This Exam P sample reference tests Joint Discrete Distributions. Summing the joint weights over the triangular support gives normalization total 50. The weighted total of the difference Y-X is 55, so its expectation is 55/50=1.10, choice E.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe closest listed value, 0.19, can tempt someone who replaces the weighted numerator by the unweighted gap sum 10: that mixed calculation is 10/50=0.20. Each gap must retain its factor x+2y.
BThe value 0.28 is obtained by ignoring the restriction x≤y and normalizing over the full 4-by-4 rectangle; that incorrect support gives 20/72=0.2778.
CA value near 0.76 results from omitting the y=3 row. On the truncated support through y=2, the weighted expectation is 15/20=0.75.
DThe value 1.00 is the average gap if the ten admissible pairs are treated as equally likely. Their probabilities are proportional to x+2y, not uniform.
Original practice · fully worked
Original variant: a linear score under a rectangular joint mass function
During one diagnostic scan, let H be the high-priority count in {0,1,2} and let L be the low-priority count in {0,1}. The probability assigned to cell (h,l) is proportional to h+l+1 across all six cells. Calculate E[2H-L].
A 1/15
B 3/5
C 19/15
D 29/15
E 38/15
Variant answer in brief
The six joint weights sum to 15. Their weighted total for 2X-Y is 29, so E[2X-Y]=29/15, choice D.
Setup
Setup
Introduce the normalizing constant over the six-point rectangular support.
p(h,l)=c(h+l+1),h∈{0,1,2},l∈{0,1}
Model
Model
List the row weights and normalize them.
h=0∑2l=0∑1(h+l+1)=1+2+2+3+3+4=15
c=151
Compute
Compute
Multiply the requested linear score by each corresponding weight.
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