{ "deckName": "Exam FAM — Severity & Frequency Models", "examCode": "Exam FAM", "cards": [ { "front": "Give the pdf, cdf, survival function, mean and variance of the **exponential** distribution with mean $\\theta$.", "back": "$f(x)=\\dfrac{1}{\\theta}e^{-x/\\theta}$, $F(x)=1-e^{-x/\\theta}$, $S(x)=e^{-x/\\theta}$ for $x>0$.\n$E[X]=\\theta$ and $\\operatorname{Var}(X)=\\theta^{2}$, so the standard deviation is also $\\theta$.\nThe exponential is the building-block severity: a single **scale** parameter $\\theta$, constant hazard rate $\\frac{1}{\\theta}$, and the memoryless property.", "tag": "Severity distributions" }, { "front": "State the raw moments $E[X^{k}]$ of an **exponential** with mean $\\theta$, and use them to find the variance.", "back": "$E[X^{k}]=k!\\,\\theta^{k}$.\nSo $E[X]=\\theta$, $E[X^{2}]=2\\theta^{2}$, $E[X^{3}]=6\\theta^{3}$.\nVariance $=E[X^{2}]-(E[X])^{2}=2\\theta^{2}-\\theta^{2}=\\theta^{2}$.\nExample: with $\\theta=500$, $E[X^{3}]=6(500)^{3}=750{,}000{,}000$.", "tag": "Moments & CV" }, { "front": "Give the mean, variance, and raw moments of the **gamma** distribution with shape $\\alpha$ and scale $\\theta$.", "back": "$E[X]=\\alpha\\theta$, $\\operatorname{Var}(X)=\\alpha\\theta^{2}$.\nRaw moments: $E[X^{k}]=\\theta^{k}\\dfrac{\\Gamma(\\alpha+k)}{\\Gamma(\\alpha)}=\\theta^{k}(\\alpha)(\\alpha+1)\\cdots(\\alpha+k-1)$ for integer $k$.\n$\\alpha$ is the **shape** (it controls skewness); $\\theta$ is the **scale**. When $\\alpha=1$ the gamma reduces to the exponential.", "tag": "Severity distributions" }, { "front": "For a **gamma** severity with $\\alpha=3$, $\\theta=500$, compute the mean, variance, standard deviation and coefficient of variation.", "back": "$E[X]=\\alpha\\theta=3(500)=1500$.\n$\\operatorname{Var}(X)=\\alpha\\theta^{2}=3(500)^{2}=750{,}000$, so $\\text{SD}=\\sqrt{750{,}000}\\approx 866.03$.\n$\\text{CV}=\\dfrac{\\text{SD}}{E[X]}=\\dfrac{866.03}{1500}\\approx 0.5774$.\nCheck: for a gamma, $\\text{CV}=\\dfrac{1}{\\sqrt{\\alpha}}=\\dfrac{1}{\\sqrt{3}}\\approx 0.5774$.", "tag": "Moments & CV" }, { "front": "Give the pdf, survival function, and mean/variance of the **two-parameter Pareto** with shape $\\alpha$ and scale $\\theta$.", "back": "$f(x)=\\dfrac{\\alpha\\theta^{\\alpha}}{(x+\\theta)^{\\alpha+1}}$, $S(x)=\\left(\\dfrac{\\theta}{x+\\theta}\\right)^{\\alpha}$, $F(x)=1-\\left(\\dfrac{\\theta}{x+\\theta}\\right)^{\\alpha}$ for $x>0$.\n$E[X]=\\dfrac{\\theta}{\\alpha-1}$ (needs $\\alpha>1$); $E[X^{2}]=\\dfrac{2\\theta^{2}}{(\\alpha-1)(\\alpha-2)}$ (needs $\\alpha>2$).\n$\\alpha$ is the **shape** governing tail thickness; $\\theta$ is the **scale**.", "tag": "Severity distributions" }, { "front": "For a **Pareto** severity with $\\alpha=3$, $\\theta=2000$, compute the mean, $E[X^{2}]$, variance, and coefficient of variation.", "back": "$E[X]=\\dfrac{\\theta}{\\alpha-1}=\\dfrac{2000}{2}=1000$.\n$E[X^{2}]=\\dfrac{2\\theta^{2}}{(\\alpha-1)(\\alpha-2)}=\\dfrac{2(2000)^{2}}{(2)(1)}=4{,}000{,}000$.\n$\\operatorname{Var}(X)=4{,}000{,}000-1000^{2}=3{,}000{,}000$, $\\text{SD}=\\sqrt{3{,}000{,}000}\\approx 1732.05$.\n$\\text{CV}=\\dfrac{1732.05}{1000}\\approx 1.7321>1$ — a sign of a heavy-tailed severity.", "tag": "Moments & CV" }, { "front": "For a **Pareto** with $\\alpha=3$, $\\theta=2000$, find $P(X>4000)$.", "back": "Use the survival function $S(x)=\\left(\\dfrac{\\theta}{x+\\theta}\\right)^{\\alpha}$.\n$S(4000)=\\left(\\dfrac{2000}{4000+2000}\\right)^{3}=\\left(\\dfrac{2000}{6000}\\right)^{3}=\\left(\\dfrac{1}{3}\\right)^{3}=\\dfrac{1}{27}\\approx 0.037037$.\nSo about $3.70\\%$ of losses exceed $4000$.", "tag": "Severity distributions" }, { "front": "Give the cdf, mean, and $E[X^{2}]$ of the **lognormal** distribution with parameters $\\mu$ and $\\sigma$.", "back": "If $\\ln X\\sim N(\\mu,\\sigma^{2})$ then $F(x)=\\Phi\\!\\left(\\dfrac{\\ln x-\\mu}{\\sigma}\\right)$.\n$E[X]=e^{\\mu+\\sigma^{2}/2}$ and $E[X^{k}]=e^{k\\mu+k^{2}\\sigma^{2}/2}$, so $E[X^{2}]=e^{2\\mu+2\\sigma^{2}}$.\nHere $\\mu,\\sigma$ are the **log-scale** parameters, not the mean/SD of $X$ itself. The lognormal is moderately heavy-tailed (all moments exist, but it has no MGF).", "tag": "Severity distributions" }, { "front": "For a **lognormal** with $\\mu=7$, $\\sigma=1$, compute the mean, variance, and coefficient of variation.", "back": "$E[X]=e^{\\mu+\\sigma^{2}/2}=e^{7.5}\\approx 1808.04$.\n$E[X^{2}]=e^{2\\mu+2\\sigma^{2}}=e^{16}\\approx 8{,}886{,}111$.\n$\\operatorname{Var}(X)=8{,}886{,}111-1808.04^{2}\\approx 5{,}617{,}093$, so $\\text{SD}\\approx 2370.04$.\n$\\text{CV}=\\dfrac{2370.04}{1808.04}\\approx 1.3108$. Check: $\\text{CV}=\\sqrt{e^{\\sigma^{2}}-1}=\\sqrt{e^{1}-1}\\approx 1.3108$.", "tag": "Moments & CV" }, { "front": "For a **lognormal** with $\\mu=7$, $\\sigma=1$, find $P(X\\le 3000)$. (Use $\\Phi(1.01)\\approx 0.8438$.)", "back": "Standardize on the log scale: $z=\\dfrac{\\ln x-\\mu}{\\sigma}=\\dfrac{\\ln 3000-7}{1}$.\n$\\ln 3000\\approx 8.00637$, so $z\\approx 8.00637-7=1.00637$.\n$P(X\\le 3000)=\\Phi(1.0064)\\approx 0.8429$.\nAbout $84.3\\%$ of losses fall at or below $3000$.", "tag": "Severity distributions" }, { "front": "Give the cdf, survival, and mean of the **Weibull** distribution with shape $\\tau$ and scale $\\theta$.", "back": "$F(x)=1-e^{-(x/\\theta)^{\\tau}}$, $S(x)=e^{-(x/\\theta)^{\\tau}}$ for $x>0$.\n$E[X]=\\theta\\,\\Gamma\\!\\left(1+\\tfrac{1}{\\tau}\\right)$ and $E[X^{k}]=\\theta^{k}\\,\\Gamma\\!\\left(1+\\tfrac{k}{\\tau}\\right)$.\n$\\tau$ is the **shape**: $\\tau=1$ gives the exponential, $\\tau>1$ gives an increasing hazard (lighter tail), $\\tau<1$ a decreasing hazard (heavier tail). $\\theta$ is the **scale**.", "tag": "Severity distributions" }, { "front": "For a **Weibull** with $\\tau=2$, $\\theta=1000$, compute the mean, variance, and coefficient of variation. (Use $\\Gamma(1.5)=0.886227$, $\\Gamma(2)=1$.)", "back": "$E[X]=\\theta\\,\\Gamma(1+\\tfrac12)=1000(0.886227)\\approx 886.23$.\n$E[X^{2}]=\\theta^{2}\\,\\Gamma(1+1)=1000^{2}(1)=1{,}000{,}000$.\n$\\operatorname{Var}(X)=1{,}000{,}000-886.23^{2}\\approx 214{,}602$, $\\text{SD}\\approx 463.25$.\n$\\text{CV}=\\dfrac{463.25}{886.23}\\approx 0.5227<1$ — the $\\tau=2$ Weibull is light-tailed relative to the exponential.", "tag": "Moments & CV" }, { "front": "How does a pure **scale** parameter $\\theta$ act on a severity distribution, and which families is $\\theta$ a scale parameter for?", "back": "If $\\theta$ is a scale parameter, multiplying losses by a constant $c$ (e.g. inflation) just replaces $\\theta$ with $c\\theta$ and leaves the **shape** parameters unchanged: $E[X^{k}]$ scales by $c^{k}$ and the CV is unchanged.\nIn the exponential, gamma, Pareto, and Weibull, $\\theta$ is a scale parameter. For the lognormal, a scale change of $c$ shifts $\\mu\\to\\mu+\\ln c$ while $\\sigma$ (the shape) is unchanged.", "tag": "Severity distributions" }, { "front": "Distinguish a **scale** parameter from a **shape** parameter, and why does the **coefficient of variation** isolate shape?", "back": "A **scale** parameter stretches the distribution along the loss axis without changing its form; a **shape** parameter changes the form (skewness, tail). Multiplying $X$ by $c$ multiplies the mean and SD both by $c$, so $\\text{CV}=\\dfrac{\\text{SD}}{\\text{mean}}$ is **scale-invariant** and depends only on the shape parameters.\nHence gamma CV $=\\frac{1}{\\sqrt{\\alpha}}$, exponential CV $=1$, lognormal CV $=\\sqrt{e^{\\sigma^{2}}-1}$ — each free of the scale.", "tag": "Moments & CV" }, { "front": "Define the **coefficient of variation** and **skewness** of a loss random variable, and give their values for the exponential.", "back": "$\\text{CV}=\\dfrac{\\sqrt{\\operatorname{Var}(X)}}{E[X]}$ (relative dispersion).\nSkewness $\\gamma_{1}=\\dfrac{E[(X-\\mu)^{3}]}{\\sigma^{3}}=\\dfrac{E[X^{3}]-3\\mu E[X^{2}]+2\\mu^{3}}{\\sigma^{3}}$ (asymmetry).\nFor the exponential with mean $\\theta$: $\\text{CV}=1$ and skewness $=2$. For a gamma, $\\text{CV}=\\frac{1}{\\sqrt{\\alpha}}$ and skewness $=\\frac{2}{\\sqrt{\\alpha}}$, both shrinking as $\\alpha$ grows.", "tag": "Moments & CV" }, { "front": "Compute the **skewness** of an exponential with mean $\\theta$ from its raw moments.", "back": "Use $E[X]=\\theta$, $E[X^{2}]=2\\theta^{2}$, $E[X^{3}]=6\\theta^{3}$, and $\\sigma=\\theta$.\nThird central moment $=E[X^{3}]-3\\mu E[X^{2}]+2\\mu^{3}=6\\theta^{3}-3\\theta(2\\theta^{2})+2\\theta^{3}=6\\theta^{3}-6\\theta^{3}+2\\theta^{3}=2\\theta^{3}$.\nSkewness $=\\dfrac{2\\theta^{3}}{\\sigma^{3}}=\\dfrac{2\\theta^{3}}{\\theta^{3}}=2$ — positive and independent of $\\theta$ (a shape feature).", "tag": "Moments & CV" }, { "front": "A gamma severity has mean $1000$ and variance $250{,}000$. Find its shape $\\alpha$, scale $\\theta$, and coefficient of variation.", "back": "$E[X]=\\alpha\\theta=1000$ and $\\operatorname{Var}(X)=\\alpha\\theta^{2}=250{,}000$.\nDivide: $\\dfrac{\\alpha\\theta^{2}}{\\alpha\\theta}=\\theta=\\dfrac{250{,}000}{1000}=250$.\nThen $\\alpha=\\dfrac{1000}{\\theta}=\\dfrac{1000}{250}=4$.\n$\\text{CV}=\\dfrac{\\sqrt{250{,}000}}{1000}=\\dfrac{500}{1000}=0.5$ (equivalently $\\frac{1}{\\sqrt{4}}$).", "tag": "Moments & CV" }, { "front": "Define the **hazard rate** (failure rate) $h(x)$ and state it for the exponential, Weibull, and Pareto.", "back": "$h(x)=\\dfrac{f(x)}{S(x)}=-\\dfrac{d}{dx}\\ln S(x)$ — the instantaneous loss/failure intensity given survival to $x$.\n**Exponential:** $h(x)=\\frac{1}{\\theta}$, constant.\n**Weibull:** $h(x)=\\dfrac{\\tau}{\\theta}\\left(\\dfrac{x}{\\theta}\\right)^{\\tau-1}$ — increasing if $\\tau>1$, decreasing if $\\tau<1$.\n**Pareto:** $h(x)=\\dfrac{\\alpha}{x+\\theta}$, decreasing in $x$.\nA **decreasing** hazard signals a **heavier** tail; an **increasing** hazard a lighter one.", "tag": "Tail weight" }, { "front": "Define the **mean excess (mean residual life)** function $e(d)$ and give it for the exponential and the two-parameter Pareto.", "back": "$e(d)=E[X-d\\mid X>d]=\\dfrac{\\int_{d}^{\\infty}S(x)\\,dx}{S(d)}=\\dfrac{E[X]-E[X\\wedge d]}{S(d)}$ — the expected loss above $d$ given the loss exceeds $d$.\n**Exponential:** $e(d)=\\theta$, constant (memorylessness).\n**Two-parameter Pareto:** $e(d)=\\dfrac{d+\\theta}{\\alpha-1}$, increasing **linearly** in $d$.\nA mean excess that **increases** with $d$ indicates a heavy tail.", "tag": "Tail weight" }, { "front": "For a **Pareto** with $\\alpha=3$, $\\theta=2000$, find the mean excess loss $e(4000)$ and interpret it against the exponential.", "back": "$e(d)=\\dfrac{d+\\theta}{\\alpha-1}$, so $e(4000)=\\dfrac{4000+2000}{3-1}=\\dfrac{6000}{2}=3000$.\nGiven a loss already exceeds $4000$, the expected amount **above** $4000$ is $3000$.\nBecause $e(d)$ rises with $d$ (here from $e(0)=1000$ to $e(4000)=3000$), the Pareto is heavy-tailed; for an exponential $e(d)=\\theta$ would be flat.", "tag": "Tail weight" }, { "front": "How does **existence of moments** classify tail weight, and rank exponential, gamma, lognormal, and Pareto?", "back": "The more moments that exist, the **lighter** the tail. A distribution with all positive moments finite is lighter-tailed than one where high moments diverge.\n**Pareto($\\alpha$):** only $E[X^{k}]$ for $k<\\alpha$ exist — heaviest.\n**Lognormal:** all moments finite but **no** MGF — heavy, but lighter than Pareto.\n**Gamma / exponential:** all moments finite **and** an MGF exists — lightest of these.\nSo (heavy → light): Pareto $\\succ$ lognormal $\\succ$ gamma/exponential. (A Weibull with $\\tau>1$ is even lighter than the exponential.)", "tag": "Tail weight" }, { "front": "Use the **limit of the ratio of survival functions** to compare the tails of a Pareto and an exponential.", "back": "Compare tail weight via $\\displaystyle\\lim_{x\\to\\infty}\\dfrac{S_{1}(x)}{S_{2}(x)}$: if the limit is $\\infty$, distribution $1$ has the heavier tail; if $0$, the lighter.\nPareto $S_{1}(x)=\\left(\\frac{\\theta}{x+\\theta}\\right)^{\\alpha}$ decays **polynomially**; exponential $S_{2}(x)=e^{-x/\\theta'}$ decays **exponentially**.\n$\\displaystyle\\lim_{x\\to\\infty}\\dfrac{(\\theta/(x+\\theta))^{\\alpha}}{e^{-x/\\theta'}}=\\infty$ (exponential beats any power), so the Pareto tail is **heavier**.", "tag": "Tail weight" }, { "front": "State the defining recursion of the **$(a,b,0)$ class** and which four distributions belong to it.", "back": "A counting distribution on $k=0,1,2,\\dots$ is in the $(a,b,0)$ class if it satisfies $\\dfrac{p_{k}}{p_{k-1}}=a+\\dfrac{b}{k}$ for $k=1,2,3,\\dots$, with $p_{0}$ set by normalization. The **only** members are the **Poisson**, **binomial**, **negative binomial**, and **geometric** (the geometric is the negative binomial with $r=1$). The pair $(a,b)$ uniquely identifies which one.", "tag": "(a,b,0) class" }, { "front": "Give the $(a,b,0)$ parameters $a,b$, the mean, and the variance for the **Poisson($\\lambda$)**.", "back": "$a=0$, $b=\\lambda$ (so $\\frac{p_{k}}{p_{k-1}}=\\frac{\\lambda}{k}$).\n$p_{k}=\\dfrac{e^{-\\lambda}\\lambda^{k}}{k!}$, with $p_{0}=e^{-\\lambda}$.\n$E[N]=\\lambda$ and $\\operatorname{Var}(N)=\\lambda$, so the **variance-to-mean ratio is exactly $1$** — the Poisson's signature.", "tag": "(a,b,0) class" }, { "front": "Give the $(a,b,0)$ parameters, mean, and variance for the **binomial($m,q$)** and the **geometric($\\beta$)**.", "back": "**Binomial($m,q$):** $a=-\\dfrac{q}{1-q}$, $b=(m+1)\\dfrac{q}{1-q}$. $E[N]=mq$, $\\operatorname{Var}(N)=mq(1-q)$ — variance $<$ mean ($a<0$).\n**Geometric($\\beta$):** $a=\\dfrac{\\beta}{1+\\beta}$, $b=0$. $E[N]=\\beta$, $\\operatorname{Var}(N)=\\beta(1+\\beta)$ — variance $>$ mean.\nThe sign of $a$ alone tells the family: $a<0$ binomial, $a=0$ Poisson, $a>0$ negative binomial/geometric.", "tag": "(a,b,0) class" }, { "front": "Give the $(a,b,0)$ parameters, mean, variance, and $p_{0}$ for the **negative binomial($r,\\beta$)**.", "back": "$a=\\dfrac{\\beta}{1+\\beta}$, $b=(r-1)\\dfrac{\\beta}{1+\\beta}$, and $p_{0}=(1+\\beta)^{-r}$.\n$E[N]=r\\beta$ and $\\operatorname{Var}(N)=r\\beta(1+\\beta)$, so the **variance-to-mean ratio is $1+\\beta>1$**.\nWith $r=1$ it collapses to the geometric ($b=0$). The negative binomial is the natural model for **over-dispersed** counts.", "tag": "(a,b,0) class" }, { "front": "State the **variance-to-mean ratio test** for distinguishing the $(a,b,0)$ distributions, and apply it to a sample with mean $3.0$ and variance $7.5$.", "back": "Compare $\\dfrac{\\operatorname{Var}(N)}{E[N]}$:\n**$<1$** → binomial; **$=1$** → Poisson; **$>1$** → negative binomial (geometric if the ratio equals $1+\\beta$ with $r=1$).\nSample: $\\dfrac{7.5}{3.0}=2.5>1$ ⇒ **negative binomial**. Match moments: $1+\\beta=2.5\\Rightarrow\\beta=1.5$, then $r\\beta=3\\Rightarrow r=\\dfrac{3}{1.5}=2$.\nSo the fitted model is negative binomial with $r=2,\\ \\beta=1.5$.", "tag": "(a,b,0) class" }, { "front": "Successive probabilities of a count satisfy $\\dfrac{p_{1}}{p_{0}}=1.2$, $\\dfrac{p_{2}}{p_{1}}=0.9$, $\\dfrac{p_{3}}{p_{2}}=0.8$. Identify the $(a,b,0)$ distribution and its parameters.", "back": "Fit $\\dfrac{p_{k}}{p_{k-1}}=a+\\dfrac{b}{k}$. Using $k=1$ and $k=2$:\n$a+b=1.2$ and $a+\\tfrac{b}{2}=0.9$. Subtract: $\\tfrac{b}{2}=0.3\\Rightarrow b=0.6$, then $a=0.6$.\nCheck $k=3$: $a+\\tfrac{b}{3}=0.6+0.2=0.8$ ✓.\nSince $a=0.6>0$, it is **negative binomial**: $a=\\frac{\\beta}{1+\\beta}=0.6\\Rightarrow\\beta=1.5$, and $b=(r-1)\\frac{\\beta}{1+\\beta}=0.6\\Rightarrow r-1=1\\Rightarrow r=2$.", "tag": "(a,b,0) class" }, { "front": "Explain the **linear-plot** method: how plotting $k\\cdot\\dfrac{p_{k}}{p_{k-1}}$ (or $k\\cdot\\dfrac{n_{k}}{n_{k-1}}$) against $k$ identifies the $(a,b,0)$ distribution.", "back": "Multiply the recursion by $k$: $k\\dfrac{p_{k}}{p_{k-1}}=ak+b$ — a **straight line** in $k$ with slope $a$ and intercept $b$. Estimate the ratios from observed counts $n_{k}$ and regress $k\\,\\dfrac{n_{k}}{n_{k-1}}$ on $k$.\n**Slope $a$:** $<0$ binomial, $=0$ Poisson, $>0$ negative binomial.\nExample (negative binomial $a=0.6,b=0.6$): the points are $k=1\\!:1.2$, $k=2\\!:1.8$, $k=3\\!:2.4$, $k=4\\!:3.0$ — a line of slope $0.6$, intercept $0.6$.", "tag": "(a,b,0) class" }, { "front": "For a **Poisson** with $\\lambda=2$, compute $p_{0},p_{1},p_{2},p_{3}$ using the $(a,b,0)$ recursion.", "back": "$a=0$, $b=2$, so $p_{k}=p_{k-1}\\dfrac{2}{k}$ starting from $p_{0}=e^{-2}\\approx 0.135335$.\n$p_{1}=p_{0}\\cdot\\tfrac{2}{1}=2e^{-2}\\approx 0.270671$.\n$p_{2}=p_{1}\\cdot\\tfrac{2}{2}=p_{1}\\approx 0.270671$.\n$p_{3}=p_{2}\\cdot\\tfrac{2}{3}\\approx 0.180447$.\n(Sum so far $\\approx 0.857$; the rest is the upper tail.)", "tag": "(a,b,0) class" }, { "front": "For a **negative binomial** with $r=2$, $\\beta=1.5$, compute $p_{0}$ and $p_{1}$ and verify the mean and variance.", "back": "$p_{0}=(1+\\beta)^{-r}=(2.5)^{-2}=\\dfrac{1}{6.25}=0.16$.\n$a=\\dfrac{\\beta}{1+\\beta}=0.6$, $b=(r-1)\\dfrac{\\beta}{1+\\beta}=0.6$, so $\\dfrac{p_{1}}{p_{0}}=a+b=1.2$ and $p_{1}=0.16(1.2)=0.192$.\nMean $=r\\beta=2(1.5)=3$; variance $=r\\beta(1+\\beta)=3(2.5)=7.5$ (ratio $2.5=1+\\beta$). ✓", "tag": "(a,b,0) class" }, { "front": "A claim count has $E[N]=1.5$ and $\\operatorname{Var}(N)=1.05$. Identify the $(a,b,0)$ model and find its parameters.", "back": "Ratio $\\dfrac{1.05}{1.5}=0.7<1$ ⇒ **binomial** ($a<0$).\nMatch moments: $mq=1.5$ and $mq(1-q)=1.05$, so $1-q=\\dfrac{1.05}{1.5}=0.7\\Rightarrow q=0.3$.\nThen $m=\\dfrac{1.5}{0.3}=5$.\nFitted model: binomial with $m=5$, $q=0.3$ (so $a=-\\frac{0.3}{0.7}\\approx-0.4286$, $b=(6)\\frac{0.3}{0.7}\\approx 2.5714$).", "tag": "(a,b,0) class" }, { "front": "Define the **$(a,b,1)$ class** and how it generalizes $(a,b,0)$.", "back": "The $(a,b,1)$ class uses the **same** recursion $\\dfrac{p_{k}}{p_{k-1}}=a+\\dfrac{b}{k}$ but only for $k\\ge 2$, leaving $p_{0}$ (hence $p_{1}$) free to be reset. This frees the probability mass at $0$ from the usual $(a,b,0)$ value.\nTwo important sub-types: **zero-truncated** ($p_{0}^{T}=0$, no zeros allowed) and **zero-modified** ($p_{0}^{M}$ chosen arbitrarily). They model count data with too many — or too few — zeros than the base distribution allows.", "tag": "(a,b,1) class" }, { "front": "Give the **zero-truncated** probabilities $p_{k}^{T}$ in terms of the base $(a,b,0)$ probabilities $p_{k}$, and the mean.", "back": "Set the probability of zero to $0$ and rescale the rest: $p_{k}^{T}=\\dfrac{p_{k}}{1-p_{0}}$ for $k=1,2,3,\\dots$.\nThe mean inflates by the same factor: $E[N^{T}]=\\dfrac{E[N]}{1-p_{0}}$.\nVariance: $\\operatorname{Var}(N^{T})=\\dfrac{E[N^{2}]}{1-p_{0}}-\\left(\\dfrac{E[N]}{1-p_{0}}\\right)^{2}$. Zero-truncation is used when a zero count is impossible (e.g. only policies with at least one claim are observed).", "tag": "(a,b,1) class" }, { "front": "For a **zero-truncated Poisson** with $\\lambda=2$, compute $p_{1}^{T}$ and the truncated mean.", "back": "Base Poisson: $p_{0}=e^{-2}\\approx 0.135335$, so $1-p_{0}\\approx 0.864665$. And $p_{1}=2e^{-2}\\approx 0.270671$.\n$p_{1}^{T}=\\dfrac{p_{1}}{1-p_{0}}=\\dfrac{0.270671}{0.864665}\\approx 0.313035$.\nTruncated mean $=\\dfrac{\\lambda}{1-p_{0}}=\\dfrac{2}{0.864665}\\approx 2.31304$ — larger than $2$ because the zeros are removed.", "tag": "(a,b,1) class" }, { "front": "Give the **zero-modified** probabilities $p_{k}^{M}$ in terms of the base $p_{k}$ and the chosen $p_{0}^{M}$, and the mean.", "back": "Choose $p_{0}^{M}$ freely, then scale the positive probabilities: $p_{k}^{M}=\\dfrac{1-p_{0}^{M}}{1-p_{0}}\\,p_{k}$ for $k=1,2,3,\\dots$.\nEquivalently $p_{k}^{M}=(1-p_{0}^{M})\\,p_{k}^{T}$, so a zero-modified is a mixture of a point mass at $0$ and the zero-truncated distribution.\nMean: $E[N^{M}]=\\dfrac{1-p_{0}^{M}}{1-p_{0}}\\,E[N]$. (Zero-truncation is the special case $p_{0}^{M}=0$.)", "tag": "(a,b,1) class" }, { "front": "A **zero-modified Poisson** is built from a base Poisson with $\\lambda=2$ but with $p_{0}^{M}=0.4$. Find $p_{1}^{M}$ and the modified mean.", "back": "Base: $p_{0}=e^{-2}\\approx 0.135335$, $1-p_{0}\\approx 0.864665$, $p_{1}=2e^{-2}\\approx 0.270671$.\nScaling factor $c=\\dfrac{1-p_{0}^{M}}{1-p_{0}}=\\dfrac{0.6}{0.864665}\\approx 0.693911$.\n$p_{1}^{M}=c\\,p_{1}=0.693911(0.270671)\\approx 0.187821$.\nMean $=c\\,\\lambda=0.693911(2)\\approx 1.38782$ — below the base $2$ because extra mass was placed at $0$.", "tag": "(a,b,1) class" }, { "front": "A **zero-modified Poisson** has base $\\lambda=1.5$ and $p_{0}^{M}=0.5$. Find $p_{2}^{M}$.", "back": "Base: $p_{0}=e^{-1.5}\\approx 0.223130$, $1-p_{0}\\approx 0.776870$.\n$p_{1}=1.5e^{-1.5}\\approx 0.334695$, and $p_{2}=p_{1}\\cdot\\dfrac{1.5}{2}\\approx 0.251021$.\nFactor $c=\\dfrac{1-0.5}{1-p_{0}}=\\dfrac{0.5}{0.776870}\\approx 0.643608$.\n$p_{2}^{M}=c\\,p_{2}=0.643608(0.251021)\\approx 0.161560$.", "tag": "(a,b,1) class" }, { "front": "Why is the **zero-modified** model so useful in claim-frequency work, and what happens to a zero-modified distribution as $p_{0}^{M}\\to 0$?", "back": "Real frequency data often have **excess zeros** (many policies file no claims) or, after conditioning, **no zeros** — neither of which a plain Poisson/negative binomial can match because their $p_{0}$ is fixed by the mean. Zero-modification decouples $P(N=0)$ from the rest of the shape, so you fit the zero rate and the positive-count shape separately.\nAs $p_{0}^{M}\\to 0$ the zero-modified distribution becomes the **zero-truncated** distribution (all mass moves to $k\\ge 1$).", "tag": "(a,b,1) class" }, { "front": "Define a **continuous mixture** of a counting distribution and state the resulting mean and variance via conditioning.", "back": "Let $N\\mid\\Lambda$ have a distribution indexed by a random parameter $\\Lambda$ with its own (mixing) distribution. The unconditional pmf is $P(N=k)=\\int P(N=k\\mid\\Lambda=\\lambda)\\,g(\\lambda)\\,d\\lambda$.\nBy the laws of total expectation/variance:\n$E[N]=E\\!\\left[E[N\\mid\\Lambda]\\right]$, and\n$\\operatorname{Var}(N)=E\\!\\left[\\operatorname{Var}(N\\mid\\Lambda)\\right]+\\operatorname{Var}\\!\\left(E[N\\mid\\Lambda]\\right)$.\nMixing **adds** the variance term $\\operatorname{Var}(E[N\\mid\\Lambda])$, producing **over-dispersion** and a heavier tail than the un-mixed model.", "tag": "Mixtures" }, { "front": "State the key result that the **gamma mixture of a Poisson is a negative binomial**, including the parameter map.", "back": "If $N\\mid\\Lambda\\sim\\text{Poisson}(\\Lambda)$ and the mixing $\\Lambda\\sim\\text{Gamma}(\\alpha,\\theta)$, then unconditionally $N\\sim\\text{Negative Binomial}(r=\\alpha,\\ \\beta=\\theta)$.\nIntuitively, the random Poisson mean injects extra variability, so the result is over-dispersed (variance $>$ mean), exactly the negative binomial's hallmark. This is the canonical motivation for the negative binomial as a claim-count model.", "tag": "Mixtures" }, { "front": "Suppose $N\\mid\\Lambda\\sim\\text{Poisson}(\\Lambda)$ with $\\Lambda\\sim\\text{Gamma}(\\alpha=2,\\theta=1.5)$. Find $E[N]$ and $\\operatorname{Var}(N)$ two ways and confirm over-dispersion.", "back": "**By the negative-binomial map** ($r=2,\\beta=1.5$): $E[N]=r\\beta=3$, $\\operatorname{Var}(N)=r\\beta(1+\\beta)=3(2.5)=7.5$.\n**By conditioning:** $E[N]=E[\\Lambda]=\\alpha\\theta=3$. $\\operatorname{Var}(N)=E[\\operatorname{Var}(N\\mid\\Lambda)]+\\operatorname{Var}(E[N\\mid\\Lambda])=E[\\Lambda]+\\operatorname{Var}(\\Lambda)=\\alpha\\theta+\\alpha\\theta^{2}=3+2(1.5)^{2}=3+4.5=7.5$.\nVariance $7.5>$ mean $3$ ⇒ over-dispersed, ratio $2.5=1+\\beta$. ✓", "tag": "Mixtures" }, { "front": "Describe the **exponential mixed over its mean** and the resulting tail.", "back": "Let the severity be $X\\mid\\Theta\\sim\\text{Exponential}(\\text{mean }\\Theta)$ with the mean $\\Theta$ random (e.g. inverse-gamma distributed). Mixing the (light-tailed, all-moments) exponential over a distribution of means produces a **heavier-tailed** severity — the classic case yields a **Pareto**.\nWhy: each conditional exponential is light, but blending many scales — some very large — fattens the upper tail. This is the continuous-mixture route to the Pareto and a model for heterogeneous risks.", "tag": "Mixtures" }, { "front": "Why does **mixing always increase variance** and tail weight relative to the conditional distribution?", "back": "From $\\operatorname{Var}(N)=\\underbrace{E[\\operatorname{Var}(N\\mid\\Lambda)]}_{\\text{average within-group var}}+\\underbrace{\\operatorname{Var}(E[N\\mid\\Lambda])}_{\\ge 0}$, the second term is non-negative, so the mixture's variance is at least the average conditional variance. The extra term reflects **heterogeneity** across the mixing parameter.\nConsequence: a gamma-mixed Poisson (negative binomial) has variance $>$ mean even though each conditional Poisson has variance $=$ mean; an exponential mixed over its scale acquires a heavier (e.g. Pareto) tail. Mixing is the standard mechanism for building over-dispersed, heavy-tailed actuarial models.", "tag": "Mixtures" } ] }