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Exam FAM: Aggregate Loss Models Flashcards

Aggregate loss modeling for SOA Exam FAM: the collective risk model $S=X_1+\dots+X_N$ with frequency independent of i.i.d. severities, the compound mean and variance formulas, the MGF/PGF composition $M_S(t)=P_N(M_X(t))$, the compound Poisson and sums of independent compound Poissons, the Panjer recursion for $(a,b,0)$ and $(a,b,1)$ frequencies with discrete severity, the individual risk model over a fixed portfolio with Bernoulli claim indicators, normal and lognormal approximations for probabilities and percentiles, and stop-loss premiums $E[(S-d)_+]$ with the $E[S]-E[S\wedge d]$ relation and integer-retention recursion — all with fully worked numeric examples.

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  1. Collective risk model
    State the **collective risk model** for aggregate losses $S$ and its independence assumptions.
    $S = X_1 + X_2 + \dots + X_N$, 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. Key assumptions: 1. The $X_i$ are mutually independent and identically distributed. 2. $N$ is independent of the $X_i$. By convention $S=0$ when $N=0$. The model treats the whole portfolio's losses in aggregate rather than policy-by-policy.
  2. Moments of S
    Give the formula for $E[S]$ in the collective risk model and explain it.
    $E[S] = E[N]\,E[X]$. This 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 $E[X]$.
  3. Moments of S
    Give the compound-variance formula $Var[S]$ in the collective risk model and identify the two pieces.
    $Var[S] = E[N]\,Var[X] + Var[N]\,E[X]^2$. The 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 $Var[S]=E[Var(S\mid N)]+Var(E[S\mid N])$.
  4. Moments of S
    State the relation between the MGF of $S$, the PGF of $N$, and the MGF of $X$ in the collective risk model.
    $M_S(t) = P_N\bigl(M_X(t)\bigr)$, 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. Derivation: $M_S(t)=E[e^{tS}]=E\bigl[E[e^{tS}\mid N]\bigr]=E\bigl[M_X(t)^{N}\bigr]=P_N(M_X(t))$. Equivalently $M_S(t)=M_N\bigl(\ln M_X(t)\bigr)$.
  5. Moments of S
    Compute $E[S]$ and $Var[S]$ when $N$ has mean $3$ and variance $5$, and severities have $E[X]=200$, $Var[X]=30{,}000$.
    $E[S]=E[N]E[X]=3(200)=\$600$. $Var[S]=E[N]Var[X]+Var[N]E[X]^2 = 3(30{,}000)+5(200)^2$ $= 90{,}000 + 5(40{,}000) = 90{,}000 + 200{,}000 = 290{,}000$. Standard deviation $=\sqrt{290{,}000}\approx \$538.52$.
  6. Moments of S
    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.
    First $Var[X]=E[X^2]-E[X]^2 = 400{,}000-250{,}000=150{,}000$. $E[S]=E[N]E[X]=2.5(500)=\$1{,}250$. General formula: $Var[S]=E[N]Var[X]+Var[N]E[X]^2 = 2.5(150{,}000)+2.5(250{,}000)$ $=375{,}000+625{,}000=1{,}000{,}000$. Poisson shortcut: $Var[S]=\lambda E[X^2]=2.5(400{,}000)=1{,}000{,}000$. Both agree.
  7. Compound Poisson
    Define the **compound Poisson** distribution and state when it arises.
    $S$ 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 $S\sim\text{CP}(\lambda, F_X)$. It 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.
  8. Compound Poisson
    Give the mean, variance, and third central moment of a compound Poisson $S$ with rate $\lambda$ and severity $X$.
    $E[S]=\lambda\,E[X]$. $Var[S]=\lambda\,E[X^2]$. Third central moment $E[(S-E[S])^3]=\lambda\,E[X^3]$. In 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.
  9. Compound Poisson
    Derive the compound-Poisson variance $Var[S]=\lambda E[X^2]$ from the general compound formula.
    For Poisson frequency $E[N]=Var[N]=\lambda$. Substituting into $Var[S]=E[N]Var[X]+Var[N]E[X]^2$: $Var[S]=\lambda\,Var[X]+\lambda\,E[X]^2 = \lambda\bigl(Var[X]+E[X]^2\bigr)=\lambda\,E[X^2]$, since $Var[X]+E[X]^2 = E[X^2]$ by definition. The frequency and severity variabilities merge into one term driven by the raw second moment.
  10. Compound Poisson
    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 $Var[S]$.
    Severity moments: $E[X]=\frac{1+2+3+4+5}{5}=3$. $E[X^2]=\frac{1+4+9+16+25}{5}=\frac{55}{5}=11$. $E[S]=\lambda E[X]=4(3)=12$. $Var[S]=\lambda E[X^2]=4(11)=44$. Standard deviation $=\sqrt{44}\approx 6.633$.
  11. Compound Poisson
    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 $S$.
    $E[S]=\lambda E[X]=10(1{,}000)=\$10{,}000$. $Var[S]=\lambda E[X^2]=10(2{,}000{,}000)=20{,}000{,}000$. Standard deviation $=\sqrt{20{,}000{,}000}\approx 4{,}472.14$. Coefficient of variation $=\dfrac{\sqrt{Var[S]}}{E[S]}=\dfrac{4{,}472.14}{10{,}000}\approx 0.4472$.
  12. Compound Poisson
    State the **sum rule** for independent compound Poisson random variables.
    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 $\lambda = \lambda_1+\dots+\lambda_m$ and severity distribution $F(x)=\dfrac{1}{\lambda}\sum_{j=1}^{m}\lambda_j\,F_j(x)$, a **$\lambda$-weighted mixture** of the individual severities. Probabilities of each severity value are mixed in proportion to each portfolio's rate.
  13. Compound Poisson
    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 $Var[S]$.
    Combined rate $\lambda=3+2=5$. The combined severity is the $\lambda$-weighted mixture: $P(X=100)=\frac{3}{5}=0.6$, $P(X=250)=\frac{2}{5}=0.4$. Severity moments: $E[X]=0.6(100)+0.4(250)=60+100=160$; $E[X^2]=0.6(100^2)+0.4(250^2)=0.6(10{,}000)+0.4(62{,}500)=6{,}000+25{,}000=31{,}000$. $E[S]=\lambda E[X]=5(160)=\$800$. $Var[S]=\lambda E[X^2]=5(31{,}000)=155{,}000$.
  14. Compound Poisson
    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?
    **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. Equivalently, 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.
  15. Compound Poisson
    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.
    By thinning, large claims are compound Poisson with rate $\lambda_L=\lambda p_L=8(0.25)=2$ and constant severity $500$. $E[S_L]=\lambda_L E[X_L]=2(500)=\$1{,}000$. $E[X_L^2]=500^2=250{,}000$, so $Var[S_L]=\lambda_L E[X_L^2]=2(250{,}000)=500{,}000$. Standard deviation $=\sqrt{500{,}000}\approx \$707.11$.
  16. Panjer recursion
    What is the **$(a,b,0)$ class** of frequency distributions, and which distributions belong to it?
    A counting distribution is in the $(a,b,0)$ class if its probabilities satisfy the recursion $\dfrac{p_k}{p_{k-1}} = a + \dfrac{b}{k}, \quad k=1,2,3,\dots$ for constants $a,b$, starting from $p_0$. Exactly three distributions qualify: - **Poisson** ($a=0$, $b=\lambda$), - **Binomial** ($a=-\frac{q}{1-q}$, $b=(m+1)\frac{q}{1-q}$), - **Negative binomial** ($a=\frac{\beta}{1+\beta}$, $b=(r-1)\frac{\beta}{1+\beta}$); geometric is the $r=1$ case.
  17. Panjer recursion
    Give the $(a,b)$ parameters for the Poisson, binomial, and negative binomial frequencies used in Panjer's recursion.
    **Poisson**$(\lambda)$: $a=0$, $b=\lambda$. **Binomial**$(m,q)$: $a=-\dfrac{q}{1-q}$, $b=(m+1)\dfrac{q}{1-q}$. **Negative binomial**$(r,\beta)$: $a=\dfrac{\beta}{1+\beta}$, $b=(r-1)\dfrac{\beta}{1+\beta}$. **Geometric**$(\beta)$ is negative binomial with $r=1$: $a=\dfrac{\beta}{1+\beta}$, $b=0$.
  18. Panjer recursion
    State the **Panjer recursion** for the aggregate distribution $f_S$ when frequency is $(a,b,0)$ and severity is discrete on $0,1,2,\dots$
    For an $(a,b,0)$ frequency and severity probabilities $f_X(j)$ on the non-negative integers, $f_S(s) = \dfrac{1}{1-a\,f_X(0)}\sum_{j=1}^{s}\Bigl(a+\dfrac{b\,j}{s}\Bigr) f_X(j)\, f_S(s-j), \quad s=1,2,3,\dots$ The 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 $1$.
  19. Panjer recursion
    Give the **starting value** $f_S(0)$ for the Panjer recursion, and the Poisson special case.
    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 $0$. - **Poisson**$(\lambda)$: $f_S(0)=e^{-\lambda(1-f_X(0))}$; if $f_X(0)=0$ this is $e^{-\lambda}=P(N=0)$. - **Binomial**$(m,q)$: $f_S(0)=\bigl(1-q+q f_X(0)\bigr)^{m}$. - **Negative binomial**$(r,\beta)$: $f_S(0)=\bigl(1+\beta(1-f_X(0))\bigr)^{-r}$. When severity is positive (no mass at $0$), $f_S(0)=P(N=0)$.
  20. Panjer recursion
    Set up Panjer for a **Poisson**$(\lambda)$ frequency: state the simplified recursion.
    For Poisson, $a=0$ and $b=\lambda$, so $a+\frac{bj}{s}=\frac{\lambda j}{s}$ and the recursion becomes $f_S(s)=\dfrac{\lambda}{s}\sum_{j=1}^{s} j\,f_X(j)\,f_S(s-j), \quad s\ge 1$, starting 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.
  21. Panjer recursion
    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 $0$).
    $f_S(0)=e^{-\lambda}=e^{-2}\approx 0.135335$. Poisson recursion $f_S(s)=\frac{\lambda}{s}\sum_{j=1}^{s} j f_X(j) f_S(s-j)$. $f_S(1)=\frac{2}{1}\bigl[1\cdot f_X(1)\cdot f_S(0)\bigr]=2(0.6)(0.135335)\approx 0.162402$. $f_S(2)=\frac{2}{2}\bigl[1\cdot f_X(1) f_S(1) + 2\cdot f_X(2) f_S(0)\bigr]$ $=1\cdot\bigl[0.6(0.162402)+2(0.4)(0.135335)\bigr]=0.097441+0.108268\approx 0.205709$.
  22. Panjer recursion
    Continue the previous Panjer example ($\lambda=2$, $f_X(1)=0.6$, $f_X(2)=0.4$): find $f_S(3)$.
    Recall $f_S(0)\approx0.135335$, $f_S(1)\approx0.162402$, $f_S(2)\approx0.205709$. $f_S(3)=\frac{2}{3}\bigl[1\cdot f_X(1) f_S(2) + 2\cdot f_X(2) f_S(1)\bigr]$ (only $j=1,2$ since $f_X(j)=0$ for $j\ge3$). $=\frac{2}{3}\bigl[0.6(0.205709)+2(0.4)(0.162402)\bigr]$ $=\frac{2}{3}\bigl[0.123425+0.129922\bigr]=\frac{2}{3}(0.253347)\approx 0.168898$.
  23. Panjer recursion
    Why does Panjer's recursion require the **severity to be on a discrete (arithmetic) grid**, and how is a continuous severity handled?
    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.
  24. Panjer recursion
    What is the $(a,b,1)$ class, and how does the Panjer recursion change for it?
    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: $f_S(s)=\dfrac{\bigl[p_1-(a+b)p_0\bigr]f_X(s)+\sum_{j=1}^{s}\bigl(a+\frac{bj}{s}\bigr)f_X(j)f_S(s-j)}{1-a f_X(0)}$. Use it for zero-truncated or zero-modified frequencies. The $(a,b,0)$ recursion is the special case $p_1=(a+b)p_0$.
  25. Individual risk model
    State the **individual risk model** for aggregate losses over a fixed portfolio.
    $S = X_1 + X_2 + \dots + X_n$, 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.
  26. Individual risk model
    Contrast the **individual** and **collective** risk models.
    **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). **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. The collective model aggregates by claim; the individual model aggregates by policy.
  27. Individual risk model
    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 $Var[X_i]$.
    $E[X_i] = q_i\,E[B_i]$. $Var[X_i] = q_i\,E[B_i^2] - \bigl(q_i E[B_i]\bigr)^2 = q_i\,Var[B_i] + q_i(1-q_i)\,E[B_i]^2$. The 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 $Var[X_i]=q_i(1-q_i)b_i^2$.
  28. Individual risk model
    In the individual risk model, how are $E[S]$ and $Var[S]$ obtained from the per-policy moments?
    By independence of the policies, $E[S]=\sum_{i=1}^{n}E[X_i]$ and $Var[S]=\sum_{i=1}^{n}Var[X_i]$. No 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.
  29. Individual risk model
    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 $Var[S]$.
    Per-policy (constant claim amount): $E[X_i]=q_i b_i$, $Var[X_i]=q_i(1-q_i)b_i^2$. $E[X_1]=0.10(1000)=100$; $E[X_2]=0.05(2000)=100$; $E[X_3]=0.20(500)=100$. $E[S]=300$. $Var[X_1]=0.10(0.90)(1000^2)=90{,}000$. $Var[X_2]=0.05(0.95)(2000^2)=0.0475(4{,}000{,}000)=190{,}000$. $Var[X_3]=0.20(0.80)(500^2)=0.16(250{,}000)=40{,}000$. $Var[S]=90{,}000+190{,}000+40{,}000=320{,}000$.
  30. Individual risk model
    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.
    Each life: $E[X_i]=q b=0.03(10{,}000)=300$; $Var[X_i]=q(1-q)b^2=0.03(0.97)(10{,}000^2)=0.0291(10^8)=2{,}910{,}000$. With $n=100$ independent lives: $E[S]=100(300)=\$30{,}000$. $Var[S]=100(2{,}910{,}000)=291{,}000{,}000$. Standard deviation $=\sqrt{291{,}000{,}000}\approx \$17{,}058.72$.
  31. Individual risk model
    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.
    $X=IB$ with $P(I=1)=0.08$. $E[X]=qE[B]=0.08(4{,}000)=\$320$. Use $Var[X]=qVar[B]+q(1-q)E[B]^2$: $=0.08(6{,}000{,}000)+0.08(0.92)(4{,}000^2)$ $=480{,}000+0.0736(16{,}000{,}000)=480{,}000+1{,}177{,}600=1{,}657{,}600$. Standard deviation $=\sqrt{1{,}657{,}600}\approx \$1{,}287.48$.
  32. Approximations & stop-loss
    How is the **normal approximation** used to estimate $P(S>x)$ for aggregate losses?
    Approximate $S$ by a normal with the same mean and variance: $S\approx N\bigl(E[S],\,Var[S]\bigr)$. Then $P(S>x)\approx P\!\left(Z>\dfrac{x-E[S]}{\sqrt{Var[S]}}\right)=1-\Phi\!\left(\dfrac{x-E[S]}{\sqrt{Var[S]}}\right)$. It 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.
  33. Approximations & stop-loss
    Aggregate losses $S$ have $E[S]=\$10{,}000$ and $Var[S]=4{,}000{,}000$. Using the normal approximation, find $P(S>13{,}000)$. (Use $\Phi(1.5)=0.9332$.)
    Standard deviation $=\sqrt{4{,}000{,}000}=2{,}000$. $z=\dfrac{13{,}000-10{,}000}{2{,}000}=\dfrac{3{,}000}{2{,}000}=1.5$. $P(S>13{,}000)\approx 1-\Phi(1.5)=1-0.9332=0.0668$. So there is about a $6.7\%$ chance aggregate losses exceed $\$13{,}000$.
  34. Approximations & stop-loss
    Find the **$95$th percentile** of aggregate losses with $E[S]=\$50{,}000$ and standard deviation $\$8{,}000$ using the normal approximation. (Use $z_{0.95}=1.645$.)
    The normal $95$th percentile is $E[S]+z_{0.95}\,\sigma_S$. $=50{,}000 + 1.645(8{,}000)=50{,}000+13{,}160=\$63{,}160$. This is the capital level such that aggregate losses stay below it with $95\%$ probability under the normal approximation.
  35. Approximations & stop-loss
    How is the **lognormal approximation** to $S$ set up, and why is it sometimes preferred over the normal?
    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 $\mu=\ln E[S]-\tfrac{\sigma^2}{2}$. Then $P(S\le x)\approx\Phi\!\left(\dfrac{\ln x-\mu}{\sigma}\right)$. It is preferred when $S$ is **right-skewed and positive** (small portfolios, heavy-tailed severity), where the symmetric normal fits poorly.
  36. Approximations & stop-loss
    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 $\Phi(1.34)=0.9099$.)
    $\sigma^2=\ln\!\bigl(1+\frac{Var[S]}{E[S]^2}\bigr)=\ln\!\bigl(1+\frac{640{,}000}{1{,}000{,}000}\bigr)=\ln(1.64)\approx 0.494696$, so $\sigma\approx 0.703346$. $\mu=\ln E[S]-\frac{\sigma^2}{2}=\ln 1000 - 0.247348 = 6.907755-0.247348=6.660407$. $z=\dfrac{\ln 2000-\mu}{\sigma}=\dfrac{7.600902-6.660407}{0.703347}=\dfrac{0.940495}{0.703347}\approx 1.337$. $P(S>2000)\approx 1-\Phi(1.34)\approx 1-0.9099=0.0901\approx 0.090$.
  37. Approximations & stop-loss
    Define the **stop-loss premium** with retention (deductible) $d$ and give two equivalent expressions.
    The stop-loss (net) premium is the expected loss above the retention $d$: $E[(S-d)_+]=\displaystyle\int_{d}^{\infty}(s-d)\,f_S(s)\,ds = \int_{d}^{\infty}\bigl[1-F_S(s)\bigr]\,ds$ (continuous case), or for discrete $S$, $E[(S-d)_+]=\sum_{s>d}(s-d)f_S(s)$. Here $(S-d)_+=\max(S-d,0)$. It is the pure premium for stop-loss reinsurance attaching at $d$.
  38. Approximations & stop-loss
    State the key identity linking the stop-loss premium $E[(S-d)_+]$ to the **limited expected value** $E[S\wedge d]$.
    $E[(S-d)_+] = E[S] - E[S\wedge d]$, where $S\wedge d=\min(S,d)$ 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.
  39. Approximations & stop-loss
    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 $d=1$.
    $E[(S-1)_+]=\sum_{s>1}(s-1)f_S(s)=(2-1)(0.15)+(3-1)(0.05)=0.15+0.10=0.25$. Check 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 $=0.75-0.50=0.25$. ✓
  40. Approximations & stop-loss
    State the **integer-retention recursion** for stop-loss premiums when $S$ is on the non-negative integers.
    Stepping the retention up by one unit, $E[(S-(d+1))_+] = E[(S-d)_+] - \bigl[1-F_S(d)\bigr]$, where $1-F_S(d)=P(S>d)$ 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.
  41. Approximations & stop-loss
    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 $d=0,1,2,3$.
    Survival probabilities: $S_S(0)=P(S>0)=0.5$, $S_S(1)=0.2$, $S_S(2)=0.05$, $S_S(3)=0$. Recursion $E[(S-(d+1))_+]=E[(S-d)_+]-P(S>d)$, starting at $E[(S-0)_+]=E[S]=0.75$. $d=1$: $0.75-0.5=0.25$. $d=2$: $0.25-0.2=0.05$. $d=3$: $0.05-0.05=0.00$. So stop-loss premiums are $0.75,\,0.25,\,0.05,\,0.00$ at $d=0,1,2,3$.
  42. Approximations & stop-loss
    A compound Poisson $S$ has $\lambda=3$ and severity always $\$100$ per claim. Find the stop-loss premium at retention $d=\$100$. (Use $e^{-3}\approx 0.049787$.)
    $S$ takes values $0,100,200,\dots$ with $P(S=100k)=P(N=k)=e^{-3}\frac{3^{k}}{k!}$. $E[S]=\lambda(100)=300$. $E[(S-100)_+]=E[S]-E[S\wedge100]$. Here $E[S\wedge100]=100\cdot P(S\ge100)=100\,P(N\ge1)=100(1-e^{-3})$. $1-e^{-3}\approx 1-0.049787=0.950213$, so $E[S\wedge100]\approx 95.0213$. $E[(S-100)_+]\approx 300-95.0213=\$204.98$.
  43. Approximations & stop-loss
    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 $\phi(1)=0.241971$, $\Phi(1)=0.8413$.)
    For $S\sim N(\mu,\sigma^2)$ with $z=\frac{d-\mu}{\sigma}$: $E[(S-d)_+]=\sigma\,\phi(z)-(d-\mu)\,\bigl[1-\Phi(z)\bigr]$. Here $\mu=10{,}000$, $\sigma=2{,}000$, $d=12{,}000$, so $z=\frac{2{,}000}{2{,}000}=1$. $=2{,}000(0.241971)-2{,}000(1-0.8413)$ $=483.942-2{,}000(0.1587)=483.942-317.40=\$166.54$.
  44. Approximations & stop-loss
    Why can two different aggregate models share the same $E[S]$ and $Var[S]$ yet give different **stop-loss premiums and tail probabilities**?
    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.