This Exam P sample reference tests Conditional Variance. For S uniform from 5 to 20, E[S]=12.5, Var(S)=18.75, and E[S²]=175. Total variance gives E[9S²]+Var(3S)=1,575+168.75=1,743.75, choice E.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 525 is 3 E[S²]. It uses the conditional mean multiplier 3 where the conditional variance requires its square 9, and it omits variation between scales.
BThe value 1,462.5 follows from the invalid expression (3E[S])²+3 Var(S)=1,406.25+56.25. It squares the mean in place of a variance and under-scales the second term.
CThe value 1,575 is E[Var(X|S)]=9 E[S²] only. It omits Var(E[X|S]) caused by the random conditional mean.
DRounding Var(S)=18.75 to 19 and then adding the incorrectly scaled term 3(19) to 1,575 gives the listed 1,632. The between-scale term is Var(3S)=9 Var(S), not 3 Var(S).
Original practice · fully worked
Original variant: covariance induced by a production regime
A production regime R takes value 1 with probability 0.25, value 3 with probability 0.35, and value 6 otherwise. Conditional on R, the defect count Y is Poisson with mean 2R. Calculate Cov(R,Y).
A 4.11
B 7.40
C 8.22
D 16.44
E 23.84
Variant answer in brief
The regime has E[R]=3.7, E[R²]=17.8, and Var(R)=4.11. Because E[Y|R]=2R, total covariance gives Cov(R,Y)=Cov(R,2R)=2 Var(R)=8.22, choice C.
Setup
Setup
Compute the first two moments of the discrete regime variable.
E[R]=1(0.25)+3(0.35)+6(0.40)=3.7
E[R2]=1(0.25)+9(0.35)+36(0.40)=17.8
Var(R)=17.8−3.72=4.11
Model
Model
Conditioning makes R fixed, so the conditional covariance with Y is zero; only the changing conditional mean contributes.
E[Y∣R]=2R
Cov(R,Y)=Cov(R,E[Y∣R])
Compute
Compute
Replace the conditional mean by its linear expression.
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