ASA exams

P

Probability · One discussion board for SOA and CAS

5 discussions

Study strategy

P: Choose the random variable before choosing a distribution

Common mistakes arise from modelling the wrong count or waiting time. Define what is counted, the observation window and any conditioning. A distribution name should follow those choices, rather than the appearance of a familiar keyword.

Worked example or practice scenario

In a toy homogeneous Poisson process with rate 2 per hour, N(3) has mean 6. The wait until the first event has exponential rate 2 and mean .5 hour. The wait until the third event has a different distribution and mean 1.5 hours.

Try this next

Write three different questions about the same process: a count in a fixed window, no event before a time, and the third arrival time. Check that each answer’s units match the variable. Do not use these assumptions for changing-rate processes without adjustment.

Reading sources

Official syllabus and resources

ActNet editorial guide · October 1, 2026 · Original illustrative examples.

0 repliesOct 1

Explanation

P: Conditional densities require a new support

A joint density can be constant without the variables being independent. A triangular support changes the range available to one variable when the other is fixed. Draw the region, obtain the marginal density and normalise within the conditional support.

Worked example or practice scenario

Let f(x,y)=2 on 0<y<x<1. Then f_X(x)=2x. Given X=x, Y is uniform on (0,x), so E[Y|X=x]=x/2. The conditional support is not (0,1).

Try this next

Integrate the conditional density over (0,x) to check it equals one. Then compute E[Y]=E[X]/2=1/3. If your answer uses a constant conditional mean of 1/2, revisit the geometry.

Reading sources

Official syllabus and resources

ActNet editorial guide · October 1, 2026 · Original illustrative examples.

0 repliesOct 1

Explanation

P: Total variance includes variation between groups

Conditioning can simplify a distribution, but taking only the average conditional variance loses the variation between group means. The law of total variance separates within-group variation from between-group variation. Identify both before substituting numbers.

Worked example or practice scenario

Two equally likely groups have conditional means 2 and 6, and conditional variances 1 and 9. E[Var(X|G)] = 5. The variance of the conditional mean is 4, so Var(X)=9. The overall mean is 4.

Try this next

Recover the same result from E[X²]−E[X]²: conditional second moments are 5 and 45. Explain what would happen if both groups had the same mean, and why that eliminates only the between-group component.

Reading sources

Official syllabus and resources

ActNet editorial guide · October 1, 2026 · Original illustrative examples.

0 repliesOct 1

Explanation

P: Independence and mutually exclusive events are different

Independence means learning one event occurred does not change the probability of the other. Mutual exclusivity means they cannot occur together. Nonzero mutually exclusive events cannot be independent. Test the definition with the joint probability, not with how unrelated the event names sound.

Worked example or practice scenario

If P(A)=.3 and P(B)=.4, independent events have P(A∩B)=.12 and P(A∪B)=.58. Mutually exclusive events with the same marginal probabilities instead have joint probability zero and union probability .7.

Try this next

For each interpretation, compute P(A|B). It is .3 in the independent case and zero in the mutually exclusive case. Add a third example where events overlap but are dependent.

Reading sources

Official syllabus and resources

ActNet editorial guide · October 1, 2026 · Original illustrative examples.

0 repliesOct 1

Explanation

P: Bayes’ rule starts with the base rate

A conditional claim probability does not tell you the probability that a claimant belongs to a risk group. Build joint probabilities first, then condition on the event actually observed. This also makes the denominator visible instead of treating Bayes’ rule as a formula to memorise.

Worked example or practice scenario

In an original population, 10% are high risk. Claim probabilities are 20% for high risk and 5% for other risks. Per 1,000 people, expected claims are 20 and 45. Thus P(high risk | claim) = 20/65 = 30.77%, not 80%.

Try this next

Change the high-risk share to 2%, leaving claim probabilities fixed. Expected claimant counts become 4 and 49 per 1,000, so the posterior is 4/53. Explain why the result falls before calculating it.

Reading sources

Official syllabus and resources

ActNet editorial guide · October 1, 2026 · Original illustrative examples.

0 repliesOct 1