This Exam P sample reference tests Bernoulli Distribution. For Bernoulli indicators, E[XY] is the probability that both indicators equal one. Solving the correlation identity gives E[XY]=0.005+0.30sqrt(0.10·0.90·0.05·0.95)=0.024615, which rounds to 0.025 and selects choice C.
Represent the two one-year claim counts by Bernoulli indicators X and Y. Their means and standard deviations follow from the two marginal probabilities.
E[X]=0.10,SD(X)=0.10(0.90)=0.30
E[Y]=0.05,SD(Y)=0.05(0.95)
Model
Model
Insert the unknown joint moment into the definition of correlation. Because X and Y are indicators, XY equals one exactly when both claims occur.
ρ=SD(X)SD(Y)E[XY]−E[X]E[Y]
E[XY]=Pr(X=1,Y=1)
Compute
Compute
Rearrange the correlation equation and substitute the two Bernoulli variances.
Pr(X=1,Y=1)=0.10(0.05)+0.300.10(0.90)0.05(0.95)
Pr(X=1,Y=1)=0.0246150452…
Answer
Answer
The joint claim probability rounds to 0.025.
0.025(C)
Test the method on unseen problems
Probability Multivariate & Risk Models Practice Workbook
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.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis multiplies the marginals, 0.10(0.05)=0.005, as though the indicators were independent. Their positive correlation makes the joint probability larger.
BA joint probability of 0.006 would produce correlation (0.006-0.005)/√(0.10(0.90)0.05(0.95))=0.0153, not the supplied 0.30.
DA joint probability of 0.033 would imply correlation about 0.428. This results from adding too much covariance to the independent baseline.
EThe value 0.045 equals 0.05-0.005, the Y-only cell under an independence calculation. It is neither the both-claim cell nor compatible with correlation 0.30.
Original practice · fully worked
Original variant: recover alert correlation from a union
During a production cycle, indicator A records a temperature alert and indicator B records a vibration alert. Their probabilities are P(A=1)=0.40 and P(B=1)=0.50, while the probability of at least one alert is 0.65. Calculate the correlation between A and B.
A -0.2041
B 0.0000
C 0.0500
D 0.2041
E 0.2500
Answer and calculation
Inclusion-exclusion gives P(A=1,B=1)=0.40+0.50-0.65=0.25. The covariance is 0.25-0.20=0.05, and standardizing it gives 0.05/√(0.24·0.25)=0.2041, so choice D.
Setup
Setup
Use the union probability to recover the one-one cell of the two-indicator table.
Pr(A=1,B=1)=0.40+0.50−0.65=0.25
Model
Model
Compute the covariance and both Bernoulli variances from the recovered joint and the supplied marginals.
Cov(A,B)=0.25−(0.40)(0.50)=0.05
Var(A)=0.40(0.60)=0.24,Var(B)=0.50(0.50)=0.25
Compute
Compute
Standardize the covariance by the product of the two standard deviations.
Corr(A,B)=(0.24)(0.25)0.05
Corr(A,B)=0.2041241452…
Answer
Answer
The alert indicators have correlation approximately 0.2041.
