Free SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) Aggregate Models Practice Questions

Aggregate loss models on SOA Exam ASTAM test compound frequency-severity models, the Panjer recursive method, fast Fourier transform (FFT) approaches, and stop-loss reinsurance pricing.

155 questions 62 easy 56 medium 37 hard 2026 syllabus

Sample Questions

Question 1 Easy
Which of the following distributions belongs to the (a,b,0)(a, b, 0) class with a>0a > 0 and b≥0b \geq 0?
Solution
D is correct.

The (a,b,0)(a,b,0) class consists of counting distributions whose probabilities satisfy
pkpk−1=a+bk,k=1,2,3,…\frac{p_k}{p_{k-1}} = a + \frac{b}{k}, \qquad k = 1, 2, 3, \ldots
with p0>0p_0 > 0.

For the negative binomial with parameters r>0r > 0 and β>0\beta > 0,
pk=(r+k−1k)(β1+β)k(11+β)r,p_k = \binom{r + k - 1}{k} \left(\frac{\beta}{1+\beta}\right)^{k} \left(\frac{1}{1+\beta}\right)^{r},
so
pkpk−1=r+k−1k⋅β1+β=β1+β+(r−1)β/(1+β)k.\frac{p_k}{p_{k-1}} = \frac{r + k - 1}{k}\cdot\frac{\beta}{1+\beta} = \frac{\beta}{1+\beta} + \frac{(r-1)\beta/(1+\beta)}{k}.
Hence
a=β1+β>0,b=(r−1)β1+β.a = \frac{\beta}{1+\beta} > 0, \qquad b = \frac{(r-1)\beta}{1+\beta}.
With r≥1r \geq 1 we get b≥0b \geq 0, and b=0b = 0 exactly in the geometric case r=1r = 1. So the negative binomial is the (a,b,0)(a,b,0) member with a>0a > 0 and b≥0b \geq 0.

For comparison, the Poisson gives a=0a = 0, b=λb = \lambda, and the binomial gives a=−q/(1−q)<0a = -q/(1-q) < 0, b=(m+1)q/(1−q)b = (m+1)q/(1-q); both are in the class but neither has a>0a > 0.
Question 2 Medium
An exponential severity with mean θ=500\theta = 500 is discretized using the method of rounding with span h=200h = 200. Compute the masses at grid points 0, 200, and 400.
Solution
B is correct.

Under the method of rounding with span h=200h=200, grid point k⋅hk \cdot h collects all probability mass within [(k−0.5)h,(k+0.5)h)[(k-0.5)h, (k+0.5)h). For k=0k=0: the interval is (−100,100)(-100, 100), but since support is [0,∞)[0,\infty), this is [0,100)[0, 100): p0=F(100)−F(0)=(1−e−100/500)−0=1−e−0.2≈0.1813p_0 = F(100) - F(0) = (1-e^{-100/500}) - 0 = 1 - e^{-0.2} \approx 0.1813. For k=1k=1 (grid point 200): interval [100,300)[100, 300): p200=F(300)−F(100)=e−0.2−e−0.6≈0.8187−0.5488=0.2700p_{200} = F(300) - F(100) = e^{-0.2} - e^{-0.6} \approx 0.8187 - 0.5488 = 0.2700. For k=2k=2 (grid point 400): interval [300,500)[300, 500): p400=F(500)−F(300)=e−0.6−e−1.0≈0.5488−0.3679=0.1809p_{400} = F(500) - F(300) = e^{-0.6} - e^{-1.0} \approx 0.5488 - 0.3679 = 0.1809.
Question 3 Hard
For a compound Poisson SS with λ=2\lambda = 2 and severity P(X=1)=P(X=2)=0.5P(X=1) = P(X=2) = 0.5, use the Panjer recursion to compute g0g_0, g1g_1, and g2g_2, then find P(S≤2)P(S \leq 2).
Solution
E is correct.

For compound Poisson with a=0a=0, b=λ=2b=\lambda=2, and f1=f2=0.5f_1 = f_2 = 0.5: g0=e−λ=e−2g_0 = e^{-\lambda} = e^{-2}. For s=1s=1: g1=b1f1g0=21(0.5)(e−2)=e−2.g_1 = \frac{b}{1}f_1 g_0 = \frac{2}{1}(0.5)(e^{-2}) = e^{-2}. For s=2s=2: g2=b2f1g1+2b2f2g0=22(0.5)(e−2)+2⋅22(0.5)(e−2)=0.5e−2+e−2=1.5e−2.g_2 = \frac{b}{2}f_1 g_1 + \frac{2b}{2}f_2 g_0 = \frac{2}{2}(0.5)(e^{-2}) + \frac{2 \cdot 2}{2}(0.5)(e^{-2}) = 0.5e^{-2} + e^{-2} = 1.5e^{-2}. Therefore P(S≤2)=e−2(1+1+1.5)=3.5e−2≈0.4726P(S \leq 2) = e^{-2}(1 + 1 + 1.5) = 3.5e^{-2} \approx 0.4726.
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