Linear Combinations of Independent Random Variables
Standardization
This Exam P sample reference tests Normal Distribution. The standardized mean of a twelve-period total is the one-period standardized mean multiplied by the square root of twelve. Dividing the 60th-percentile normal score by that factor gives a one-period positive probability of 0.529151, so choice B is correct.
Write one period's result as a normal variable with mean mu and standard deviation sigma. Independence makes the twelve-period total normal with additive means and variances.
X∼N(μ,σ2)
S=i=1∑12Xi∼N(12μ,12σ2)
Model
Model
Convert the supplied total-period probability into its standard-normal score.
Pr(S>0)=Φ(12σ12μ)=0.60
12σμ=Φ−1(0.60)
Compute
Compute
Recover the one-period standardized mean and return through the normal distribution function.
σμ=120.2533471031=0.0731350091
Pr(X>0)=Φ(σμ)=Φ(0.0731350091)=0.5291506585
Answer
Answer
The one-period positive probability rounds to 0.53.
0.53(B)
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 divides the annual z-score by 12 instead of by √(12): Φ(0.253347/12)=0.5084, which rounds to 0.51.
CThis simply carries the twelve-period probability 0.60 over to one period, ignoring how independent normal variances aggregate.
DThis multiplies rather than divides the total z-score by √(12), giving Φ(0.87756)=0.8100.
EThis treats a positive total as equivalent to all twelve individual periods being positive and solves q¹²=0.60, giving q=0.9583.
Original practice · fully worked
Original variant: correlated quality-score contrast
Quality scores X and Y are jointly normal. Their means are 10 and 4, their standard deviations are 2 and 3, and their correlation is 0.25. A review index is W=2X-Y. Calculate P(W>20).
A 0.1794
B 0.2119
C 0.2362
D 0.4166
E 0.8206
Answer and calculation
The covariance is 0.25(2)(3)=1.5. Thus W has mean 16 and variance 4(4)+9-4(1.5)=19, so its standardized threshold is 4/√(19) and the upper tail is 0.179398, selecting choice A.
Setup
Setup
Convert the reported correlation to covariance and write the review index.
Cov(X,Y)=0.25(2)(3)=1.5
W=2X−Y
Model
Model
A linear combination of jointly normal variables is normal. Its variance includes the signed covariance term.
E[W]=2(10)−4=16
Var(W)=22Var(X)+Var(Y)−4Cov(X,Y)
Compute
Compute
Evaluate the index variance and standardize the cutoff.
Var(W)=4(4)+9−4(1.5)=19
Pr(W>20)=1−Φ(1920−16)=0.1793976789
Answer
Answer
The review index exceeds 20 with probability 0.1794.
