This Exam P sample reference tests Variance. Let b=P(XY=1), so P(XY=2)=3b and P(XY=0)=1-4b. The variance is 13b-49b², maximized at b=13/98; therefore P(XY=0)=23/49, choice C.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AIf P(Z=0)=11/25, normalization would give b=0.14. The variance derivative there is 13-98(0.14)=-0.72, so this is not the maximizer.
BThe value 23/50 is a premature decimal-style denominator change. The exact optimizer b=13/98 gives 1-4b=23/49, not 23/50.
DThe value 26/49 is 1-23/49=P(Z>0). It is the complement of the requested zero-product event.
EIf P(Z=0)=14/25, then b=0.11 and the derivative is 2.22 rather than zero; the variance is still increasing there.
Original practice · fully worked
Original variant: minimize a blended estimate's variance
Two unbiased estimates X and Y target the same calibration constant. Their variances are 9 and 4, and their covariance is 1. For W_a=aX+(1-a)Y, determine the real value of a that minimizes Var(W_a).
A 0
B 7/24
C 3/11
D 4/13
E 1/3
Variant answer in brief
Expanding the weighted-sum variance gives 11a²-6a+4. Its derivative vanishes at a=3/11, and the positive quadratic coefficient makes this the minimum, choice C.
Setup
Setup
Apply the variance formula to the weighted combination.
Var(Wa)=a2Var(X)+(1−a)2Var(Y)+2a(1−a)Cov(X,Y)
Model
Model
Substitute the three supplied second-moment quantities and collect powers of a.
Var(Wa)=9a2+4(1−a)2+2a(1−a)
Var(Wa)=11a2−6a+4
Compute
Compute
Differentiate the convex quadratic and solve its first-order condition.
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