac{1}{x}\,dx = \ln|x| + C; \qquad \int_a^b \frac{1}{x}\,dx = \ln b - \ln a = \ln\frac{b}{a}

Key actuarial integrals — Gamma function

\int_0^\infty x^{n-1} e^{-x}\,dx = \Gamma(n), \qquad \int_0^\infty x^{n-1} e^{-\lambda x}\,dx = \frac{\Gamma(n)}{\lambda^n}

Moments via survival function

E[X^n] = n\int_0^\infty x^{n-1}[1-F(x)]\,dx = n\int_0^\infty x^{n-1}S(x)\,dx

1. Integration by Parts

Formula ($u\,dv = uv - v\,du$)

\int u\,dv = uv - \int v\,du

\int_a^b u\,v'\,dx = [uv]_a^b - \int_a^b v\,u'\,dx

📋 Example: $E[X]$ for $X \sim \text{Exp}(\lambda)$ via by-parts

$E[X] = \int_0^\infty x \lambda e^{-\lambda x}dx$. Let $u=x$, $dv=\lambda e^{-\lambda x}dx$.
$= [x\cdot(-e^{-\lambda x})]_0^\infty + \int_0^\infty e^{-\lambda x}dx = 0 + \frac{1}{\lambda} = \frac{1}{\lambda}$.
Much easier: $E[X] = \int_0^\infty S(x)dx = \int_0^\infty e^{-\lambda x}dx = 1/\lambda$.

2. Substitution

Change of variable $u = g(x)$

\int f(g(x))g'(x)\,dx = \int f(u)\,du

📋 Example: Exponential moments

$E[X^2] = \int_0^\infty x^2 \lambda e^{-\lambda x}dx$. Let $u = \lambda x$, $x = u/\lambda$, $dx = du/\lambda$:
$= \frac{1}{\lambda^2}\int_0^\infty u^2 e^{-u}du = \frac{\Gamma(3)}{\lambda^2} = \frac{2}{\lambda^2}$.

3. Improper Integrals

Definition

\int_a^\infty f(x)\,dx = \lim_{b \to \infty} \int_a^b f(x)\,dx

Key convergence result

\int_0^\infty x^n e^{-\lambda x}\,dx = \frac{n!}{\lambda^{n+1}}, \quad \lambda > 0,\ n = 0,1,2,\ldots

Tip: Use L'Hôpital's rule for $\lim_{x\to\infty} x^n e^{-\lambda x} = 0$ (exponential decay beats any polynomial).

1. Continuous Cash Flows

Let $a(t)$ be the rate of payment at time $t$, and $A(t)$ the total payment accumulated up to time $t$.

Fundamental relations

A(t) = \int_0^t a(s)\,ds, \qquad dA(t) = a(t)\,dt

A(s+ds) - A(s) \approx a(s)\,ds \quad \text{(payment over small interval)}

2. Force of Interest

Definition

\delta(t) = \lim_{dt\to 0}\frac{A(t+dt)-A(t)}{A(t)\,dt} = \frac{A'(t)}{A(t)} = \frac{d\ln A(t)}{dt}

Present value of continuous cash flow from 0 to $t$

\bar{A}(t) = \int_0^t e^{-\int_0^s \delta(y)\,dy}\,a(s)\,ds

Accumulated value at time $t$

A(t) = \int_0^t e^{\int_s^t \delta(y)\,dy}\,a(s)\,ds

Connection to ODE: $A(t+dt) - A(t) \approx A(t)\delta(t)dt$, so $A'(t) = \delta(t)A(t)$. This is exactly the ODE $f'(t) = a(t)f(t)$ with $b(t)=0$!

Changing the Order of Integration

The strategy: draw the region of integration in the $xy$-plane, identify all $(x,y)$ points in the region, then re-express boundaries for the swapped order.

Classic triangular region

\int_0^a\int_x^a f(x,y)\,dy\,dx = \int_0^a\int_0^y f(x,y)\,dx\,dy

Square root boundary

\int_0^1\int_{x^2}^{\sqrt{x}} f(x,y)\,dy\,dx = \int_0^1\int_{y^2}^{\sqrt{y}} f(x,y)\,dx\,dy

Quadratic upper bound (splitting required)

\int_1^2\int_0^{x^2} f(x,y)\,dy\,dx = \int_0^1\int_2^2 f(x,y)\,dx\,dy + \int_1^4\int_{\sqrt{y}}^{2} f(x,y)\,dx\,dy

📋 Why this matters in actuarial work

$E[XY] = \int_0^\infty \int_0^\infty xy\,f(x,y)\,dx\,dy$. If the joint density has a triangular support (e.g. $0 \leq x \leq y$), swapping the order of integration often converts a hard integral into a product of standard ones.

Step-by-step method:
1. Write out the original region as inequalities: e.g. $0 \leq x \leq a$, $x \leq y \leq a$.
2. Combine: $0 \leq x \leq y \leq a$.
3. Re-read as: outer variable $y$ from 0 to $a$, inner variable $x$ from 0 to $y$.
4. Split into pieces if the boundary changes character.

Statistics Basics

Chi-squared, t, F and Z — origins, what they test, and key test statistics for actuarial inference.

1. Sample Mean

The sample mean $\bar{X}$ estimates $\mu$. It is unbiased, consistent, and approximately Normal for large $n$ by the CLT.

Definition & properties

\bar{X} = \frac{1}{n}\sum_{i=1}^n X_i, \qquad E[\bar{X}]=\mu, \qquad \operatorname{Var}(\bar{X})=\frac{\sigma^2}{n}

\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}\sim N(0,1) \text{ exactly when }X_i\sim N(\mu,\sigma^2)\text{; approximately for large }n\text{ by CLT}

Actuarial use

$\bar{X}$ estimates expected claim cost per policy. The standard error $\sigma/\sqrt{n}$ quantifies precision — larger portfolios give tighter estimates.

2. Sample Variance

$S^2$ estimates $\sigma^2$. The divisor $n-1$ (not $n$) makes it unbiased — one degree of freedom is lost estimating $\mu$ with $\bar{X}$.

Definition (unbiased)

S^2 = \frac{1}{n-1}\sum_{i=1}^n(X_i-\bar{X})^2 = \frac{1}{n-1}\Bigl(\sum X_i^2 - n\bar{X}^2\Bigr)

E[S^2]=\sigma^2 \qquad \frac{(n-1)S^2}{\sigma^2}\sim\chi^2(n-1) \text{ when }X_i\sim N(\mu,\sigma^2)

\bar{X}\perp S^2 \text{ when }X_i\overset{\text{i.i.d.}}{\sim}N(\mu,\sigma^2)\quad\text{(unique to the Normal)}

Why $n-1$? Using $n$ gives the MLE but is biased: $E\bigl[\tfrac{1}{n}\sum(X_i-\bar{X})^2\bigr]=\tfrac{n-1}{n}\sigma^2$. Bessel's correction divides by $n-1$ to fix this.

3. Sample Quantiles & Empirical CDF

Order statistics & quantiles

X_{(1)}\leq X_{(2)}\leq\cdots\leq X_{(n)}, \qquad \hat{Q}(p)\approx X_{(\lceil np\rceil)}

\text{Median}=\hat{Q}(0.5),\quad Q_1=\hat{Q}(0.25),\quad Q_3=\hat{Q}(0.75),\quad\text{IQR}=Q_3-Q_1

Empirical CDF

\hat{F}(x)=\frac{1}{n}\sum_{i=1}^n\mathbf{1}(X_i\leq x)\xrightarrow{p}F(x)\quad\text{(Glivenko-Cantelli)}

Actuarial use

$\hat{Q}(0.99)$ of claim sizes = empirical 99th percentile = non-parametric VaR. The empirical CDF is compared against a fitted parametric $F$ to assess model quality.

4. Key Sampling Distributions (Summary)

When $X_i\overset{\text{i.i.d.}}{\sim}N(\mu,\sigma^2)$

\bar{X}\sim N\!\left(\mu,\tfrac{\sigma^2}{n}\right), \qquad \frac{(n-1)S^2}{\sigma^2}\sim\chi^2(n-1)

\frac{\bar{X}-\mu}{S/\sqrt{n}}\sim t(n-1)\;(\sigma\text{ unknown}), \qquad \frac{S_1^2/\sigma_1^2}{S_2^2/\sigma_2^2}\sim F(n_1-1,n_2-1)

1. Method of Moments (MoM)

Match theoretical moments $E[X^k]$ to sample moments $\tfrac{1}{n}\sum X_i^k$. Solve the resulting system for the parameters. Simple and always gives closed-form estimates.

Procedure

\text{Set }E[X^k]=\frac{1}{n}\sum_{i=1}^nX_i^k\text{ for }k=1,\ldots,p\text{ and solve for }\hat{\theta}_1,\ldots,\hat{\theta}_p

Example — Gamma($\alpha,\beta$) rate form

$E[X]=\alpha/\beta$ and $\operatorname{Var}(X)=\alpha/\beta^2$. Given sample $\bar{x}$ and $s^2$:
$\hat{\beta}=\bar{x}/s^2$, $\;\hat{\alpha}=\bar{x}^2/s^2$. Closed form — no optimisation needed.

Example — Exponential($\lambda$)

$E[X]=1/\lambda$. Set $1/\hat{\lambda}=\bar{X}$, giving $\hat{\lambda}_{\text{MoM}}=1/\bar{X}$. Coincides with MLE here.

Pros & cons: MoM is easy to compute and always gives a closed form. But it is generally less efficient than MLE (higher variance) and the estimates may occasionally fall outside the valid parameter space.

2. Maximum Likelihood Estimation (MLE)

Choose parameter values making the observed data most probable. Asymptotically unbiased, consistent, and achieves the Cramér-Rao lower bound — the gold standard in actuarial statistics.

Likelihood & log-likelihood

L(\theta)=\prod_{i=1}^nf(x_i;\theta), \qquad \ell(\theta)=\ln L(\theta)=\sum_{i=1}^n\ln f(x_i;\theta)

Score equation & asymptotic normality

\hat{\theta}_{\text{MLE}}:\;\frac{\partial\ell}{\partial\theta}=0

\sqrt{n}(\hat{\theta}_{\text{MLE}}-\theta_0)\xrightarrow{d}N\!\left(0,\,\mathcal{I}(\theta_0)^{-1}\right),\quad\mathcal{I}(\theta)=-E[\ell''(\theta)]\text{ (Fisher information)}

Invariance property

\text{If }\hat{\theta}\text{ is MLE of }\theta,\text{ then }g(\hat{\theta})\text{ is MLE of }g(\theta)

Example — Exponential($\lambda$)

$\ell(\lambda)=n\ln\lambda-\lambda\sum x_i$. Score: $n/\lambda-\sum x_i=0\Rightarrow\hat{\lambda}=1/\bar{x}$.
By invariance, MLE of mean $=1/\hat{\lambda}=\bar{x}$.

Example — Normal($\mu,\sigma^2$)

Score equations give $\hat{\mu}=\bar{x}$ and $\hat{\sigma}^2=\tfrac{1}{n}\sum(x_i-\bar{x})^2$ (biased — uses $n$ not $n-1$, but consistent).

3. Cramér-Rao Lower Bound

Variance lower bound for unbiased estimators

\operatorname{Var}(\hat{\theta})\geq\frac{1}{\mathcal{I}(\theta)}=\frac{1}{-E[\ell''(\theta)]}

\text{An efficient estimator achieves this bound. MLE achieves it asymptotically.}

4. MoM vs MLE — Quick Comparison

Comparison table

MoM MLE Computation Closed form May need numerical opt. Efficiency Generally lower Asymptotically efficient Bias Can be biased Asymptotically unbiased Best for Quick initial estimates Standard fitting

	MoM	MLE
Computation	Closed form	May need numerical opt.
Efficiency	Generally lower	Asymptotically efficient
Bias	Can be biased	Asymptotically unbiased
Best for	Quick initial estimates	Standard fitting

How the Four Distributions Relate

All four statistical test distributions are built from the standard Normal $Z \sim N(0,1)$. Understanding this family tree is the key to knowing which test to use and why.

The family tree

Z \sim N(0,1) \xrightarrow{\text{square}} Z^2 \sim \chi^2(1) \xrightarrow{\text{sum}} \chi^2(k)

t(\nu) = \frac{Z}{\sqrt{\chi^2(\nu)/\nu}} \qquad F(d_1,d_2) = \frac{\chi^2(d_1)/d_1}{\chi^2(d_2)/d_2}

t^2(\nu) \sim F(1,\nu) \qquad \chi^2(k) \xrightarrow{k\to\infty} N(0,1) \qquad t(\nu) \xrightarrow{\nu\to\infty} N(0,1)

Quick reference — which test for which question

Test Needs Tests Z-test $\sigma$ known, $n$ large One/two means; proportion t-test $\sigma$ unknown, Normal data One/two means; paired differences $\chi^2$-test Categorical/Normal data Variance; goodness-of-fit; independence F-test Two Normal populations Equality of variances; ANOVA; regression

Test	Needs	Tests
Z-test	$\sigma$ known, $n$ large	One/two means; proportion
t-test	$\sigma$ unknown, Normal data	One/two means; paired differences
$\chi^2$-test	Categorical/Normal data	Variance; goodness-of-fit; independence
F-test	Two Normal populations	Equality of variances; ANOVA; regression

Golden rule: Always identify what you are testing (mean? variance? fit? independence?), whether $\sigma$ is known, and whether the data is Normal. These three questions determine the correct test every time.

Where It Comes From

The chi-squared distribution arises entirely from squaring and summing standard Normal random variables. There is no deeper mystery — it is simply $\chi^2(k) = Z_1^2 + Z_2^2 + \cdots + Z_k^2$ where each $Z_i \sim N(0,1)$ independently.

Origin

\chi^2(k) = \sum_{i=1}^k Z_i^2, \quad Z_i \overset{\text{i.i.d.}}{\sim} N(0,1)

E[\chi^2(k)] = k \qquad \text{Var}[\chi^2(k)] = 2k

Key relationships

\chi^2(k) = \text{Gamma}\!\left(\frac{k}{2},\, \frac{1}{2}\right) \quad \text{(rate parametrisation)}

\chi^2(k_1) + \chi^2(k_2) = \chi^2(k_1+k_2) \quad \text{(additivity — independent)}

\chi^2(k) \xrightarrow{d} N(k, 2k) \quad \text{as } k \to \infty \text{ by CLT}

Sample variance connection

\frac{(n-1)S^2}{\sigma^2} \sim \chi^2(n-1), \quad S^2 = \frac{1}{n-1}\sum_{i=1}^n (X_i - \bar{X})^2

\text{(requires } X_i \overset{\text{i.i.d.}}{\sim} N(\mu, \sigma^2)\text{; loses 1 df for estimating }\mu\text{)}

Test 1 — Variance Test (one population)

Test whether the true variance $\sigma^2$ equals a specified value $\sigma_0^2$, given a sample from a Normal population.

Hypotheses

H_0: \sigma^2 = \sigma_0^2 \qquad H_1: \sigma^2 \neq \sigma_0^2 \text{ (or one-sided)}

Test statistic

\chi^2 = \frac{(n-1)S^2}{\sigma_0^2} \sim \chi^2(n-1) \quad \text{under } H_0

Decision rule (two-tailed at level $\alpha$)

\text{Reject } H_0 \text{ if } \chi^2 < \chi^2_{\alpha/2}(n-1) \text{ or } \chi^2 > \chi^2_{1-\alpha/2}(n-1)

Actuarial example

An insurer claims claim processing times have $\sigma^2 = 25$ mins$^2$. A sample of $n=20$ gives $S^2 = 35$.
$\chi^2 = 19 \times 35/25 = 26.6$ on df=19.
$\chi^2_{0.975}(19)=32.85$, $\chi^2_{0.025}(19)=8.91$. Since $8.91 < 26.6 < 32.85$, fail to reject $H_0$ at 5%.

Test 2 — Goodness-of-Fit Test

Test whether observed frequencies match a theoretical distribution. The test statistic compares observed counts $O_i$ to expected counts $E_i$ across $k$ categories.

Hypotheses

H_0: \text{data follows the specified distribution} \qquad H_1: \text{it does not}

Pearson chi-squared test statistic

\chi^2 = \sum_{i=1}^k \frac{(O_i - E_i)^2}{E_i} \sim \chi^2(k - 1 - p) \quad \text{under } H_0

\text{where } p = \text{number of parameters estimated from data}

Decision rule

\text{Reject } H_0 \text{ if } \chi^2 > \chi^2_{1-\alpha}(k-1-p)

\text{Rule of thumb: require }E_i\geq 5\text{ in every cell before applying the }\chi^2\text{ approximation}

Cell merging procedure — step by step

\textbf{Step 1: }\text{Compute all expected counts }E_i=n\cdot p_i

\textbf{Step 2: }\text{Identify all cells with }E_i{<}5

\textbf{Step 3: }\text{Merge each such cell with an adjacent cell (typically the nearest tail cell)}

\textbf{Step 4: }\text{Recompute merged }E_i=\sum_{\text{merged}} E_j\text{ and }O_i=\sum_{\text{merged}} O_j

\textbf{Step 5: }\text{Repeat until all cells have }E_i\geq 5\text{. Update }k\text{ and df}=k-1-p

Worked merging example

Fitting Poisson($\hat{\lambda}=0.5$) to counts 0, 1, 2, 3, 4+ with $n=100$:
$E_0=60.7,\;E_1=30.3,\;E_2=7.6,\;E_3=1.3,\;E_{4+}=0.1$
Cells 3 and 4+ both have $E_i{<}5$ → merge: $E_{3+}=1.3+0.1=1.4$, still ${<}5$
Merge again with cell 2: $E_{2+}=7.6+1.4=9.0\geq 5$ ✓
Final cells: 0, 1, 2+ → $k=3$, df $=3-1-1=1$. Compare to $\chi^2_{0.95}(1)=3.84$.

Actuarial example

Fitting Poisson($\hat{\lambda}=2$) to 5 claim-count categories, $n=200$. Estimated $p=1$ parameter.
df $= 5 - 1 - 1 = 3$. Compute $\sum(O_i-E_i)^2/E_i$; compare to $\chi^2_{0.95}(3) = 7.815$.
If $\chi^2 > 7.815$, reject Poisson at 5% — the model is a poor fit.

Test 3 — Independence Test (Contingency Table)

Test whether two categorical variables are independent. Data is arranged in an $r \times c$ contingency table with row totals $R_i$, column totals $C_j$ and grand total $n$.

Expected frequency under independence

E_{ij} = \frac{R_i \cdot C_j}{n}

Test statistic

\chi^2 = \sum_{i=1}^r \sum_{j=1}^c \frac{(O_{ij} - E_{ij})^2}{E_{ij}} \sim \chi^2\!\left((r-1)(c-1)\right) \quad \text{under } H_0

Decision rule

\text{Reject independence if } \chi^2 > \chi^2_{1-\alpha}\!\left((r-1)(c-1)\right)

Actuarial example

Testing whether claim type (3 types) is independent of policyholder age band (4 bands) — a $3 \times 4$ table.
df $= (3-1)(4-1) = 6$. Compare $\chi^2$ statistic to $\chi^2_{0.95}(6) = 12.59$.
If $\chi^2 > 12.59$, claim type and age band are NOT independent — useful for pricing segmentation.

Where It Comes From

The Z-test uses the standard Normal distribution directly. It applies when the population standard deviation $\sigma$ is known, or when the sample is large enough that $S \approx \sigma$ by the LLN. The test statistic is simply a standardised mean — it measures how many standard errors the sample mean is from the hypothesised value.

Origin — standardised sample mean

\bar{X} \sim N\!\left(\mu, \frac{\sigma^2}{n}\right) \implies Z = \frac{\bar{X} - \mu_0}{\sigma/\sqrt{n}} \sim N(0,1) \quad \text{under } H_0

When to use Z vs t: Use Z when $\sigma$ is known or $n \gtrsim 30$ (CLT kicks in). Use t when $\sigma$ is unknown and $n$ is small — the extra uncertainty in estimating $\sigma$ widens the distribution.

Test 1 — One-Sample Mean Test

Hypotheses & test statistic

H_0: \mu = \mu_0 \qquad Z = \frac{\bar{X} - \mu_0}{\sigma/\sqrt{n}} \sim N(0,1)

Decision rules

\text{Two-tailed: Reject } H_0 \text{ if } |Z| > z_{1-\alpha/2}

\text{Upper-tailed: Reject } H_0 \text{ if } Z > z_{1-\alpha}

\text{Lower-tailed: Reject } H_0 \text{ if } Z < -z_{1-\alpha}

\text{Common critical values: } z_{0.95}=1.645,\; z_{0.975}=1.960,\; z_{0.995}=2.576

Actuarial example

Known $\sigma = £200$. Sample of $n=100$ claims gives $\bar{X} = £1{,}050$. Test $H_0: \mu = £1{,}000$.
$Z = (1050 - 1000)/(200/\sqrt{100}) = 50/20 = 2.5$.
Since $|2.5| > 1.96$, reject $H_0$ at 5%. Mean claim cost has significantly increased.

Test 2 — Two-Sample Mean Test

Test statistic ($\sigma_1, \sigma_2$ known)

Z = \frac{(\bar{X}_1 - \bar{X}_2) - \delta_0}{\sqrt{\sigma_1^2/n_1 + \sigma_2^2/n_2}} \sim N(0,1) \quad \text{under } H_0: \mu_1 - \mu_2 = \delta_0

Test 3 — Proportion Test

Test statistic for $H_0: p = p_0$

Z = \frac{\hat{p} - p_0}{\sqrt{p_0(1-p_0)/n}} \sim N(0,1) \quad \text{under } H_0

\text{Valid for large } n \text{ (CLT): need } np_0 \geq 5 \text{ and } n(1-p_0) \geq 5

Actuarial example

Industry claim rate is $p_0 = 0.08$. Insurer observes 12 claims in $n=100$ policies ($\hat{p}=0.12$).
$Z = (0.12-0.08)/\sqrt{0.08 \times 0.92/100} = 0.04/0.02714 = 1.474$.
Since $1.474 < 1.645$, fail to reject $H_0$ at 5% — no significant difference from industry rate.

Confidence Intervals

CI for $\mu$ ($\sigma$ known)

\bar{X} \pm z_{1-\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}

Duality: CI and hypothesis test

\text{Reject } H_0: \mu = \mu_0 \iff \mu_0 \notin \left(\bar{X} - z_{1-\alpha/2}\frac{\sigma}{\sqrt{n}},\; \bar{X} + z_{1-\alpha/2}\frac{\sigma}{\sqrt{n}}\right)

Where It Comes From

When $\sigma$ is unknown, we estimate it with $S$. Replacing $\sigma$ by $S$ in the Z-statistic introduces extra randomness — the statistic is no longer Normal. William Gosset (publishing as "Student") showed in 1908 that it follows a $t$-distribution with heavier tails than the Normal, reflecting the uncertainty in estimating $\sigma$.

Origin — $T$ is $Z$ divided by $\sqrt{\chi^2/\nu}$

T = \frac{Z}{\sqrt{\chi^2(\nu)/\nu}}, \quad Z \perp \chi^2(\nu) \implies T \sim t(\nu)

For a sample mean with unknown $\sigma$

T = \frac{\bar{X} - \mu_0}{S/\sqrt{n}} \sim t(n-1) \quad \text{under } H_0

\text{where } S^2 = \frac{1}{n-1}\sum(X_i-\bar{X})^2 \text{ estimates } \sigma^2 \text{, using } n-1 \text{ df}

Key property: As $\nu \to \infty$, $t(\nu) \to N(0,1)$. For $\nu \geq 30$ the $t$ and $Z$ critical values are nearly identical. The $t$ distribution has heavier tails — it is more conservative (harder to reject $H_0$), which is appropriate when estimating $\sigma$ from a small sample.

Test 1 — One-Sample t-Test

Hypotheses & test statistic

H_0: \mu = \mu_0 \qquad T = \frac{\bar{X} - \mu_0}{S/\sqrt{n}} \sim t(n-1)

Decision rule & CI

\text{Two-tailed: Reject } H_0 \text{ if } |T| > t_{1-\alpha/2}(n-1)

\text{CI for }\mu: \quad \bar{X} \pm t_{1-\alpha/2}(n-1) \cdot \frac{S}{\sqrt{n}}

Actuarial example

$n=16$ claim settlements: $\bar{X} = £1{,}200$, $S = £300$. Test $H_0: \mu = £1{,}000$.
$T = (1200-1000)/(300/4) = 200/75 = 2.667$ on df $=15$.
$t_{0.975}(15) = 2.131$. Since $2.667 > 2.131$, reject $H_0$ at 5% — mean significantly above £1,000.

Test 2 — Two-Sample t-Test (Equal Variances)

Pooled variance estimate

S_p^2 = \frac{(n_1-1)S_1^2 + (n_2-1)S_2^2}{n_1+n_2-2}

Test statistic for $H_0: \mu_1 = \mu_2$

T = \frac{\bar{X}_1 - \bar{X}_2}{S_p\sqrt{1/n_1 + 1/n_2}} \sim t(n_1+n_2-2) \quad \text{under } H_0

Test 3 — Paired t-Test

When observations are naturally paired (e.g. before/after for the same policyholder), work with the differences $D_i = X_{1i} - X_{2i}$.

Test statistic for $H_0: \mu_D = 0$

T = \frac{\bar{D}}{S_D/\sqrt{n}} \sim t(n-1), \quad \bar{D} = \frac{1}{n}\sum D_i,\quad S_D^2 = \frac{\sum(D_i-\bar{D})^2}{n-1}

Actuarial example

Comparing 10 policyholders' claim costs before ($X_1$) and after ($X_2$) a safety intervention. Compute differences $D_i = X_{1i} - X_{2i}$, then test whether $\bar{D} = 0$.
The paired test is more powerful than two-sample t because it removes between-policyholder variability.

Where It Comes From

The F-distribution is the ratio of two independent chi-squared variables each divided by their degrees of freedom. Because sample variances $S^2$ follow scaled chi-squared distributions, the ratio of two sample variances follows an F-distribution — making it ideal for comparing variability.

Origin — ratio of two chi-squared variables

F(d_1, d_2) = \frac{\chi^2(d_1)/d_1}{\chi^2(d_2)/d_2}, \quad \chi^2(d_1) \perp \chi^2(d_2)

Key properties

E[F] = \frac{d_2}{d_2-2} \text{ for } d_2 > 2 \qquad \text{(always slightly above 1)}

\frac{1}{F(d_1,d_2)} \sim F(d_2,d_1) \qquad T^2(\nu) \sim F(1,\nu)

\text{Upper critical value: } F_{1-\alpha}(d_1,d_2) = \frac{1}{F_{\alpha}(d_2,d_1)}

Test 1 — Equality of Two Variances

Before running a two-sample t-test with pooled variance, you should verify the equal-variance assumption using the F-test.

Hypotheses & test statistic

H_0: \sigma_1^2 = \sigma_2^2 \qquad F = \frac{S_1^2}{S_2^2} \sim F(n_1-1, n_2-1) \quad \text{under } H_0

\text{Convention: put the larger } S^2 \text{ in the numerator (one-tailed test)}

Decision rule

\text{Reject } H_0 \text{ if } F > F_{1-\alpha/2}(n_1-1, n_2-1)

Actuarial example

Two portfolios: $n_1=21$, $S_1^2=1600$ and $n_2=16$, $S_2^2=900$.
$F = 1600/900 = 1.778$ on $F(20,15)$. $F_{0.95}(20,15) \approx 2.33$.
Since $1.778 < 2.33$, fail to reject equal variances — the pooled t-test is appropriate.

Test 2 — ANOVA (One-Way)

Analysis of Variance tests whether $k$ group means are equal by comparing the variance between groups to the variance within groups. If between-group variation is much larger than within-group variation, the means differ.

Decomposition

SS_{\text{Total}} = SS_{\text{Between}} + SS_{\text{Within}}

SS_{\text{Between}} = \sum_{j=1}^k n_j(\bar{X}_j - \bar{X})^2, \quad SS_{\text{Within}} = \sum_{j=1}^k \sum_{i=1}^{n_j}(X_{ij}-\bar{X}_j)^2

Test statistic for $H_0: \mu_1 = \mu_2 = \cdots = \mu_k$

F = \frac{SS_{\text{Between}}/(k-1)}{SS_{\text{Within}}/(n-k)} = \frac{MS_{\text{Between}}}{MS_{\text{Within}}} \sim F(k-1, n-k) \quad \text{under } H_0

Decision rule

\text{Reject } H_0 \text{ if } F > F_{1-\alpha}(k-1, n-k)

Actuarial example

Testing whether mean claim amounts differ across $k=4$ risk territories with a total of $n=100$ observations.
$F \sim F(3, 96)$ under $H_0$. Critical value $F_{0.95}(3,96) \approx 2.70$.
If $F > 2.70$, at least one territory has a significantly different mean — pricing by territory is justified.

Test 3 — Overall Regression Significance

In linear regression with $p$ predictors and $n$ observations, the F-test checks whether the model as a whole explains a significant amount of variance.

Test statistic for $H_0$: all $p$ slope coefficients $= 0$

F = \frac{R^2/p}{(1-R^2)/(n-p-1)} \sim F(p, n-p-1) \quad \text{under } H_0

R^2 = \frac{SS_{\text{Regression}}}{SS_{\text{Total}}} = 1 - \frac{SS_{\text{Residual}}}{SS_{\text{Total}}}

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