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Exam FAM — Aggregate Loss Models Practice Flashcards

Thirty exam-realistic multiple-choice problems on SOA Exam FAM aggregate loss models — collective-risk mean and variance, the compound-Poisson $Var[S]=\lambda E[X^2]$ shortcut and superposition/thinning of independent compound Poissons, Panjer recursion for $f_S(0),f_S(1),f_S(2)$ under Poisson, negative binomial, binomial, and geometric frequencies, the individual risk model with Bernoulli claim indicators, normal and lognormal approximations for $\Pr(S>c)$ and percentiles, and stop-loss premiums via $E[(S-d)_+]=E[S]-E[S\wedge d]$ — each with a fully worked solution and distractors tied to common candidate errors.

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Browse all 30 problems as a list
  1. Collective risk model
    In the collective risk model $S=X_1+\dots+X_N$, the frequency $N$ has $E[N]=4$ and $Var[N]=6$, and the severities are i.i.d. with $E[X]=300$ and $Var[X]=50{,}000$. Calculate $Var[S]$. (A) $\$200{,}000$ (B) $\$540{,}000$ (C) $\$560{,}000$ (D) $\$740{,}000$ (E) $\$920{,}000$
    **Answer: (D).** The compound variance formula is $Var[S]=E[N]Var[X]+Var[N]E[X]^2$. $Var[S]=4(50{,}000)+6(300)^2 = 200{,}000 + 6(90{,}000) = 200{,}000 + 540{,}000 = \$740{,}000$. Dropping the frequency term $Var[N]E[X]^2$ leaves only $E[N]Var[X]=\$200{,}000$ (distractor A); dropping the severity term leaves $\$540{,}000$ (B). Misapplying the Poisson shortcut $\lambda E[X^2]$ with $E[N]=4$ and $E[X^2]=Var[X]+E[X]^2=140{,}000$ gives $4(140{,}000)=\$560{,}000$ (C) — wrong because $N$ is not Poisson here ($Var[N]\neq E[N]$).
  2. Collective risk model
    Aggregate losses follow the collective risk model. The number of claims $N$ has mean $5$ and variance $5$, and individual claim sizes are i.i.d. with $E[X]=\$2{,}000$ and $E[X^2]=5{,}500{,}000$. Calculate $E[S]$ and $Var[S]$. (A) $E[S]=\$10{,}000$, $Var[S]=7{,}500{,}000$ (B) $E[S]=\$10{,}000$, $Var[S]=27{,}500{,}000$ (C) $E[S]=\$10{,}000$, $Var[S]=20{,}000{,}000$ (D) $E[S]=\$2{,}000$, $Var[S]=27{,}500{,}000$ (E) $E[S]=\$10{,}000$, $Var[S]=55{,}000{,}000$
    **Answer: (B).** $E[S]=E[N]E[X]=5(2{,}000)=\$10{,}000$. Since $E[N]=Var[N]=5$, the frequency is Poisson, so the shortcut $Var[S]=\lambda E[X^2]$ applies: $Var[S]=5(5{,}500{,}000)=27{,}500{,}000$. Check with the general formula: $Var[X]=E[X^2]-E[X]^2=5{,}500{,}000-4{,}000{,}000=1{,}500{,}000$, so $Var[S]=5(1{,}500{,}000)+5(2{,}000)^2=7{,}500{,}000+20{,}000{,}000=27{,}500{,}000$. ✓ (Using $\lambda Var[X]=7{,}500{,}000$ instead of $\lambda E[X^2]$ gives distractor A; using $Var[N]E[X]^2=20{,}000{,}000$ alone gives C.)
  3. Collective risk model
    In the collective risk model, the moment generating function of $S$ is expressed through the frequency and severity. Which expression is correct? (A) $M_S(t)=M_N(t)\,M_X(t)$ (B) $M_S(t)=P_N\bigl(M_X(t)\bigr)$ (C) $M_S(t)=M_X\bigl(P_N(t)\bigr)$ (D) $M_S(t)=P_X\bigl(M_N(t)\bigr)$ (E) $M_S(t)=E[N]\,M_X(t)$
    **Answer: (B).** Conditioning on $N$, $M_S(t)=E[e^{tS}]=E\bigl[E[e^{tS}\mid N]\bigr]=E\bigl[M_X(t)^{N}\bigr]=P_N\bigl(M_X(t)\bigr)$, where $P_N(z)=E[z^{N}]$ is the frequency's probability generating function and $M_X(t)$ is the severity MGF. The composition substitutes the severity MGF into the frequency PGF — not a product (distractor A would be the MGF of a *fixed* sum of independent variables, not a random sum). Equivalently $M_S(t)=M_N\bigl(\ln M_X(t)\bigr)$.
  4. Collective risk model
    Aggregate losses follow the collective risk model with geometric frequency $N$ having parameter $\beta=3$ (so $E[N]=\beta=3$ and $Var[N]=\beta(1+\beta)=12$). Severities are i.i.d. with $E[X]=\$200$ and $E[X^2]=80{,}000$. Calculate $Var[S]$. (A) $\$120{,}000$ (B) $\$240{,}000$ (C) $\$480{,}000$ (D) $\$600{,}000$ (E) $\$720{,}000$
    **Answer: (D).** First $Var[X]=E[X^2]-E[X]^2=80{,}000-40{,}000=40{,}000$. $Var[S]=E[N]Var[X]+Var[N]E[X]^2 = 3(40{,}000)+12(200)^2 = 120{,}000 + 12(40{,}000) = 120{,}000 + 480{,}000 = \$600{,}000$. Using only the severity term $E[N]Var[X]=\$120{,}000$ gives (A); the geometric is **not** Poisson, so a $\lambda E[X^2]$-style shortcut with $\lambda=3$ would wrongly give $3(80{,}000)=\$240{,}000$ (B), ignoring the larger frequency variance $Var[N]=12$.
  5. Moments of S
    The frequency $N$ has $E[N]=2.5$ and $Var[N]=3.0$. Severities are i.i.d. with $E[X]=400$ and $Var[X]=120{,}000$. Calculate $E[S]$ and the standard deviation of $S$. (A) $E[S]=\$1{,}000$, $SD[S]\approx \$836.66$ (B) $E[S]=\$1{,}000$, $SD[S]\approx \$883.18$ (C) $E[S]=\$1{,}000$, $SD[S]\approx \$547.72$ (D) $E[S]=\$1{,}000$, $SD[S]\approx \$692.82$ (E) $E[S]=\$1{,}000$, $SD[S]\approx \$632.46$
    **Answer: (B).** $E[S]=E[N]E[X]=2.5(400)=\$1{,}000$. $Var[S]=E[N]Var[X]+Var[N]E[X]^2 = 2.5(120{,}000)+3.0(400)^2 = 300{,}000 + 3.0(160{,}000) = 300{,}000 + 480{,}000 = 780{,}000$. $SD[S]=\sqrt{780{,}000}\approx \$883.18$. Using only the severity term $\sqrt{300{,}000}\approx\$547.72$ gives (C); using only the frequency term $\sqrt{480{,}000}\approx\$692.82$ gives (D); mistakenly taking $Var[N]=2.5$ (treating $N$ as Poisson) gives $\sqrt{2.5(120{,}000)+2.5(160{,}000)}=\sqrt{700{,}000}\approx\$836.66$ (A).
  6. Moments of S
    The number of claims $N$ is negative binomial with $r=3$ and $\beta=2$, so $E[N]=r\beta=6$ and $Var[N]=r\beta(1+\beta)=18$. Severities are i.i.d. with $E[X]=400$ and $Var[X]=90{,}000$. Calculate $Var[S]$. (A) $\$540{,}000$ (B) $\$2{,}880{,}000$ (C) $\$3{,}420{,}000$ (D) $\$3{,}960{,}000$ (E) $\$8{,}640{,}000$
    **Answer: (C).** $Var[S]=E[N]Var[X]+Var[N]E[X]^2 = 6(90{,}000)+18(400)^2 = 540{,}000 + 18(160{,}000) = 540{,}000 + 2{,}880{,}000 = \$3{,}420{,}000$. The two pieces are the severity term $\$540{,}000$ (distractor A) and the frequency term $\$2{,}880{,}000$ (B); reporting either alone is the common slip. Forgetting that the negative binomial has $Var[N]=r\beta(1+\beta)$ and using $Var[N]=E[N]=6$ would give $6(90{,}000)+6(160{,}000)=\$1{,}500{,}000$, which is not offered.
  7. Moments of S
    The frequency $N$ is binomial with $m=10$ and $q=0.3$, so $E[N]=mq=3$ and $Var[N]=mq(1-q)=2.1$. Severities are i.i.d. with $E[X]=\$1{,}000$ and $E[X^2]=1{,}500{,}000$. Calculate $Var[S]$. (A) $\$1{,}500{,}000$ (B) $\$2{,}100{,}000$ (C) $\$3{,}600{,}000$ (D) $\$4{,}500{,}000$ (E) $\$5{,}100{,}000$
    **Answer: (C).** First $Var[X]=E[X^2]-E[X]^2=1{,}500{,}000-1{,}000{,}000=500{,}000$. $Var[S]=E[N]Var[X]+Var[N]E[X]^2 = 3(500{,}000)+2.1(1{,}000)^2 = 1{,}500{,}000 + 2{,}100{,}000 = \$3{,}600{,}000$. The severity term is $\$1{,}500{,}000$ (A) and the frequency term $\$2{,}100{,}000$ (B). A frequent error treats $N$ as Poisson and uses $\lambda E[X^2]=3(1{,}500{,}000)=\$4{,}500{,}000$ (D) — invalid because the binomial is underdispersed ($Var[N]=2.1<E[N]=3$).
  8. Moments of S
    In a collective risk model, $E[N]=8$, $Var[N]=8$, and each claim is exactly $\$500$ (a degenerate severity). Calculate $Var[S]$. (A) $\$0$ (B) $\$1{,}000{,}000$ (C) $\$2{,}000{,}000$ (D) $\$4{,}000{,}000$ (E) $\$16{,}000{,}000$
    **Answer: (C).** A constant severity has $Var[X]=0$ and $E[X]=500$, so the severity term vanishes and only frequency randomness remains: $Var[S]=E[N]Var[X]+Var[N]E[X]^2 = 8(0)+8(500)^2 = 8(250{,}000)=\$2{,}000{,}000$. The Poisson shortcut agrees: $\lambda E[X^2]=8(500^2)=8(250{,}000)=\$2{,}000{,}000$. Concluding $Var[S]=0$ because each claim is fixed (A) ignores that the *count* is still random; using $\$1{,}000{,}000$ comes from $\lambda E[X]^2/2$-type slips.