Worked solution · Exam P
This Exam P sample reference tests Binomial Distribution. The two treatment counts are independent binomials with variances 1.35 and 3.6. Squaring the benefit coefficients gives total variance 4(1.35)+9(3.6)=37.8, choice B.
Independent solution
Setup
Let R and C count the two treatment types across the group.
Model
Compute each count variance and express the aggregate payment.
Compute
Independence removes the covariance term; scale variances by squared payment coefficients.
Answer
The total payment variance is 37.8.
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30 original multivariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.
Error diagnosis
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Original practice · fully worked
Three event indicators I1, I2, and I3 each have event probability 0.20. Every pair has covariance 0.02. A payment of 100 is made for each occurring event. Calculate the variance of the aggregate payment.
The indicator-count variance is 3(0.16)+2(3)(0.02)=0.60. Multiplying by 100² gives aggregate payment variance 6000, choice E.
Setup
Write the aggregate as 100 times the sum of three indicators.
Model
Include each of the three pairwise covariances twice in the variance of a sum.
Compute
Evaluate the count variance and apply the squared payment scale.
Answer
The aggregate payment variance is 6000.
Keep practicing
These references share syllabus concepts or sit nearby in the official sequence. All 140 Exam P multivariate random variables practice problems.
Drill the method: Exam P Independence problems
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