21
More General Distributions
•
But, problem here; let \alpha = 3/2:
E(
)
= 3
a
– 3 = 3/2
E(
)
= 3
a
– 3 +
a
– 2 = 1
Hence the more complex graph is more likely !
Solution: adjust E(H) to be the max of E(H
0
) for H
0
µ
H