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Exam MAS-I: Frequency, Severity & Aggregate Models Flashcards

The loss-modeling core of CAS Exam MAS-I: the $(a,b,0)$ and $(a,b,1)$ frequency classes and the variance-to-mean test, continuous severity distributions with their means, variances and limited expected values, coverage modifications (deductibles, limits, coinsurance) on a per-loss and per-payment basis, the collective-risk aggregate model with its mean and variance, the compound-Poisson identity and the Panjer recursion, and the normal and lognormal approximations with stop-loss premiums — built from concept cards and fully worked numeric examples.

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  1. (a,b,0) & (a,b,1)
    State the defining recursion of the $(a,b,0)$ frequency class and name its four members.
    A counting distribution is in the $(a,b,0)$ class if its probabilities satisfy $\frac{p_k}{p_{k-1}} = a + \frac{b}{k}, \quad k=1,2,3,\dots$ for constants $a$ and $b$, with $p_0$ set so the probabilities sum to $1$. The only members are: - **Poisson** ($a=0$, $b=\lambda$), - **Binomial** ($a<0$), - **Negative binomial** ($0<a<1$), - **Geometric** ($0<a<1$, $b=0$ — the special case $r=1$ of the negative binomial).
  2. (a,b,0) & (a,b,1)
    Give the $(a,b,0)$ parameters $a$ and $b$ for the **Poisson**, **binomial**, and **negative binomial** distributions.
    **Poisson($\lambda$):** $a=0,\; b=\lambda$. **Binomial($m,q$):** $a=-\frac{q}{1-q},\; b=(m+1)\frac{q}{1-q}$. **Negative binomial($r,\beta$):** $a=\frac{\beta}{1+\beta},\; b=(r-1)\frac{\beta}{1+\beta}$. The **geometric($\beta$)** is the $r=1$ negative binomial, giving $a=\frac{\beta}{1+\beta},\; b=0$.
  3. (a,b,0) & (a,b,1)
    How does the **variance-to-mean ratio** identify which $(a,b,0)$ distribution a data set follows?
    Compare $\frac{\text{Var}(N)}{E[N]}$: - **$=1$** → **Poisson** (mean equals variance). - **$<1$** → **binomial** (underdispersed; variance below mean). - **$>1$** → **negative binomial / geometric** (overdispersed; variance above mean). This is the fastest screening tool: compute the sample mean and variance of the counts and read off the family.
  4. (a,b,0) & (a,b,1)
    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.
    Variance-to-mean $=\frac{0.58}{0.40}=1.45>1$, so use the **negative binomial**. Moment match: $E[N]=r\beta=0.40$ and $\text{Var}(N)=r\beta(1+\beta)=0.58$. Divide: $\frac{0.58}{0.40}=1+\beta \Rightarrow \beta = 0.45$. Then $r = \frac{0.40}{\beta} = \frac{0.40}{0.45} \approx 0.889$. So $\hat\beta = 0.45$, $\hat r \approx 0.889$.
  5. (a,b,0) & (a,b,1)
    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.
    The recursion at $k=1,2$ gives $a+b = 0.6$ and $a + \frac{b}{2} = 0.45$. Subtract: $\frac{b}{2}=0.15 \Rightarrow b = 0.30$, then $a = 0.6-0.30 = 0.30$. Since $0<a<1$, this is a **negative binomial**. Recover parameters from $a=\frac{\beta}{1+\beta}=0.30 \Rightarrow \beta = \frac{0.30}{0.70}\approx 0.4286$, and $b=(r-1)a \Rightarrow r-1 = \frac{0.30}{0.30}=1$, so $r = 2$.
  6. (a,b,0) & (a,b,1)
    Define the $(a,b,1)$ class and distinguish **zero-truncated** from **zero-modified** distributions.
    The $(a,b,1)$ class uses the same recursion $\frac{p_k}{p_{k-1}}=a+\frac{b}{k}$ but only for $k\ge 2$, leaving $p_0$ (and hence $p_1$) free. - **Zero-truncated (ZT):** $p_0^M = 0$ — the count $0$ is impossible (e.g. number of claims given at least one claim). - **Zero-modified (ZM):** $p_0^M$ is an arbitrary chosen mass in $[0,1)$, inflating or deflating the probability of zero relative to the base distribution. The ZT case is the special ZM case with $p_0^M=0$.
  7. (a,b,0) & (a,b,1)
    Give the rescaling formulas relating zero-modified probabilities $p_k^M$ to the base $(a,b,0)$ probabilities $p_k$.
    Let $p_k$ be the base $(a,b,0)$ probabilities. For $k\ge 1$ the zero-modified probabilities are $p_k^M = \frac{1-p_0^M}{1-p_0}\,p_k.$ The zero-truncated case sets $p_0^M=0$, giving $p_k^T = \frac{p_k}{1-p_0}$ for $k\ge 1$. The scaling factor $\frac{1-p_0^M}{1-p_0}$ redistributes the removed/added mass at $0$ across the positive counts so everything still sums to $1$.
  8. (a,b,0) & (a,b,1)
    A base Poisson has $\lambda = 1.5$. Build the **zero-truncated** Poisson and find $P[N=1]$ and $P[N=2]$.
    Base: $p_0 = e^{-1.5}\approx 0.223130$, $p_1 = 1.5e^{-1.5}\approx 0.334695$, $p_2 = \frac{1.5^2}{2}e^{-1.5}\approx 0.251021$. Truncation factor $\frac{1}{1-p_0}=\frac{1}{1-0.223130}=\frac{1}{0.776870}\approx 1.287218$. $p_1^T = 0.334695(1.287218)\approx 0.430823$. $p_2^T = 0.251021(1.287218)\approx 0.323117$. These now sum (with the higher terms) to $1$ over $k\ge 1$.
  9. (a,b,0) & (a,b,1)
    A base Poisson has $\lambda = 2$. Construct a **zero-modified** version with $p_0^M = 0.5$ and find $P[N=1]$.
    Base: $p_0 = e^{-2}\approx 0.135335$, $p_1 = 2e^{-2}\approx 0.270671$. Scaling factor $=\frac{1-p_0^M}{1-p_0}=\frac{1-0.5}{1-0.135335}=\frac{0.5}{0.864665}\approx 0.578259$. $p_1^M = 0.578259 \times 0.270671 \approx 0.156527$. The extra mass placed at $0$ (raising $p_0$ from $0.135$ to $0.5$) is removed proportionally from all positive counts.
  10. Severity & moments
    State the mean and variance of the **exponential**, **gamma**, and **Pareto** severity distributions (using the Loss Models parameterizations).
    **Exponential($\theta$):** $f(x)=\frac{1}{\theta}e^{-x/\theta}$, $E[X]=\theta$, $\text{Var}(X)=\theta^2$. **Gamma($\alpha,\theta$):** $E[X]=\alpha\theta$, $\text{Var}(X)=\alpha\theta^2$. **Pareto($\alpha,\theta$):** $E[X]=\frac{\theta}{\alpha-1}$ (for $\alpha>1$), $E[X^2]=\frac{2\theta^2}{(\alpha-1)(\alpha-2)}$ (for $\alpha>2$), so $\text{Var}(X)=\frac{\alpha\theta^2}{(\alpha-1)^2(\alpha-2)}$.
  11. Severity & moments
    State the mean, variance, and key moment of the **lognormal** and **Weibull** severity distributions.
    **Lognormal($\mu,\sigma$):** $\ln X \sim N(\mu,\sigma^2)$. The $k$-th raw moment is $E[X^k]=\exp\!\big(k\mu+\tfrac{1}{2}k^2\sigma^2\big)$. So $E[X]=e^{\mu+\sigma^2/2}$ and $E[X^2]=e^{2\mu+2\sigma^2}$, giving $\text{Var}(X)=e^{2\mu+\sigma^2}\big(e^{\sigma^2}-1\big)$. **Weibull($\tau,\theta$):** $S(x)=\exp\!\big(-(x/\theta)^\tau\big)$, with $E[X]=\theta\,\Gamma\!\big(1+\tfrac{1}{\tau}\big)$ and $E[X^2]=\theta^2\,\Gamma\!\big(1+\tfrac{2}{\tau}\big)$.
  12. Severity & moments
    Define the **limited expected value** $E[X\wedge u]$ and give its integral form in terms of the survival function.
    $X\wedge u = \min(X,u)$ caps the loss at $u$. Its expectation is $E[X\wedge u]=\int_0^u x\,f(x)\,dx + u\,S(u) = \int_0^u S(x)\,dx.$ The last form (integrating the survival function from $0$ to $u$) is usually the easiest. As $u\to\infty$, $E[X\wedge u]\to E[X]$.
  13. Severity & moments
    Give the closed-form limited expected value $E[X\wedge u]$ for the **exponential** and the **Pareto** distributions.
    **Exponential($\theta$):** $E[X\wedge u]=\theta\big(1-e^{-u/\theta}\big)$. **Pareto($\alpha,\theta$):** for $\alpha\neq 1$, $E[X\wedge u]=\frac{\theta}{\alpha-1}\left[1-\left(\frac{\theta}{\theta+u}\right)^{\alpha-1}\right].$ Both approach the full mean ($\theta$ and $\frac{\theta}{\alpha-1}$) as $u\to\infty$.
  14. Severity & moments
    Losses follow an **exponential** with $\theta = 1000$. Compute $E[X\wedge 500]$ and the variance of $X$.
    Variance: $\text{Var}(X)=\theta^2 = 1000^2 = 1{,}000{,}000$. Limited expected value: $E[X\wedge 500]=\theta\big(1-e^{-500/1000}\big)=1000\big(1-e^{-0.5}\big)$. Since $e^{-0.5}\approx 0.606531$, $E[X\wedge 500]=1000(0.393469)\approx \$393.47$.
  15. Severity & moments
    Losses follow a **Pareto** with $\alpha = 3$, $\theta = 2000$. Find $E[X]$, $\text{Var}(X)$, and $E[X\wedge 1000]$.
    $E[X]=\frac{\theta}{\alpha-1}=\frac{2000}{2}=1000$. $E[X^2]=\frac{2\theta^2}{(\alpha-1)(\alpha-2)}=\frac{2(2000)^2}{(2)(1)}=4{,}000{,}000$, so $\text{Var}(X)=4{,}000{,}000 - 1000^2 = 3{,}000{,}000$. $E[X\wedge 1000]=\frac{\theta}{\alpha-1}\Big[1-\big(\tfrac{\theta}{\theta+u}\big)^{\alpha-1}\Big]=1000\Big[1-\big(\tfrac{2000}{3000}\big)^{2}\Big]=1000\big(1-0.4444\big)\approx \$555.56$.
  16. Severity & moments
    A severity is **lognormal** with $\mu = 7$, $\sigma = 1.2$. Find $E[X]$ and $\text{Var}(X)$.
    $E[X]=e^{\mu+\sigma^2/2}=e^{7+0.72}=e^{7.72}$. Since $e^{7}\approx 1096.633$ and $e^{0.72}\approx 2.054433$, $E[X]\approx 1096.633(2.054433)\approx \$2{,}253.00$. $\text{Var}(X)=E[X]^2\big(e^{\sigma^2}-1\big)$ since $\text{Var}(X)=e^{2\mu+\sigma^2}\big(e^{\sigma^2}-1\big)$. With $e^{\sigma^2}=e^{1.44}\approx 4.220696$, $\text{Var}(X)\approx 2253.00^2(4.220696-1)\approx 5{,}076{,}009(3.220696)\approx 1.635\times 10^{7}$.
  17. Coverage modifications
    Define the ordinary **deductible** $d$ and write the per-loss expected payment $E[(X-d)_+]$ in terms of limited expected values.
    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 $E[(X-d)_+]=E[X]-E[X\wedge d].$ This is the total expected loss minus the part capped at the deductible — equivalently $\int_d^\infty S(x)\,dx$.
  18. Coverage modifications
    Distinguish the **per-loss** and **per-payment** expected cost of a deductible $d$, and give the per-payment formula.
    **Per-loss** averages over every loss, counting non-payments (loss $\le d$) as $0$: $E[(X-d)_+]=E[X]-E[X\wedge d]$. **Per-payment** conditions on a payment actually occurring (loss $>d$), dividing by the probability of a payment $S(d)$: $E[(X-d)_+\mid X>d]=\frac{E[X]-E[X\wedge d]}{S(d)}.$ The per-payment cost is always $\ge$ the per-loss cost, since $S(d)\le 1$.
  19. Coverage modifications
    Write the expected payment **per loss** under a policy with deductible $d$, maximum covered loss $u$, and coinsurance $\alpha$.
    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 $\alpha\big(E[X\wedge u]-E[X\wedge d]\big).$ Here $u$ 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.
  20. Coverage modifications
    How does an ordinary deductible $d$ change the **number of payments** relative to the number of losses?
    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 $N$. If losses are **Poisson($\lambda$)**, the payment count is **Poisson($\lambda v$)**. If losses are **negative binomial($r,\beta$)**, payments are **negative binomial($r,\beta v$)**. If losses are **binomial($m,q$)**, payments are **binomial($m,qv$)**. The frequency family is preserved; only the relevant parameter is scaled by $v$.
  21. Coverage modifications
    Losses are **exponential** with $\theta = 800$. A policy has an ordinary deductible of $200$. Find the expected payment **per loss** and **per payment**.
    $E[X]=\theta=800$. $E[X\wedge 200]=800\big(1-e^{-200/800}\big)=800\big(1-e^{-0.25}\big)$. With $e^{-0.25}\approx 0.778801$, $E[X\wedge 200]=800(0.221199)\approx 176.96$. **Per loss:** $E[(X-200)_+]=800-176.96=\$623.04$. **Per payment:** divide by $S(200)=e^{-0.25}\approx 0.778801$: $\frac{623.04}{0.778801}\approx \$800.00$. (For the exponential, the memoryless property makes the per-payment cost equal the full mean $\theta$.)
  22. Coverage modifications
    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.
    Use $E[X\wedge u]=\frac{\theta}{\alpha-1}\big[1-(\tfrac{\theta}{\theta+u})^{2}\big]$ with $\frac{\theta}{\alpha-1}=1000$. $E[X\wedge 5000]=1000\big[1-(\tfrac{2000}{7000})^2\big]=1000(1-0.081633)\approx 918.37$. $E[X\wedge 500]=1000\big[1-(\tfrac{2000}{2500})^2\big]=1000(1-0.64)=360.00$. Layer $=918.37-360.00=558.37$; apply coinsurance: $0.80(558.37)\approx \$446.70$.
  23. Coverage modifications
    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?
    Probability a loss pierces the deductible: $v=S(500)=e^{-500/1000}=e^{-0.5}\approx 0.606531$. Thinning a Poisson keeps it Poisson with rate $\lambda v$: number of payments $\sim$ **Poisson($6 \times 0.606531$)** $=$ **Poisson($3.639$)**. So the expected number of payments per year is $\approx 3.64$.
  24. Aggregate moments
    State the collective-risk (compound) model for aggregate losses $S$ and the conditions on $N$ and the $X_i$.
    $S = X_1 + X_2 + \cdots + X_N = \sum_{i=1}^{N} X_i$, where: - $N$ is the **claim count** (frequency), - the $X_i$ are i.i.d. **severities**, independent of $N$, - when $N=0$, $S=0$. Frequency and severity are modeled separately and combined. $S$ is called a **compound** distribution (e.g. compound Poisson if $N$ is Poisson).
  25. Aggregate moments
    State the mean and variance of the aggregate loss $S=\sum_{i=1}^N X_i$ in the collective-risk model.
    $E[S]=E[N]\,E[X]$ (product of the frequency mean and severity mean). $\text{Var}(S)=E[N]\,\text{Var}(X)+\text{Var}(N)\,E[X]^2.$ The 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).
  26. Aggregate moments
    Derive $\text{Var}(S)$ for a compound distribution using conditioning on $N$ (law of total variance).
    Condition on $N$. With i.i.d. severities independent of $N$: $E[S\mid N]=N\,E[X]$ and $\text{Var}(S\mid N)=N\,\text{Var}(X)$. Law of total variance: $\text{Var}(S)=E\big[\text{Var}(S\mid N)\big]+\text{Var}\big(E[S\mid N]\big)$ $=E[N]\,\text{Var}(X)+\text{Var}(N)\,E[X]^2.$ The two pieces are the expected within-$N$ variance plus the variance of the conditional mean.
  27. Aggregate moments
    Frequency is **Poisson($\lambda = 4$)** and severity has $E[X]=500$, $\text{Var}(X)=250{,}000$. Find $E[S]$ and $\text{Var}(S)$.
    $E[S]=E[N]E[X]=4(500)=\$2{,}000$. For a Poisson, $E[N]=\text{Var}(N)=4$, so $\text{Var}(S)=E[N]\text{Var}(X)+\text{Var}(N)E[X]^2 = 4(250{,}000)+4(500)^2$ $=1{,}000{,}000 + 4(250{,}000)=1{,}000{,}000 + 1{,}000{,}000 = 2{,}000{,}000$. Standard deviation $=\sqrt{2{,}000{,}000}\approx \$1{,}414.21$.
  28. Aggregate moments
    Frequency is **negative binomial** with $r=3$, $\beta=2$ and severity is **exponential** with $\theta=400$. Find $E[S]$ and $\text{Var}(S)$.
    Frequency: $E[N]=r\beta=6$, $\text{Var}(N)=r\beta(1+\beta)=3(2)(3)=18$. Severity: $E[X]=400$, $\text{Var}(X)=\theta^2=160{,}000$. $E[S]=6(400)=\$2{,}400$. $\text{Var}(S)=E[N]\text{Var}(X)+\text{Var}(N)E[X]^2 = 6(160{,}000)+18(400)^2$ $=960{,}000 + 18(160{,}000)=960{,}000+2{,}880{,}000=3{,}840{,}000$.
  29. Compound Poisson & Panjer
    State the special variance identity for a **compound Poisson** aggregate, and why it simplifies.
    If $N\sim$ Poisson($\lambda$), then $E[N]=\text{Var}(N)=\lambda$, so the general formula collapses to $\text{Var}(S)=\lambda\,\text{Var}(X)+\lambda\,E[X]^2 = \lambda\big(\text{Var}(X)+E[X]^2\big)=\lambda\,E[X^2].$ Also $E[S]=\lambda E[X]$. 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.
  30. Compound Poisson & Panjer
    A **compound Poisson** has $\lambda = 10$ and severity uniform on $(0,2000)$. Find $E[S]$ and $\text{Var}(S)$.
    Uniform$(0,2000)$: $E[X]=1000$, and $E[X^2]=\frac{2000^2}{3}=\frac{4{,}000{,}000}{3}\approx 1{,}333{,}333$. $E[S]=\lambda E[X]=10(1000)=\$10{,}000$. $\text{Var}(S)=\lambda E[X^2]=10\big(1{,}333{,}333\big)\approx 13{,}333{,}333$. Standard deviation $\approx \sqrt{13{,}333{,}333}\approx \$3{,}651.48$.
  31. Compound Poisson & Panjer
    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 $0,1,2,\dots$.
    For $(a,b,0)$ frequency and severity probabilities $f_j = P[X=j]$ on the non-negative integers, $g_s = \frac{1}{1-a\,f_0}\sum_{j=1}^{s}\Big(a+\frac{b\,j}{s}\Big)f_j\,g_{s-j}, \quad s=1,2,\dots$ The 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.
  32. Compound Poisson & Panjer
    Give the Panjer starting value $g_0=P[S=0]$ for a compound Poisson, and explain when $S=0$.
    $S=0$ occurs when the total of all claims is $0$. For a **compound Poisson($\lambda$)** with severity probability of a zero claim $f_0=P[X=0]$, $g_0 = e^{-\lambda(1-f_0)}.$ If 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.
  33. Compound Poisson & Panjer
    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 $g_2$.
    Poisson: $a=0$, $b=\lambda=2$, and $f_0=0$ so $1-a f_0 = 1$. $g_0 = e^{-\lambda(1-f_0)} = e^{-2}\approx 0.135335$. $g_1 = \sum_{j=1}^{1}\big(0+\tfrac{2j}{1}\big)f_j g_{1-j} = (2)(0.6)g_0 = 1.2(0.135335)\approx 0.162402$. $g_2 = \sum_{j=1}^{2}\big(\tfrac{2j}{2}\big)f_j g_{2-j} = (1)f_1 g_1 + (2)f_2 g_0$ $= (0.6)(0.162402) + (2)(0.4)(0.135335) \approx 0.097441 + 0.108268 = 0.205709$.
  34. Compound Poisson & Panjer
    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 $g_1$.
    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 $f_0=0$, $f_1=1$. $g_0 = P[N=0]=(1-q)^m=0.8^3=0.512$. Since $f_0=0$, $1-a f_0=1$. $g_1 = \sum_{j=1}^{1}\big(a+\tfrac{b j}{1}\big)f_j g_0 = (a+b)f_1 g_0 = (-0.25+1.0)(1)(0.512)=0.75(0.512)=0.384.$ (Check: $S$ here equals $N$, and $P[N=1]=\binom{3}{1}0.2(0.8)^2=0.384$. ✓)
  35. Compound Poisson & Panjer
    Why is the **Panjer recursion** preferred over direct convolution for computing the aggregate distribution?
    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 $n$.
  36. Approximations
    Describe the **normal approximation** to the aggregate loss $S$ and when it is reasonable.
    Treat $S$ as approximately $N\big(E[S],\,\text{Var}(S)\big)$ and standardize: $P[S\le s]\approx \Phi\!\left(\frac{s-E[S]}{\sqrt{\text{Var}(S)}}\right).$ It 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.
  37. Approximations
    Aggregate losses have $E[S]=10{,}000$ and $\text{Var}(S)=4{,}000{,}000$. Using the **normal approximation**, find $P[S>13{,}000]$.
    $\text{SD}(S)=\sqrt{4{,}000{,}000}=2000$. Standardize: $z=\frac{13{,}000-10{,}000}{2000}=\frac{3000}{2000}=1.50$. $P[S>13{,}000]\approx 1-\Phi(1.50)=1-0.9332=0.0668.$ So about a $6.7\%$ chance aggregate losses exceed $\$13{,}000$.
  38. Approximations
    A **compound Poisson** has $\lambda = 50$ and severity exponential with $\theta = 200$. Use the normal approximation to find $P[S < 12{,}000]$.
    $E[X]=200$, $E[X^2]=2\theta^2=2(200)^2=80{,}000$. $E[S]=\lambda E[X]=50(200)=10{,}000$. $\text{Var}(S)=\lambda E[X^2]=50(80{,}000)=4{,}000{,}000$, so $\text{SD}(S)=2000$. $z=\frac{12{,}000-10{,}000}{2000}=1.00$, so $P[S<12{,}000]\approx \Phi(1.00)=0.8413.$
  39. Approximations
    Describe the **lognormal approximation** to aggregate losses and how its parameters are fitted.
    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 $E[S]=e^{\mu+\sigma^2/2}, \qquad E[S^2]=e^{2\mu+2\sigma^2}.$ Then $\sigma^2 = \ln\!\frac{E[S^2]}{E[S]^2}=\ln\!\big(1+\frac{\text{Var}(S)}{E[S]^2}\big)$ and $\mu = \ln E[S]-\tfrac{1}{2}\sigma^2$. Unlike the normal, the lognormal is right-skewed and positive, capturing the heavy upper tail better for moderate claim counts.
  40. Approximations
    Aggregate losses have $E[S]=10{,}000$ and $\text{Var}(S)=25{,}000{,}000$. Fit a **lognormal** by moment matching and find $P[S>20{,}000]$.
    $\sigma^2 = \ln\!\big(1+\frac{\text{Var}(S)}{E[S]^2}\big)=\ln\!\big(1+\frac{25{,}000{,}000}{100{,}000{,}000}\big)=\ln(1.25)\approx 0.223144$, so $\sigma\approx 0.472382$. $\mu = \ln(10{,}000)-\tfrac{1}{2}(0.223144)=9.210340-0.111572\approx 9.098768$. $P[S>20{,}000]=1-\Phi\!\big(\frac{\ln 20{,}000-\mu}{\sigma}\big)$. $\ln 20{,}000\approx 9.903488$, so $z=\frac{9.903488-9.098768}{0.472382}\approx 1.7034$. $P\approx 1-\Phi(1.70)=1-0.9554=0.0446.$
  41. Approximations
    Define the **stop-loss premium** $E[(S-d)_+]$ and relate it to the limited expected value of $S$.
    A stop-loss (aggregate excess) contract with retention $d$ pays $\max(S-d,0)=(S-d)_+$. The stop-loss premium (net expected cost) is $E[(S-d)_+]=E[S]-E[S\wedge d]=\int_d^\infty \big(1-F_S(x)\big)\,dx.$ For a continuous approximation, integrate the aggregate survival function above $d$; for a discrete $S$, sum $\sum_{x>d}(x-d)\,P[S=x]$.
  42. Approximations
    Aggregate losses are approximated as normal with $E[S]=10{,}000$ and $\text{SD}(S)=2000$. Find the **stop-loss premium** $E[(S-13{,}000)_+]$.
    For a normal $S$ with mean $\mu_S$ and SD $\sigma_S$, the stop-loss premium is $E[(S-d)_+]=\sigma_S\big[\phi(z)-z\,(1-\Phi(z))\big], \quad z=\frac{d-\mu_S}{\sigma_S}.$ Here $z=\frac{13{,}000-10{,}000}{2000}=1.50$. With $\phi(1.5)=\frac{1}{\sqrt{2\pi}}e^{-1.125}\approx 0.129518$ and $1-\Phi(1.5)=0.0668$: $E[(S-13{,}000)_+]=2000\big[0.129518 - 1.5(0.0668)\big]=2000(0.129518-0.10020)=2000(0.029318)\approx \$58.64$.
  43. Approximations
    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 $d=1000$.
    Stop-loss premium $E[(S-1000)_+]=\sum_{x>1000}(x-1000)P[S=x]$. Only $x=2000$ and $x=3000$ contribute: $(2000-1000)(0.15) + (3000-1000)(0.05) = 1000(0.15) + 2000(0.05) = 150 + 100 = \$250.$ Equivalently $E[(S-1000)_+]=E[S]-E[S\wedge 1000]$; both give $\$250$.
  44. Approximations
    How does the stop-loss premium change between adjacent integer retentions for a discrete $S$, and what is the recursion?
    For a severity/aggregate on a unit grid, the stop-loss premium decreases by the survival function as the retention rises by one unit: $E[(S-(d+1))_+]=E[(S-d)_+]-\big(1-F_S(d)\big)=E[(S-d)_+]-P[S>d].$ Intuitively, 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.