Question: In the preceding problem consider arbitrary contrasts ci i with ci = 0. The event & & X j Xi

In the preceding problem consider arbitrary contrasts 

 ci ξi with ci = 0. The event

&

&

X j − Xi

ξ j − ξi

&

& ≤  for all i = j (9.149)

is equivalent to the event

&

&

&

ci Xi −ci ξi

&

&

& ≤



2

|ci| for all c withci = 0, (9.150)

which therefore also has probability γ. This shows how to extend the Tukey intervals for all pairs to all contrasts.

[That (9.150) implies (9.149) is obvious. To see that (9.149) implies (9.150), let yi = xi − ξi and maximize |

ci yi| subject to |yj − yi| ≤  for all i and j. Let P and N denote the sets {i : ci > 0} and {i : ci < 0}, so that

ci yi = 

i∈P ci yi −

i∈N

|ci| yi .

Then for fixed

c, the sum ci yi is maximized by maximizing the yi’s for i ∈

P and minimizing those for i ∈ N. Since |yj − yi| ≤ , it is seen that ci yi is



maximized by yi = /2 for i ∈ P, yi = −/2 for i ∈ N. The minimization of ci yi is handled analogously.]

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