{ "deckName": "Exam FAM — Aggregate Loss Models", "examCode": "Exam FAM", "cards": [ { "front": "State the **collective risk model** for aggregate losses $S$ and its independence assumptions.", "back": "$S = X_1 + X_2 + \\dots + X_N$, a **random sum**: the claim count (frequency) $N$ is random, and the individual claim amounts (severities) $X_1,X_2,\\dots$ are i.i.d.\nKey assumptions:\n1. The $X_i$ are mutually independent and identically distributed.\n2. $N$ is independent of the $X_i$.\nBy convention $S=0$ when $N=0$. The model treats the whole portfolio's losses in aggregate rather than policy-by-policy.", "tag": "Collective risk model" }, { "front": "Give the formula for $E[S]$ in the collective risk model and explain it.", "back": "$E[S] = E[N]\\,E[X]$.\nThis is the **conditional mean** result: $E[S]=E\\bigl[E[S\\mid N]\\bigr]=E[N\\,E[X]]=E[N]E[X]$. Expected aggregate loss is just the expected number of claims times the expected size of each claim. The $X_i$ being identically distributed lets us write a single $E[X]$.", "tag": "Moments of S" }, { "front": "Give the compound-variance formula $Var[S]$ in the collective risk model and identify the two pieces.", "back": "$Var[S] = E[N]\\,Var[X] + Var[N]\\,E[X]^2$.\nThe first term $E[N]Var[X]$ is the **severity (process) variability** — randomness in claim sizes. The second term $Var[N]E[X]^2$ is the **frequency variability** — randomness in how many claims occur. It follows from the conditional-variance formula $Var[S]=E[Var(S\\mid N)]+Var(E[S\\mid N])$.", "tag": "Moments of S" }, { "front": "State the relation between the MGF of $S$, the PGF of $N$, and the MGF of $X$ in the collective risk model.", "back": "$M_S(t) = P_N\\bigl(M_X(t)\\bigr)$, where $P_N(z)=E[z^{N}]$ is the probability generating function of the frequency and $M_X(t)=E[e^{tX}]$ is the moment generating function of the severity.\nDerivation: $M_S(t)=E[e^{tS}]=E\\bigl[E[e^{tS}\\mid N]\\bigr]=E\\bigl[M_X(t)^{N}\\bigr]=P_N(M_X(t))$.\nEquivalently $M_S(t)=M_N\\bigl(\\ln M_X(t)\\bigr)$.", "tag": "Moments of S" }, { "front": "Compute $E[S]$ and $Var[S]$ when $N$ has mean $3$ and variance $5$, and severities have $E[X]=200$, $Var[X]=30{,}000$.", "back": "$E[S]=E[N]E[X]=3(200)=\\$600$.\n$Var[S]=E[N]Var[X]+Var[N]E[X]^2 = 3(30{,}000)+5(200)^2$\n$= 90{,}000 + 5(40{,}000) = 90{,}000 + 200{,}000 = 290{,}000$.\nStandard deviation $=\\sqrt{290{,}000}\\approx \\$538.52$.", "tag": "Moments of S" }, { "front": "A frequency $N$ has $E[N]=2.5$, $Var[N]=2.5$ (Poisson). Severity $X$ has $E[X]=500$ and $E[X^2]=400{,}000$. Find $E[S]$ and $Var[S]$ two ways.", "back": "First $Var[X]=E[X^2]-E[X]^2 = 400{,}000-250{,}000=150{,}000$.\n$E[S]=E[N]E[X]=2.5(500)=\\$1{,}250$.\nGeneral formula: $Var[S]=E[N]Var[X]+Var[N]E[X]^2 = 2.5(150{,}000)+2.5(250{,}000)$\n$=375{,}000+625{,}000=1{,}000{,}000$.\nPoisson shortcut: $Var[S]=\\lambda E[X^2]=2.5(400{,}000)=1{,}000{,}000$. Both agree.", "tag": "Moments of S" }, { "front": "Define the **compound Poisson** distribution and state when it arises.", "back": "$S$ is **compound Poisson** when the frequency $N\\sim\\text{Poisson}(\\lambda)$ and the severities $X_i$ are i.i.d. and independent of $N$. We write $S\\sim\\text{CP}(\\lambda, F_X)$.\nIt is the workhorse aggregate model: claim counts over a fixed exposure period are Poisson, and each claim has the same severity law. Because the Poisson has $E[N]=Var[N]=\\lambda$, the compound formulas collapse to especially clean forms.", "tag": "Compound Poisson" }, { "front": "Give the mean, variance, and third central moment of a compound Poisson $S$ with rate $\\lambda$ and severity $X$.", "back": "$E[S]=\\lambda\\,E[X]$.\n$Var[S]=\\lambda\\,E[X^2]$.\nThird central moment $E[(S-E[S])^3]=\\lambda\\,E[X^3]$.\nIn general the $k$-th cumulant of $S$ equals $\\lambda\\,E[X^{k}]$. The variance uses $E[X^2]$ (the raw second moment), **not** $Var[X]$ — a common trap.", "tag": "Compound Poisson" }, { "front": "Derive the compound-Poisson variance $Var[S]=\\lambda E[X^2]$ from the general compound formula.", "back": "For Poisson frequency $E[N]=Var[N]=\\lambda$. Substituting into $Var[S]=E[N]Var[X]+Var[N]E[X]^2$:\n$Var[S]=\\lambda\\,Var[X]+\\lambda\\,E[X]^2 = \\lambda\\bigl(Var[X]+E[X]^2\\bigr)=\\lambda\\,E[X^2]$,\nsince $Var[X]+E[X]^2 = E[X^2]$ by definition. The frequency and severity variabilities merge into one term driven by the raw second moment.", "tag": "Compound Poisson" }, { "front": "A compound Poisson $S$ has $\\lambda=4$ and severities uniform on the integers $\\{1,2,3,4,5\\}$ (each probability $0.2$). Find $E[S]$ and $Var[S]$.", "back": "Severity moments: $E[X]=\\frac{1+2+3+4+5}{5}=3$.\n$E[X^2]=\\frac{1+4+9+16+25}{5}=\\frac{55}{5}=11$.\n$E[S]=\\lambda E[X]=4(3)=12$.\n$Var[S]=\\lambda E[X^2]=4(11)=44$.\nStandard deviation $=\\sqrt{44}\\approx 6.633$.", "tag": "Compound Poisson" }, { "front": "A compound Poisson has $\\lambda=10$ and lognormal severities with $E[X]=1{,}000$, $E[X^2]=2{,}000{,}000$. Find $E[S]$, $Var[S]$, and the coefficient of variation of $S$.", "back": "$E[S]=\\lambda E[X]=10(1{,}000)=\\$10{,}000$.\n$Var[S]=\\lambda E[X^2]=10(2{,}000{,}000)=20{,}000{,}000$.\nStandard deviation $=\\sqrt{20{,}000{,}000}\\approx 4{,}472.14$.\nCoefficient of variation $=\\dfrac{\\sqrt{Var[S]}}{E[S]}=\\dfrac{4{,}472.14}{10{,}000}\\approx 0.4472$.", "tag": "Compound Poisson" }, { "front": "State the **sum rule** for independent compound Poisson random variables.", "back": "If $S_1,\\dots,S_m$ are **independent** compound Poissons with rates $\\lambda_j$ and severity distributions $F_j$, then $S=S_1+\\dots+S_m$ is again compound Poisson with\n$\\lambda = \\lambda_1+\\dots+\\lambda_m$ and severity distribution\n$F(x)=\\dfrac{1}{\\lambda}\\sum_{j=1}^{m}\\lambda_j\\,F_j(x)$,\na **$\\lambda$-weighted mixture** of the individual severities. Probabilities of each severity value are mixed in proportion to each portfolio's rate.", "tag": "Compound Poisson" }, { "front": "Two independent compound Poisson portfolios merge: portfolio A has $\\lambda_A=3$ with severity always $100$; portfolio B has $\\lambda_B=2$ with severity always $250$. Describe the combined $S$ and find $E[S]$ and $Var[S]$.", "back": "Combined rate $\\lambda=3+2=5$. The combined severity is the $\\lambda$-weighted mixture: $P(X=100)=\\frac{3}{5}=0.6$, $P(X=250)=\\frac{2}{5}=0.4$.\nSeverity moments: $E[X]=0.6(100)+0.4(250)=60+100=160$; $E[X^2]=0.6(100^2)+0.4(250^2)=0.6(10{,}000)+0.4(62{,}500)=6{,}000+25{,}000=31{,}000$.\n$E[S]=\\lambda E[X]=5(160)=\\$800$.\n$Var[S]=\\lambda E[X^2]=5(31{,}000)=155{,}000$.", "tag": "Compound Poisson" }, { "front": "In a compound Poisson, claims are classified into types $1,\\dots,m$ with probabilities $p_1,\\dots,p_m$. What is the distribution of the count of type-$j$ claims?", "back": "**Thinning / decomposition:** the number of type-$j$ claims $N_j$ is **independent Poisson** with rate $\\lambda p_j$, and the $N_j$ are mutually independent.\nEquivalently, each type forms its own independent compound Poisson with rate $\\lambda_j=\\lambda p_j$ and that type's severity distribution. This is the converse of the superposition (sum) rule and is why compound Poisson models are so modular.", "tag": "Compound Poisson" }, { "front": "A compound Poisson has $\\lambda=8$. Claims are 'small' ($\\$50$) with probability $0.75$ and 'large' ($\\$500$) with probability $0.25$. Find the mean and variance of aggregate **large** losses alone.", "back": "By thinning, large claims are compound Poisson with rate $\\lambda_L=\\lambda p_L=8(0.25)=2$ and constant severity $500$.\n$E[S_L]=\\lambda_L E[X_L]=2(500)=\\$1{,}000$.\n$E[X_L^2]=500^2=250{,}000$, so $Var[S_L]=\\lambda_L E[X_L^2]=2(250{,}000)=500{,}000$.\nStandard deviation $=\\sqrt{500{,}000}\\approx \\$707.11$.", "tag": "Compound Poisson" }, { "front": "What is the **$(a,b,0)$ class** of frequency distributions, and which distributions belong to it?", "back": "A counting distribution is in the $(a,b,0)$ class if its probabilities satisfy the recursion\n$\\dfrac{p_k}{p_{k-1}} = a + \\dfrac{b}{k}, \\quad k=1,2,3,\\dots$\nfor constants $a,b$, starting from $p_0$. Exactly three distributions qualify:\n- **Poisson** ($a=0$, $b=\\lambda$),\n- **Binomial** ($a=-\\frac{q}{1-q}$, $b=(m+1)\\frac{q}{1-q}$),\n- **Negative binomial** ($a=\\frac{\\beta}{1+\\beta}$, $b=(r-1)\\frac{\\beta}{1+\\beta}$); geometric is the $r=1$ case.", "tag": "Panjer recursion" }, { "front": "Give the $(a,b)$ parameters for the Poisson, binomial, and negative binomial frequencies used in Panjer's recursion.", "back": "**Poisson**$(\\lambda)$: $a=0$, $b=\\lambda$.\n**Binomial**$(m,q)$: $a=-\\dfrac{q}{1-q}$, $b=(m+1)\\dfrac{q}{1-q}$.\n**Negative binomial**$(r,\\beta)$: $a=\\dfrac{\\beta}{1+\\beta}$, $b=(r-1)\\dfrac{\\beta}{1+\\beta}$.\n**Geometric**$(\\beta)$ is negative binomial with $r=1$: $a=\\dfrac{\\beta}{1+\\beta}$, $b=0$.", "tag": "Panjer recursion" }, { "front": "State the **Panjer recursion** for the aggregate distribution $f_S$ when frequency is $(a,b,0)$ and severity is discrete on $0,1,2,\\dots$", "back": "For an $(a,b,0)$ frequency and severity probabilities $f_X(j)$ on the non-negative integers,\n$f_S(s) = \\dfrac{1}{1-a\\,f_X(0)}\\sum_{j=1}^{s}\\Bigl(a+\\dfrac{b\\,j}{s}\\Bigr) f_X(j)\\, f_S(s-j), \\quad s=1,2,3,\\dots$\nThe recursion builds $f_S$ from $f_S(0)$ upward, avoiding a full convolution. When severity has no mass at $0$ (i.e. $f_X(0)=0$), the leading factor is just $1$.", "tag": "Panjer recursion" }, { "front": "Give the **starting value** $f_S(0)$ for the Panjer recursion, and the Poisson special case.", "back": "In general $f_S(0)=P_N\\bigl(f_X(0)\\bigr)$, the probability generating function of $N$ evaluated at the chance a single claim is $0$.\n- **Poisson**$(\\lambda)$: $f_S(0)=e^{-\\lambda(1-f_X(0))}$; if $f_X(0)=0$ this is $e^{-\\lambda}=P(N=0)$.\n- **Binomial**$(m,q)$: $f_S(0)=\\bigl(1-q+q f_X(0)\\bigr)^{m}$.\n- **Negative binomial**$(r,\\beta)$: $f_S(0)=\\bigl(1+\\beta(1-f_X(0))\\bigr)^{-r}$.\nWhen severity is positive (no mass at $0$), $f_S(0)=P(N=0)$.", "tag": "Panjer recursion" }, { "front": "Set up Panjer for a **Poisson**$(\\lambda)$ frequency: state the simplified recursion.", "back": "For Poisson, $a=0$ and $b=\\lambda$, so $a+\\frac{bj}{s}=\\frac{\\lambda j}{s}$ and the recursion becomes\n$f_S(s)=\\dfrac{\\lambda}{s}\\sum_{j=1}^{s} j\\,f_X(j)\\,f_S(s-j), \\quad s\\ge 1$,\nstarting from $f_S(0)=e^{-\\lambda(1-f_X(0))}$. (If $f_X(0)=0$, then $f_S(0)=e^{-\\lambda}$.) This is the form most often used on the exam.", "tag": "Panjer recursion" }, { "front": "Compute $f_S(0)$, $f_S(1)$, $f_S(2)$ for a compound Poisson with $\\lambda=2$ and severity $f_X(1)=0.6$, $f_X(2)=0.4$ (no mass at $0$).", "back": "$f_S(0)=e^{-\\lambda}=e^{-2}\\approx 0.135335$.\nPoisson recursion $f_S(s)=\\frac{\\lambda}{s}\\sum_{j=1}^{s} j f_X(j) f_S(s-j)$.\n$f_S(1)=\\frac{2}{1}\\bigl[1\\cdot f_X(1)\\cdot f_S(0)\\bigr]=2(0.6)(0.135335)\\approx 0.162402$.\n$f_S(2)=\\frac{2}{2}\\bigl[1\\cdot f_X(1) f_S(1) + 2\\cdot f_X(2) f_S(0)\\bigr]$\n$=1\\cdot\\bigl[0.6(0.162402)+2(0.4)(0.135335)\\bigr]=0.097441+0.108268\\approx 0.205709$.", "tag": "Panjer recursion" }, { "front": "Continue the previous Panjer example ($\\lambda=2$, $f_X(1)=0.6$, $f_X(2)=0.4$): find $f_S(3)$.", "back": "Recall $f_S(0)\\approx0.135335$, $f_S(1)\\approx0.162402$, $f_S(2)\\approx0.205709$.\n$f_S(3)=\\frac{2}{3}\\bigl[1\\cdot f_X(1) f_S(2) + 2\\cdot f_X(2) f_S(1)\\bigr]$ (only $j=1,2$ since $f_X(j)=0$ for $j\\ge3$).\n$=\\frac{2}{3}\\bigl[0.6(0.205709)+2(0.4)(0.162402)\\bigr]$\n$=\\frac{2}{3}\\bigl[0.123425+0.129922\\bigr]=\\frac{2}{3}(0.253347)\\approx 0.168898$.", "tag": "Panjer recursion" }, { "front": "Why does Panjer's recursion require the **severity to be on a discrete (arithmetic) grid**, and how is a continuous severity handled?", "back": "The recursion sums over integer severity values $j$, so severity mass must sit on a lattice $0,h,2h,\\dots$ (after scaling, the integers). A continuous severity must first be **discretized** onto a span $h$ — e.g. by the method of rounding (mass-matching to the nearest grid point) or the method of local moment matching — turning $F_X$ into $f_X(0),f_X(h),f_X(2h),\\dots$. Smaller $h$ gives a finer, more accurate approximation at greater computational cost.", "tag": "Panjer recursion" }, { "front": "What is the $(a,b,1)$ class, and how does the Panjer recursion change for it?", "back": "The $(a,b,1)$ class lets $p_0$ be set freely (zero-modified or zero-truncated) while the ratio $\\frac{p_k}{p_{k-1}}=a+\\frac{b}{k}$ holds for $k\\ge 2$ only. The aggregate recursion gains a correction term for the first claim:\n$f_S(s)=\\dfrac{\\bigl[p_1-(a+b)p_0\\bigr]f_X(s)+\\sum_{j=1}^{s}\\bigl(a+\\frac{bj}{s}\\bigr)f_X(j)f_S(s-j)}{1-a f_X(0)}$.\nUse it for zero-truncated or zero-modified frequencies. The $(a,b,0)$ recursion is the special case $p_1=(a+b)p_0$.", "tag": "Panjer recursion" }, { "front": "State the **individual risk model** for aggregate losses over a fixed portfolio.", "back": "$S = X_1 + X_2 + \\dots + X_n$, where $n$ is a **fixed, known** number of policies (not random) and $X_i$ is the loss on policy $i$. The $X_i$ are assumed **independent** but need **not** be identically distributed — each policy can have its own loss law. Typically $X_i = I_i\\,B_i$, where $I_i$ is a claim indicator (claim or no claim) and $B_i$ is the claim amount given a claim.", "tag": "Individual risk model" }, { "front": "Contrast the **individual** and **collective** risk models.", "back": "**Individual risk model:** $S=\\sum_{i=1}^{n}X_i$ over a **fixed** number $n$ of policies; the $X_i$ are independent but generally **not identical**; each $X_i$ usually has a probability mass at $0$ (no claim).\n**Collective risk model:** $S=\\sum_{i=1}^{N}X_i$ with a **random** count $N$; the $X_i$ are i.i.d. and represent only the **positive** claim amounts.\nThe collective model aggregates by claim; the individual model aggregates by policy.", "tag": "Individual risk model" }, { "front": "In the individual risk model with $X_i = I_i B_i$ (Bernoulli indicator $I_i$ with $P(I_i=1)=q_i$, claim amount $B_i$), give $E[X_i]$ and $Var[X_i]$.", "back": "$E[X_i] = q_i\\,E[B_i]$.\n$Var[X_i] = q_i\\,E[B_i^2] - \\bigl(q_i E[B_i]\\bigr)^2 = q_i\\,Var[B_i] + q_i(1-q_i)\\,E[B_i]^2$.\nThe second form separates **claim-size variability** ($q_i Var[B_i]$) from **claim-occurrence variability** ($q_i(1-q_i)E[B_i]^2$). If the claim amount is a fixed constant $b_i$ (so $Var[B_i]=0$): $E[X_i]=q_i b_i$ and $Var[X_i]=q_i(1-q_i)b_i^2$.", "tag": "Individual risk model" }, { "front": "In the individual risk model, how are $E[S]$ and $Var[S]$ obtained from the per-policy moments?", "back": "By independence of the policies,\n$E[S]=\\sum_{i=1}^{n}E[X_i]$ and $Var[S]=\\sum_{i=1}^{n}Var[X_i]$.\nNo cross terms appear because the $X_i$ are independent. You just add the policy means and add the policy variances — even though the policies are not identically distributed.", "tag": "Individual risk model" }, { "front": "A portfolio has 3 independent policies. Policy probabilities of a claim and fixed claim amounts: $(q_1,b_1)=(0.10,\\$1{,}000)$, $(q_2,b_2)=(0.05,\\$2{,}000)$, $(q_3,b_3)=(0.20,\\$500)$. Find $E[S]$ and $Var[S]$.", "back": "Per-policy (constant claim amount): $E[X_i]=q_i b_i$, $Var[X_i]=q_i(1-q_i)b_i^2$.\n$E[X_1]=0.10(1000)=100$; $E[X_2]=0.05(2000)=100$; $E[X_3]=0.20(500)=100$. $E[S]=300$.\n$Var[X_1]=0.10(0.90)(1000^2)=90{,}000$.\n$Var[X_2]=0.05(0.95)(2000^2)=0.0475(4{,}000{,}000)=190{,}000$.\n$Var[X_3]=0.20(0.80)(500^2)=0.16(250{,}000)=40{,}000$.\n$Var[S]=90{,}000+190{,}000+40{,}000=320{,}000$.", "tag": "Individual risk model" }, { "front": "A group of 100 identical lives each has claim probability $q=0.03$; a claim pays a fixed $\\$10{,}000$. Find $E[S]$ and $Var[S]$ in the individual risk model.", "back": "Each life: $E[X_i]=q b=0.03(10{,}000)=300$; $Var[X_i]=q(1-q)b^2=0.03(0.97)(10{,}000^2)=0.0291(10^8)=2{,}910{,}000$.\nWith $n=100$ independent lives:\n$E[S]=100(300)=\\$30{,}000$.\n$Var[S]=100(2{,}910{,}000)=291{,}000{,}000$.\nStandard deviation $=\\sqrt{291{,}000{,}000}\\approx \\$17{,}058.72$.", "tag": "Individual risk model" }, { "front": "A policy pays claim amount $B$ with mean $\\$4{,}000$ and variance $\\$6{,}000{,}000$ when a claim occurs, with claim probability $q=0.08$. Find $E[X]$ and $Var[X]$ for this single policy.", "back": "$X=IB$ with $P(I=1)=0.08$.\n$E[X]=qE[B]=0.08(4{,}000)=\\$320$.\nUse $Var[X]=qVar[B]+q(1-q)E[B]^2$:\n$=0.08(6{,}000{,}000)+0.08(0.92)(4{,}000^2)$\n$=480{,}000+0.0736(16{,}000{,}000)=480{,}000+1{,}177{,}600=1{,}657{,}600$.\nStandard deviation $=\\sqrt{1{,}657{,}600}\\approx \\$1{,}287.48$.", "tag": "Individual risk model" }, { "front": "How is the **normal approximation** used to estimate $P(S>x)$ for aggregate losses?", "back": "Approximate $S$ by a normal with the same mean and variance: $S\\approx N\\bigl(E[S],\\,Var[S]\\bigr)$. Then\n$P(S>x)\\approx P\\!\\left(Z>\\dfrac{x-E[S]}{\\sqrt{Var[S]}}\\right)=1-\\Phi\\!\\left(\\dfrac{x-E[S]}{\\sqrt{Var[S]}}\\right)$.\nIt is most accurate when $E[N]$ is large (many expected claims), so $S$ is roughly symmetric. It can be poor for skewed, small-portfolio aggregates because the normal ignores skewness.", "tag": "Approximations & stop-loss" }, { "front": "Aggregate losses $S$ have $E[S]=\\$10{,}000$ and $Var[S]=4{,}000{,}000$. Using the normal approximation, find $P(S>13{,}000)$. (Use $\\Phi(1.5)=0.9332$.)", "back": "Standard deviation $=\\sqrt{4{,}000{,}000}=2{,}000$.\n$z=\\dfrac{13{,}000-10{,}000}{2{,}000}=\\dfrac{3{,}000}{2{,}000}=1.5$.\n$P(S>13{,}000)\\approx 1-\\Phi(1.5)=1-0.9332=0.0668$.\nSo there is about a $6.7\\%$ chance aggregate losses exceed $\\$13{,}000$.", "tag": "Approximations & stop-loss" }, { "front": "Find the **$95$th percentile** of aggregate losses with $E[S]=\\$50{,}000$ and standard deviation $\\$8{,}000$ using the normal approximation. (Use $z_{0.95}=1.645$.)", "back": "The normal $95$th percentile is $E[S]+z_{0.95}\\,\\sigma_S$.\n$=50{,}000 + 1.645(8{,}000)=50{,}000+13{,}160=\\$63{,}160$.\nThis is the capital level such that aggregate losses stay below it with $95\\%$ probability under the normal approximation.", "tag": "Approximations & stop-loss" }, { "front": "How is the **lognormal approximation** to $S$ set up, and why is it sometimes preferred over the normal?", "back": "Fit a lognormal $S\\approx e^{Y}$ with $Y\\sim N(\\mu,\\sigma^2)$ by **moment matching**: solve $E[S]=e^{\\mu+\\sigma^2/2}$ and $E[S^2]=e^{2\\mu+2\\sigma^2}$. Equivalently $\\sigma^2=\\ln\\!\\bigl(1+\\tfrac{Var[S]}{E[S]^2}\\bigr)$ and $\\mu=\\ln E[S]-\\tfrac{\\sigma^2}{2}$.\nThen $P(S\\le x)\\approx\\Phi\\!\\left(\\dfrac{\\ln x-\\mu}{\\sigma}\\right)$.\nIt is preferred when $S$ is **right-skewed and positive** (small portfolios, heavy-tailed severity), where the symmetric normal fits poorly.", "tag": "Approximations & stop-loss" }, { "front": "Aggregate losses have $E[S]=\\$1{,}000$ and $Var[S]=640{,}000$. Fit a lognormal by moment matching and estimate $P(S>2{,}000)$. (Use $\\Phi(1.34)=0.9099$.)", "back": "$\\sigma^2=\\ln\\!\\bigl(1+\\frac{Var[S]}{E[S]^2}\\bigr)=\\ln\\!\\bigl(1+\\frac{640{,}000}{1{,}000{,}000}\\bigr)=\\ln(1.64)\\approx 0.494696$, so $\\sigma\\approx 0.703346$.\n$\\mu=\\ln E[S]-\\frac{\\sigma^2}{2}=\\ln 1000 - 0.247348 = 6.907755-0.247348=6.660407$.\n$z=\\dfrac{\\ln 2000-\\mu}{\\sigma}=\\dfrac{7.600902-6.660407}{0.703347}=\\dfrac{0.940495}{0.703347}\\approx 1.337$.\n$P(S>2000)\\approx 1-\\Phi(1.34)\\approx 1-0.9099=0.0901\\approx 0.090$.", "tag": "Approximations & stop-loss" }, { "front": "Define the **stop-loss premium** with retention (deductible) $d$ and give two equivalent expressions.", "back": "The stop-loss (net) premium is the expected loss above the retention $d$:\n$E[(S-d)_+]=\\displaystyle\\int_{d}^{\\infty}(s-d)\\,f_S(s)\\,ds = \\int_{d}^{\\infty}\\bigl[1-F_S(s)\\bigr]\\,ds$ (continuous case),\nor for discrete $S$, $E[(S-d)_+]=\\sum_{s>d}(s-d)f_S(s)$.\nHere $(S-d)_+=\\max(S-d,0)$. It is the pure premium for stop-loss reinsurance attaching at $d$.", "tag": "Approximations & stop-loss" }, { "front": "State the key identity linking the stop-loss premium $E[(S-d)_+]$ to the **limited expected value** $E[S\\wedge d]$.", "back": "$E[(S-d)_+] = E[S] - E[S\\wedge d]$,\nwhere $S\\wedge d=\\min(S,d)$ is the loss limited at $d$. This holds because $S = (S\\wedge d) + (S-d)_+$ for every outcome, so taking expectations splits $E[S]$ into the part below the retention and the part above it. It is the fastest route to a stop-loss premium once $E[S\\wedge d]$ is known.", "tag": "Approximations & stop-loss" }, { "front": "Aggregate $S$ takes values $0,1,2,3$ with probabilities $0.5,\\,0.3,\\,0.15,\\,0.05$. Find the stop-loss premium $E[(S-1)_+]$ at retention $d=1$.", "back": "$E[(S-1)_+]=\\sum_{s>1}(s-1)f_S(s)=(2-1)(0.15)+(3-1)(0.05)=0.15+0.10=0.25$.\nCheck via $E[(S-1)_+]=E[S]-E[S\\wedge 1]$: $E[S]=0(0.5)+1(0.3)+2(0.15)+3(0.05)=0.75$; $E[S\\wedge1]=0(0.5)+1(0.5)=0.50$; difference $=0.75-0.50=0.25$. ✓", "tag": "Approximations & stop-loss" }, { "front": "State the **integer-retention recursion** for stop-loss premiums when $S$ is on the non-negative integers.", "back": "Stepping the retention up by one unit,\n$E[(S-(d+1))_+] = E[(S-d)_+] - \\bigl[1-F_S(d)\\bigr]$,\nwhere $1-F_S(d)=P(S>d)$ is the survival probability at $d$. Each increase of the deductible by $1$ reduces the stop-loss premium by the probability that $S$ exceeds the current retention. Start from $E[(S-0)_+]=E[S]$ and march upward.", "tag": "Approximations & stop-loss" }, { "front": "For $S$ with $f_S(0)=0.5,\\,f_S(1)=0.3,\\,f_S(2)=0.15,\\,f_S(3)=0.05$ (and $E[S]=0.75$), use the integer recursion to find the stop-loss premiums at $d=0,1,2,3$.", "back": "Survival probabilities: $S_S(0)=P(S>0)=0.5$, $S_S(1)=0.2$, $S_S(2)=0.05$, $S_S(3)=0$.\nRecursion $E[(S-(d+1))_+]=E[(S-d)_+]-P(S>d)$, starting at $E[(S-0)_+]=E[S]=0.75$.\n$d=1$: $0.75-0.5=0.25$.\n$d=2$: $0.25-0.2=0.05$.\n$d=3$: $0.05-0.05=0.00$.\nSo stop-loss premiums are $0.75,\\,0.25,\\,0.05,\\,0.00$ at $d=0,1,2,3$.", "tag": "Approximations & stop-loss" }, { "front": "A compound Poisson $S$ has $\\lambda=3$ and severity always $\\$100$ per claim. Find the stop-loss premium at retention $d=\\$100$. (Use $e^{-3}\\approx 0.049787$.)", "back": "$S$ takes values $0,100,200,\\dots$ with $P(S=100k)=P(N=k)=e^{-3}\\frac{3^{k}}{k!}$.\n$E[S]=\\lambda(100)=300$.\n$E[(S-100)_+]=E[S]-E[S\\wedge100]$. Here $E[S\\wedge100]=100\\cdot P(S\\ge100)=100\\,P(N\\ge1)=100(1-e^{-3})$.\n$1-e^{-3}\\approx 1-0.049787=0.950213$, so $E[S\\wedge100]\\approx 95.0213$.\n$E[(S-100)_+]\\approx 300-95.0213=\\$204.98$.", "tag": "Approximations & stop-loss" }, { "front": "Aggregate losses $S$ are approximated as $N(\\$10{,}000,\\,2{,}000^2)$. Estimate the stop-loss premium $E[(S-d)_+]$ at $d=\\$12{,}000$ using the normal stop-loss formula. (Use $\\phi(1)=0.241971$, $\\Phi(1)=0.8413$.)", "back": "For $S\\sim N(\\mu,\\sigma^2)$ with $z=\\frac{d-\\mu}{\\sigma}$:\n$E[(S-d)_+]=\\sigma\\,\\phi(z)-(d-\\mu)\\,\\bigl[1-\\Phi(z)\\bigr]$.\nHere $\\mu=10{,}000$, $\\sigma=2{,}000$, $d=12{,}000$, so $z=\\frac{2{,}000}{2{,}000}=1$.\n$=2{,}000(0.241971)-2{,}000(1-0.8413)$\n$=483.942-2{,}000(0.1587)=483.942-317.40=\\$166.54$.", "tag": "Approximations & stop-loss" }, { "front": "Why can two different aggregate models share the same $E[S]$ and $Var[S]$ yet give different **stop-loss premiums and tail probabilities**?", "back": "The mean and variance fix only the first two moments; stop-loss premiums and tail probabilities $P(S>x)$ depend on the **whole distribution shape**, especially **skewness and tail thickness**. A heavier-tailed or more right-skewed $S$ puts more mass far above the mean, raising high-retention stop-loss premiums and tail probabilities even with identical variance. This is why moment-matching approximations (normal vs lognormal) can disagree, and why the exact Panjer-built $f_S$ is preferred for tail quantities when feasible.", "tag": "Approximations & stop-loss" } ] }