{ "deckName": "Exam MAS-I — Frequency, Severity & Aggregate Models", "examCode": "Exam MAS-I", "cards": [ { "front": "State the defining recursion of the $(a,b,0)$ frequency class and name its four members.", "back": "A counting distribution is in the $(a,b,0)$ class if its probabilities satisfy\n$\\frac{p_k}{p_{k-1}} = a + \\frac{b}{k}, \\quad k=1,2,3,\\dots$\nfor constants $a$ and $b$, with $p_0$ set so the probabilities sum to $1$.\nThe only members are:\n- **Poisson** ($a=0$, $b=\\lambda$),\n- **Binomial** ($a<0$),\n- **Negative binomial** ($01$** → **negative binomial / geometric** (overdispersed; variance above mean).\nThis is the fastest screening tool: compute the sample mean and variance of the counts and read off the family.", "tag": "(a,b,0) & (a,b,1)" }, { "front": "A portfolio's claim counts have sample mean $\\bar N = 0.40$ and sample variance $s^2 = 0.58$. Which $(a,b,0)$ distribution fits, and estimate its parameters by moment matching.", "back": "Variance-to-mean $=\\frac{0.58}{0.40}=1.45>1$, so use the **negative binomial**.\nMoment match: $E[N]=r\\beta=0.40$ and $\\text{Var}(N)=r\\beta(1+\\beta)=0.58$.\nDivide: $\\frac{0.58}{0.40}=1+\\beta \\Rightarrow \\beta = 0.45$.\nThen $r = \\frac{0.40}{\\beta} = \\frac{0.40}{0.45} \\approx 0.889$.\nSo $\\hat\\beta = 0.45$, $\\hat r \\approx 0.889$.", "tag": "(a,b,0) & (a,b,1)" }, { "front": "Observed counts give relative frequencies with $\\frac{p_1}{p_0}=0.6$ and $\\frac{p_2}{p_1}=0.45$. Use the $(a,b,0)$ test to find $a$, $b$ and identify the distribution.", "back": "The recursion at $k=1,2$ gives $a+b = 0.6$ and $a + \\frac{b}{2} = 0.45$.\nSubtract: $\\frac{b}{2}=0.15 \\Rightarrow b = 0.30$, then $a = 0.6-0.30 = 0.30$.\nSince $01$), $E[X^2]=\\frac{2\\theta^2}{(\\alpha-1)(\\alpha-2)}$ (for $\\alpha>2$), so $\\text{Var}(X)=\\frac{\\alpha\\theta^2}{(\\alpha-1)^2(\\alpha-2)}$.", "tag": "Severity & moments" }, { "front": "State the mean, variance, and key moment of the **lognormal** and **Weibull** severity distributions.", "back": "**Lognormal($\\mu,\\sigma$):** $\\ln X \\sim N(\\mu,\\sigma^2)$. The $k$-th raw moment is $E[X^k]=\\exp\\!\\big(k\\mu+\\tfrac{1}{2}k^2\\sigma^2\\big)$.\nSo $E[X]=e^{\\mu+\\sigma^2/2}$ and $E[X^2]=e^{2\\mu+2\\sigma^2}$, giving $\\text{Var}(X)=e^{2\\mu+\\sigma^2}\\big(e^{\\sigma^2}-1\\big)$.\n**Weibull($\\tau,\\theta$):** $S(x)=\\exp\\!\\big(-(x/\\theta)^\\tau\\big)$, with $E[X]=\\theta\\,\\Gamma\\!\\big(1+\\tfrac{1}{\\tau}\\big)$ and $E[X^2]=\\theta^2\\,\\Gamma\\!\\big(1+\\tfrac{2}{\\tau}\\big)$.", "tag": "Severity & moments" }, { "front": "Define the **limited expected value** $E[X\\wedge u]$ and give its integral form in terms of the survival function.", "back": "$X\\wedge u = \\min(X,u)$ caps the loss at $u$. Its expectation is\n$E[X\\wedge u]=\\int_0^u x\\,f(x)\\,dx + u\\,S(u) = \\int_0^u S(x)\\,dx.$\nThe last form (integrating the survival function from $0$ to $u$) is usually the easiest. As $u\\to\\infty$, $E[X\\wedge u]\\to E[X]$.", "tag": "Severity & moments" }, { "front": "Give the closed-form limited expected value $E[X\\wedge u]$ for the **exponential** and the **Pareto** distributions.", "back": "**Exponential($\\theta$):** $E[X\\wedge u]=\\theta\\big(1-e^{-u/\\theta}\\big)$.\n**Pareto($\\alpha,\\theta$):** for $\\alpha\\neq 1$,\n$E[X\\wedge u]=\\frac{\\theta}{\\alpha-1}\\left[1-\\left(\\frac{\\theta}{\\theta+u}\\right)^{\\alpha-1}\\right].$\nBoth approach the full mean ($\\theta$ and $\\frac{\\theta}{\\alpha-1}$) as $u\\to\\infty$.", "tag": "Severity & moments" }, { "front": "Losses follow an **exponential** with $\\theta = 1000$. Compute $E[X\\wedge 500]$ and the variance of $X$.", "back": "Variance: $\\text{Var}(X)=\\theta^2 = 1000^2 = 1{,}000{,}000$.\nLimited expected value: $E[X\\wedge 500]=\\theta\\big(1-e^{-500/1000}\\big)=1000\\big(1-e^{-0.5}\\big)$.\nSince $e^{-0.5}\\approx 0.606531$, $E[X\\wedge 500]=1000(0.393469)\\approx \\$393.47$.", "tag": "Severity & moments" }, { "front": "Losses follow a **Pareto** with $\\alpha = 3$, $\\theta = 2000$. Find $E[X]$, $\\text{Var}(X)$, and $E[X\\wedge 1000]$.", "back": "$E[X]=\\frac{\\theta}{\\alpha-1}=\\frac{2000}{2}=1000$.\n$E[X^2]=\\frac{2\\theta^2}{(\\alpha-1)(\\alpha-2)}=\\frac{2(2000)^2}{(2)(1)}=4{,}000{,}000$, so $\\text{Var}(X)=4{,}000{,}000 - 1000^2 = 3{,}000{,}000$.\n$E[X\\wedge 1000]=\\frac{\\theta}{\\alpha-1}\\Big[1-\\big(\\tfrac{\\theta}{\\theta+u}\\big)^{\\alpha-1}\\Big]=1000\\Big[1-\\big(\\tfrac{2000}{3000}\\big)^{2}\\Big]=1000\\big(1-0.4444\\big)\\approx \\$555.56$.", "tag": "Severity & moments" }, { "front": "A severity is **lognormal** with $\\mu = 7$, $\\sigma = 1.2$. Find $E[X]$ and $\\text{Var}(X)$.", "back": "$E[X]=e^{\\mu+\\sigma^2/2}=e^{7+0.72}=e^{7.72}$. Since $e^{7}\\approx 1096.633$ and $e^{0.72}\\approx 2.054433$, $E[X]\\approx 1096.633(2.054433)\\approx \\$2{,}253.00$.\n$\\text{Var}(X)=E[X]^2\\big(e^{\\sigma^2}-1\\big)$ since $\\text{Var}(X)=e^{2\\mu+\\sigma^2}\\big(e^{\\sigma^2}-1\\big)$.\nWith $e^{\\sigma^2}=e^{1.44}\\approx 4.220696$, $\\text{Var}(X)\\approx 2253.00^2(4.220696-1)\\approx 5{,}076{,}009(3.220696)\\approx 1.635\\times 10^{7}$.", "tag": "Severity & moments" }, { "front": "Define the ordinary **deductible** $d$ and write the per-loss expected payment $E[(X-d)_+]$ in terms of limited expected values.", "back": "With an ordinary deductible $d$, a loss $x$ pays $\\max(x-d,0)=(x-d)_+$. The expected payment **per loss** (averaging over all losses, including those that pay $0$) is\n$E[(X-d)_+]=E[X]-E[X\\wedge d].$\nThis is the total expected loss minus the part capped at the deductible — equivalently $\\int_d^\\infty S(x)\\,dx$.", "tag": "Coverage modifications" }, { "front": "Distinguish the **per-loss** and **per-payment** expected cost of a deductible $d$, and give the per-payment formula.", "back": "**Per-loss** averages over every loss, counting non-payments (loss $\\le d$) as $0$: $E[(X-d)_+]=E[X]-E[X\\wedge d]$.\n**Per-payment** conditions on a payment actually occurring (loss $>d$), dividing by the probability of a payment $S(d)$:\n$E[(X-d)_+\\mid X>d]=\\frac{E[X]-E[X\\wedge d]}{S(d)}.$\nThe per-payment cost is always $\\ge$ the per-loss cost, since $S(d)\\le 1$.", "tag": "Coverage modifications" }, { "front": "Write the expected payment **per loss** under a policy with deductible $d$, maximum covered loss $u$, and coinsurance $\\alpha$.", "back": "With coinsurance factor $\\alpha$ (insurer pays a fraction $\\alpha$), deductible $d$, and a maximum covered loss $u$ (so the maximum payment is $\\alpha(u-d)$), the expected payment per loss is\n$\\alpha\\big(E[X\\wedge u]-E[X\\wedge d]\\big).$\nHere $u$ is the loss level at which the payment caps; the policy limit (max payment) equals $\\alpha(u-d)$. Apply coinsurance **last**, after the deductible-and-limit layer is formed.", "tag": "Coverage modifications" }, { "front": "How does an ordinary deductible $d$ change the **number of payments** relative to the number of losses?", "back": "Each loss independently produces a payment only if it exceeds $d$, with probability $v = S(d) = P[X>d]$. So the payment count is a **thinned** version of the loss count $N$.\nIf losses are **Poisson($\\lambda$)**, the payment count is **Poisson($\\lambda v$)**.\nIf losses are **negative binomial($r,\\beta$)**, payments are **negative binomial($r,\\beta v$)**.\nIf losses are **binomial($m,q$)**, payments are **binomial($m,qv$)**. The frequency family is preserved; only the relevant parameter is scaled by $v$.", "tag": "Coverage modifications" }, { "front": "Losses are **exponential** with $\\theta = 800$. A policy has an ordinary deductible of $200$. Find the expected payment **per loss** and **per payment**.", "back": "$E[X]=\\theta=800$. $E[X\\wedge 200]=800\\big(1-e^{-200/800}\\big)=800\\big(1-e^{-0.25}\\big)$. With $e^{-0.25}\\approx 0.778801$, $E[X\\wedge 200]=800(0.221199)\\approx 176.96$.\n**Per loss:** $E[(X-200)_+]=800-176.96=\\$623.04$.\n**Per payment:** divide by $S(200)=e^{-0.25}\\approx 0.778801$: $\\frac{623.04}{0.778801}\\approx \\$800.00$.\n(For the exponential, the memoryless property makes the per-payment cost equal the full mean $\\theta$.)", "tag": "Coverage modifications" }, { "front": "Losses are **Pareto** with $\\alpha = 3$, $\\theta = 2000$. A policy pays $80\\%$ coinsurance on losses between a $\\$500$ deductible and a $\\$5000$ maximum covered loss. Find the expected payment per loss.", "back": "Use $E[X\\wedge u]=\\frac{\\theta}{\\alpha-1}\\big[1-(\\tfrac{\\theta}{\\theta+u})^{2}\\big]$ with $\\frac{\\theta}{\\alpha-1}=1000$.\n$E[X\\wedge 5000]=1000\\big[1-(\\tfrac{2000}{7000})^2\\big]=1000(1-0.081633)\\approx 918.37$.\n$E[X\\wedge 500]=1000\\big[1-(\\tfrac{2000}{2500})^2\\big]=1000(1-0.64)=360.00$.\nLayer $=918.37-360.00=558.37$; apply coinsurance: $0.80(558.37)\\approx \\$446.70$.", "tag": "Coverage modifications" }, { "front": "A **Poisson** loss frequency has $\\lambda = 6$ per year. Severities are exponential with $\\theta = 1000$ and a $\\$500$ deductible is applied. What is the distribution and mean of the **number of payments** per year?", "back": "Probability a loss pierces the deductible: $v=S(500)=e^{-500/1000}=e^{-0.5}\\approx 0.606531$.\nThinning a Poisson keeps it Poisson with rate $\\lambda v$:\nnumber of payments $\\sim$ **Poisson($6 \\times 0.606531$)** $=$ **Poisson($3.639$)**.\nSo the expected number of payments per year is $\\approx 3.64$.", "tag": "Coverage modifications" }, { "front": "State the collective-risk (compound) model for aggregate losses $S$ and the conditions on $N$ and the $X_i$.", "back": "$S = X_1 + X_2 + \\cdots + X_N = \\sum_{i=1}^{N} X_i$, where:\n- $N$ is the **claim count** (frequency),\n- the $X_i$ are i.i.d. **severities**, independent of $N$,\n- when $N=0$, $S=0$.\nFrequency and severity are modeled separately and combined. $S$ is called a **compound** distribution (e.g. compound Poisson if $N$ is Poisson).", "tag": "Aggregate moments" }, { "front": "State the mean and variance of the aggregate loss $S=\\sum_{i=1}^N X_i$ in the collective-risk model.", "back": "$E[S]=E[N]\\,E[X]$ (product of the frequency mean and severity mean).\n$\\text{Var}(S)=E[N]\\,\\text{Var}(X)+\\text{Var}(N)\\,E[X]^2.$\nThe first term is the severity contribution to variance; the second captures the variability in the number of claims. These follow from conditioning on $N$ (the compound-variance / law-of-total-variance identity).", "tag": "Aggregate moments" }, { "front": "Derive $\\text{Var}(S)$ for a compound distribution using conditioning on $N$ (law of total variance).", "back": "Condition on $N$. With i.i.d. severities independent of $N$:\n$E[S\\mid N]=N\\,E[X]$ and $\\text{Var}(S\\mid N)=N\\,\\text{Var}(X)$.\nLaw of total variance:\n$\\text{Var}(S)=E\\big[\\text{Var}(S\\mid N)\\big]+\\text{Var}\\big(E[S\\mid N]\\big)$\n$=E[N]\\,\\text{Var}(X)+\\text{Var}(N)\\,E[X]^2.$\nThe two pieces are the expected within-$N$ variance plus the variance of the conditional mean.", "tag": "Aggregate moments" }, { "front": "Frequency is **Poisson($\\lambda = 4$)** and severity has $E[X]=500$, $\\text{Var}(X)=250{,}000$. Find $E[S]$ and $\\text{Var}(S)$.", "back": "$E[S]=E[N]E[X]=4(500)=\\$2{,}000$.\nFor a Poisson, $E[N]=\\text{Var}(N)=4$, so\n$\\text{Var}(S)=E[N]\\text{Var}(X)+\\text{Var}(N)E[X]^2 = 4(250{,}000)+4(500)^2$\n$=1{,}000{,}000 + 4(250{,}000)=1{,}000{,}000 + 1{,}000{,}000 = 2{,}000{,}000$.\nStandard deviation $=\\sqrt{2{,}000{,}000}\\approx \\$1{,}414.21$.", "tag": "Aggregate moments" }, { "front": "Frequency is **negative binomial** with $r=3$, $\\beta=2$ and severity is **exponential** with $\\theta=400$. Find $E[S]$ and $\\text{Var}(S)$.", "back": "Frequency: $E[N]=r\\beta=6$, $\\text{Var}(N)=r\\beta(1+\\beta)=3(2)(3)=18$.\nSeverity: $E[X]=400$, $\\text{Var}(X)=\\theta^2=160{,}000$.\n$E[S]=6(400)=\\$2{,}400$.\n$\\text{Var}(S)=E[N]\\text{Var}(X)+\\text{Var}(N)E[X]^2 = 6(160{,}000)+18(400)^2$\n$=960{,}000 + 18(160{,}000)=960{,}000+2{,}880{,}000=3{,}840{,}000$.", "tag": "Aggregate moments" }, { "front": "State the special variance identity for a **compound Poisson** aggregate, and why it simplifies.", "back": "If $N\\sim$ Poisson($\\lambda$), then $E[N]=\\text{Var}(N)=\\lambda$, so the general formula collapses to\n$\\text{Var}(S)=\\lambda\\,\\text{Var}(X)+\\lambda\\,E[X]^2 = \\lambda\\big(\\text{Var}(X)+E[X]^2\\big)=\\lambda\\,E[X^2].$\nAlso $E[S]=\\lambda E[X]$. The compact result $\\text{Var}(S)=\\lambda E[X^2]$ uses only the **second raw moment** of severity — no need to split into mean and variance.", "tag": "Compound Poisson & Panjer" }, { "front": "A **compound Poisson** has $\\lambda = 10$ and severity uniform on $(0,2000)$. Find $E[S]$ and $\\text{Var}(S)$.", "back": "Uniform$(0,2000)$: $E[X]=1000$, and $E[X^2]=\\frac{2000^2}{3}=\\frac{4{,}000{,}000}{3}\\approx 1{,}333{,}333$.\n$E[S]=\\lambda E[X]=10(1000)=\\$10{,}000$.\n$\\text{Var}(S)=\\lambda E[X^2]=10\\big(1{,}333{,}333\\big)\\approx 13{,}333{,}333$.\nStandard deviation $\\approx \\sqrt{13{,}333{,}333}\\approx \\$3{,}651.48$.", "tag": "Compound Poisson & Panjer" }, { "front": "State the **Panjer recursion** for the aggregate probabilities $g_s = P[S=s]$ when frequency is in the $(a,b,0)$ class and severity is discrete on $0,1,2,\\dots$.", "back": "For $(a,b,0)$ frequency and severity probabilities $f_j = P[X=j]$ on the non-negative integers,\n$g_s = \\frac{1}{1-a\\,f_0}\\sum_{j=1}^{s}\\Big(a+\\frac{b\\,j}{s}\\Big)f_j\\,g_{s-j}, \\quad s=1,2,\\dots$\nThe starting value is $g_0$, the probability that the aggregate is $0$; e.g. for Poisson, $g_0 = e^{-\\lambda(1-f_0)}$. The recursion replaces a costly convolution with an $O(s^2)$ pass.", "tag": "Compound Poisson & Panjer" }, { "front": "Give the Panjer starting value $g_0=P[S=0]$ for a compound Poisson, and explain when $S=0$.", "back": "$S=0$ occurs when the total of all claims is $0$. For a **compound Poisson($\\lambda$)** with severity probability of a zero claim $f_0=P[X=0]$,\n$g_0 = e^{-\\lambda(1-f_0)}.$\nIf severity is strictly positive ($f_0=0$), this reduces to $g_0 = e^{-\\lambda}=P[N=0]$ — aggregate is $0$ exactly when there are no claims.", "tag": "Compound Poisson & Panjer" }, { "front": "Run the **Panjer recursion** for a compound Poisson with $\\lambda = 2$ and discrete severity $f_1 = 0.6$, $f_2 = 0.4$ (so $f_0=0$). Find $g_0$, $g_1$, and $g_2$.", "back": "Poisson: $a=0$, $b=\\lambda=2$, and $f_0=0$ so $1-a f_0 = 1$.\n$g_0 = e^{-\\lambda(1-f_0)} = e^{-2}\\approx 0.135335$.\n$g_1 = \\sum_{j=1}^{1}\\big(0+\\tfrac{2j}{1}\\big)f_j g_{1-j} = (2)(0.6)g_0 = 1.2(0.135335)\\approx 0.162402$.\n$g_2 = \\sum_{j=1}^{2}\\big(\\tfrac{2j}{2}\\big)f_j g_{2-j} = (1)f_1 g_1 + (2)f_2 g_0$\n$= (0.6)(0.162402) + (2)(0.4)(0.135335) \\approx 0.097441 + 0.108268 = 0.205709$.", "tag": "Compound Poisson & Panjer" }, { "front": "Run the **Panjer recursion** for a compound **binomial** with $m=3$, $q=0.2$ and severity $f_1=1$ (each claim equals $1$). Find $g_0$ and $g_1$.", "back": "Binomial $(a,b,0)$ params: $a=-\\frac{q}{1-q}=-\\frac{0.2}{0.8}=-0.25$, $b=(m+1)\\frac{q}{1-q}=4(0.25)=1.0$. Here $f_0=0$, $f_1=1$.\n$g_0 = P[N=0]=(1-q)^m=0.8^3=0.512$.\nSince $f_0=0$, $1-a f_0=1$.\n$g_1 = \\sum_{j=1}^{1}\\big(a+\\tfrac{b j}{1}\\big)f_j g_0 = (a+b)f_1 g_0 = (-0.25+1.0)(1)(0.512)=0.75(0.512)=0.384.$\n(Check: $S$ here equals $N$, and $P[N=1]=\\binom{3}{1}0.2(0.8)^2=0.384$. ✓)", "tag": "Compound Poisson & Panjer" }, { "front": "Why is the **Panjer recursion** preferred over direct convolution for computing the aggregate distribution?", "back": "Direct convolution sums over all ways the claims add to $s$ and over every possible claim count $n$, which is computationally heavy and grows fast. Panjer exploits the $(a,b,0)$/$(a,b,1)$ recursive structure of the frequency to compute each $g_s$ from the previously computed $g_0,\\dots,g_{s-1}$ in a single pass, reducing the work to $O(s^2)$ for a target span of $s$ values and avoiding nested summation over $n$.", "tag": "Compound Poisson & Panjer" }, { "front": "Describe the **normal approximation** to the aggregate loss $S$ and when it is reasonable.", "back": "Treat $S$ as approximately $N\\big(E[S],\\,\\text{Var}(S)\\big)$ and standardize:\n$P[S\\le s]\\approx \\Phi\\!\\left(\\frac{s-E[S]}{\\sqrt{\\text{Var}(S)}}\\right).$\nIt works well when the expected claim count is **large** (CLT makes $S$ roughly symmetric/bell-shaped). It performs poorly for small $E[N]$ or highly skewed severity, where $S$ is right-skewed and the normal understates the upper tail.", "tag": "Approximations" }, { "front": "Aggregate losses have $E[S]=10{,}000$ and $\\text{Var}(S)=4{,}000{,}000$. Using the **normal approximation**, find $P[S>13{,}000]$.", "back": "$\\text{SD}(S)=\\sqrt{4{,}000{,}000}=2000$.\nStandardize: $z=\\frac{13{,}000-10{,}000}{2000}=\\frac{3000}{2000}=1.50$.\n$P[S>13{,}000]\\approx 1-\\Phi(1.50)=1-0.9332=0.0668.$\nSo about a $6.7\\%$ chance aggregate losses exceed $\\$13{,}000$.", "tag": "Approximations" }, { "front": "A **compound Poisson** has $\\lambda = 50$ and severity exponential with $\\theta = 200$. Use the normal approximation to find $P[S < 12{,}000]$.", "back": "$E[X]=200$, $E[X^2]=2\\theta^2=2(200)^2=80{,}000$.\n$E[S]=\\lambda E[X]=50(200)=10{,}000$.\n$\\text{Var}(S)=\\lambda E[X^2]=50(80{,}000)=4{,}000{,}000$, so $\\text{SD}(S)=2000$.\n$z=\\frac{12{,}000-10{,}000}{2000}=1.00$, so $P[S<12{,}000]\\approx \\Phi(1.00)=0.8413.$", "tag": "Approximations" }, { "front": "Describe the **lognormal approximation** to aggregate losses and how its parameters are fitted.", "back": "Approximate $S$ by a lognormal whose first two moments match $E[S]$ and $E[S^2]=\\text{Var}(S)+E[S]^2$. Solve for $\\mu,\\sigma$ from\n$E[S]=e^{\\mu+\\sigma^2/2}, \\qquad E[S^2]=e^{2\\mu+2\\sigma^2}.$\nThen $\\sigma^2 = \\ln\\!\\frac{E[S^2]}{E[S]^2}=\\ln\\!\\big(1+\\frac{\\text{Var}(S)}{E[S]^2}\\big)$ and $\\mu = \\ln E[S]-\\tfrac{1}{2}\\sigma^2$.\nUnlike the normal, the lognormal is right-skewed and positive, capturing the heavy upper tail better for moderate claim counts.", "tag": "Approximations" }, { "front": "Aggregate losses have $E[S]=10{,}000$ and $\\text{Var}(S)=25{,}000{,}000$. Fit a **lognormal** by moment matching and find $P[S>20{,}000]$.", "back": "$\\sigma^2 = \\ln\\!\\big(1+\\frac{\\text{Var}(S)}{E[S]^2}\\big)=\\ln\\!\\big(1+\\frac{25{,}000{,}000}{100{,}000{,}000}\\big)=\\ln(1.25)\\approx 0.223144$, so $\\sigma\\approx 0.472382$.\n$\\mu = \\ln(10{,}000)-\\tfrac{1}{2}(0.223144)=9.210340-0.111572\\approx 9.098768$.\n$P[S>20{,}000]=1-\\Phi\\!\\big(\\frac{\\ln 20{,}000-\\mu}{\\sigma}\\big)$. $\\ln 20{,}000\\approx 9.903488$, so $z=\\frac{9.903488-9.098768}{0.472382}\\approx 1.7034$.\n$P\\approx 1-\\Phi(1.70)=1-0.9554=0.0446.$", "tag": "Approximations" }, { "front": "Define the **stop-loss premium** $E[(S-d)_+]$ and relate it to the limited expected value of $S$.", "back": "A stop-loss (aggregate excess) contract with retention $d$ pays $\\max(S-d,0)=(S-d)_+$. The stop-loss premium (net expected cost) is\n$E[(S-d)_+]=E[S]-E[S\\wedge d]=\\int_d^\\infty \\big(1-F_S(x)\\big)\\,dx.$\nFor a continuous approximation, integrate the aggregate survival function above $d$; for a discrete $S$, sum $\\sum_{x>d}(x-d)\\,P[S=x]$.", "tag": "Approximations" }, { "front": "Aggregate losses are approximated as normal with $E[S]=10{,}000$ and $\\text{SD}(S)=2000$. Find the **stop-loss premium** $E[(S-13{,}000)_+]$.", "back": "For a normal $S$ with mean $\\mu_S$ and SD $\\sigma_S$, the stop-loss premium is\n$E[(S-d)_+]=\\sigma_S\\big[\\phi(z)-z\\,(1-\\Phi(z))\\big], \\quad z=\\frac{d-\\mu_S}{\\sigma_S}.$\nHere $z=\\frac{13{,}000-10{,}000}{2000}=1.50$. With $\\phi(1.5)=\\frac{1}{\\sqrt{2\\pi}}e^{-1.125}\\approx 0.129518$ and $1-\\Phi(1.5)=0.0668$:\n$E[(S-13{,}000)_+]=2000\\big[0.129518 - 1.5(0.0668)\\big]=2000(0.129518-0.10020)=2000(0.029318)\\approx \\$58.64$.", "tag": "Approximations" }, { "front": "Aggregate $S$ is discrete with $P[S=0]=0.5$, $P[S=1000]=0.3$, $P[S=2000]=0.15$, $P[S=3000]=0.05$. Find the stop-loss premium at retention $d=1000$.", "back": "Stop-loss premium $E[(S-1000)_+]=\\sum_{x>1000}(x-1000)P[S=x]$.\nOnly $x=2000$ and $x=3000$ contribute:\n$(2000-1000)(0.15) + (3000-1000)(0.05) = 1000(0.15) + 2000(0.05) = 150 + 100 = \\$250.$\nEquivalently $E[(S-1000)_+]=E[S]-E[S\\wedge 1000]$; both give $\\$250$.", "tag": "Approximations" }, { "front": "How does the stop-loss premium change between adjacent integer retentions for a discrete $S$, and what is the recursion?", "back": "For a severity/aggregate on a unit grid, the stop-loss premium decreases by the survival function as the retention rises by one unit:\n$E[(S-(d+1))_+]=E[(S-d)_+]-\\big(1-F_S(d)\\big)=E[(S-d)_+]-P[S>d].$\nIntuitively, raising the retention by $1$ saves exactly $P[S>d]$ in expectation (each outcome above $d$ pays $\\$1$ less). This recursion builds a full stop-loss table from $E[(S-0)_+]=E[S]$ downward.", "tag": "Approximations" } ] }