Worked solution · Exam P
This Exam P sample reference tests Conditional Distributions. This is a conditional Bernoulli variance obtained by normalizing the relevant slice of a discrete joint distribution. Within Y=1, the probability that W=1 is 3/8, so the variance is 15/64 and choice B is correct.
Independent solution
Setup
After fixing Y at 1, sum the proportional weights over the two possible values of X for each value of W.
Model
Normalize the two weights. The conditional distribution of W is Bernoulli because W can only be zero or one.
Compute
Insert the conditional success probability into the Bernoulli variance formula.
Answer
The conditional variance is 15/64.
Test the method on unseen problems
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
A quality audit records binary indicators U and V. Their joint probabilities are P(U=0,V=0)=0.18, P(U=0,V=1)=0.12, P(U=1,V=0)=0.28, and P(U=1,V=1)=0.42. The audit score is S=3U-1. Calculate Var(S | V=1).
Given V=1, the probability that U=1 is 0.42/(0.12+0.42)=7/9. Scaling a Bernoulli variable by 3 multiplies its variance by 9, so Var(S|V=1)=14/9 and choice D is correct.
Setup
Normalize the V=1 column of the joint table.
Model
A constant shift has no effect on variance, while multiplication by 3 contributes a factor of 9.
Compute
Use the conditional Bernoulli variance and apply the scaling factor.
Answer
The conditional score variance is 14/9.
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