Free CAS MAS-I (Modern Actuarial Statistics I) Statistics Practice Questions

Master statistical inference for CAS MAS-I. Questions test hypothesis testing, confidence intervals, Bayesian estimation, simulation methods, and applied statistics for property and casualty problems.

137 Questions
46 Easy
65 Medium
26 Hard
2026 Syllabus
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Sample Questions

Question 1 Easy
You are given the following information:

- Random variable XX has an exponential distribution
- Var(X)=19\text{Var}(X) = \frac{1}{9}

Calculate the median of the distribution.
Solution
For an exponential distribution with rate parameter λ\lambda, the variance is 1/λ21/\lambda^2. Given Var(X)=1/9\text{Var}(X) = 1/9, we have 1/λ2=1/91/\lambda^2 = 1/9, so λ=3\lambda = 3 and the mean is 1/λ=1/31/\lambda = 1/3. The median of an exponential distribution is ln2λ=ln23=0.693130.2310\frac{\ln 2}{\lambda} = \frac{\ln 2}{3} = \frac{0.6931}{3} \approx 0.2310. This falls in [0.20,0.25)[0.20, 0.25).

Choice A is incorrect because 0.2310 is not less than 0.15.
Choice B is incorrect because 0.2310 is not less than 0.20.
Choice
((E)) is correct because 0.200.2310<0.250.20 \leq 0.2310 < 0.25.
Choice C is incorrect because 0.2310 is less than 0.25.
Choice D is incorrect because 0.2310 is less than 0.30.
Question 2 Medium
An actuary studies mortality in a group of 80 insured lives:

| Time tjt_j | Risk set njn_j | Deaths djd_j |
|------|--------|--------|
| 1 | 80 | 8 |
| 2 | 68 | 10 |
| 3 | 52 | 7 |

Using the Nelson-Aalen estimator, calculate H^(3)\hat{H}(3).
Solution

(A) is correct.

The Nelson-Dalen estimator is:
H^(3)=d1n1+d2n2+d3n3=880+1068+752\hat{H}(3) = \frac{d_1}{n_1} + \frac{d_2}{n_2} + \frac{d_3}{n_3} = \frac{8}{80} + \frac{10}{68} + \frac{7}{52}
=0.10000+0.14706+0.13462=0.381680.382= 0.10000 + 0.14706 + 0.13462 = 0.38168 \approx 0.382

Choice E is incorrect because 0.217 omits the third term, computing only through time 2 and understating it.
Choice B is incorrect because 0.282 uses dj/(nj+dj)d_j/(n_j + d_j) instead of dj/njd_j/n_j.
Choice D is incorrect because 0.452 results from using reduced risk sets as if censoring preceded deaths.
Choice C is incorrect because 0.547 is 1S^(3)1 - \hat{S}(3) from the Kaplan-Meier, not the cumulative hazard.
Question 3 Hard
Let X1,X2,,XnX_1, X_2, \ldots, X_n be a random sample from a distribution with density:

f(xθ)=θxθ1,0<x<1,θ>0f(x \mid \theta) = \theta x^{\theta - 1}, \quad 0 < x < 1, \quad \theta > 0

Using the Fisher-Neyman factorization theorem, determine a sufficient statistic for θ\theta.
Solution

((A)) is correct.

The joint density is:
i=1nf(xiθ)=i=1nθxiθ1=θn(i=1nxi)θ1\prod_{i=1}^n f(x_i \mid \theta) = \prod_{i=1}^n \theta x_i^{\theta-1} = \theta^n \left(\prod_{i=1}^n x_i\right)^{\theta-1}

Rewriting using T=xiT = \prod x_i:
=θnexp[(θ1)i=1nlnxi]=θnTθ1= \theta^n \exp\left[(\theta - 1) \sum_{i=1}^n \ln x_i\right] = \theta^n T^{\theta-1}

This factors as g(T,θ)=θnTθ1g(T, \theta) = \theta^n T^{\theta-1} and h(x)=1h(\mathbf{x}) = 1. Since the θ\theta-dependent part depends on the data only through T=XiT = \prod X_i, by the factorization theorem Xi\prod X_i is sufficient for θ\theta. Equivalently, lnXi\sum \ln X_i is sufficient since it is a one-to-one transformation of Xi\prod X_i.

Choice E is incorrect because Xi\sum X_i does not appear in the factored likelihood; the joint density involves xi\prod x_i raised to a power, not xi\sum x_i.
Choice C is incorrect because Xiθ\sum X_i^\theta depends on the parameter θ\theta inside the statistic, which does not yield a valid sufficient statistic.
Choice D is incorrect because the support (0,1)(0,1) does not depend on θ\theta, so the maximum order statistic carries no special role; it does not capture the xi\prod x_i structure.
Choice B is incorrect because Xi2\sum X_i^2 does not appear in the factorization and is not sufficient for this power-family distribution.
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