Continuation. For testing the hypothesis of independence of the X’s, H : p0 = p1, against the alternatives K : p0 < p1, consider the run test, which rejects H when the total number of runs R = U + V is less than a constant C(m)

depending on the number m of zeros in the sequence. When R = C(m), the hypothesis is rejected with probability γ(m), where C and γ are determined by PH {R < C(m)|m} + γ(m)PH {R = C(m)|m} = α.

(i) Against any alternative of K the most powerful similar test (which is at least as powerful as the most powerful unbiased test) coincides with the run test in that it rejects H when R < C(m). Only the supplementary rule for bringing the conditional probability of rejection (given m) up to α depends on the specific alternative under consideration.

(ii) The run test is unbiased against the alternatives K.

(iii) The conditional distribution of R given m, when H is true, is16 P{R = 2r} =

2

m−1 r−1

n−1 r−1



m+n m

 , P{R = 2r + 1} =

m−1 r−1

n−1 r



+ m−1 r

n−1 r−1



m+n m

 .

[(i): Unbiasedness implies that the conditional probability of rejection given m is

α for all m. The most powerful conditional level-α test rejects H for those sample sequences for which (u, v) = (p0/p1)v(q1/q0)u is too large. Since p0 < p1 and q1 < q0 and since |v − u| can only take on the values 0 and 1, it follows that

(1, 1) > (1, 2), (2, 1) > (2, 2) > (2, 3), (3, 2) > ··· .

Thus only the relation between (i,i + 1) and (i + 1,i) depends on the specific alternative, and this establishes the desired result.

(ii): That the above conditional test is unbiased for each m is seen by writing its power as

β(p0, p1|m) = (1 − γ)P{R < C(m)|m} + γP{R ≤ C(m)|m},

since by (i) the rejection regions R < C(m) and R < C(m) + 1 are both UMP at their respective conditional levels.
(iii): When H is true, the conditional probability given m of any set of m zeros and n ones is 1/
m+n m 
. The number of ways of dividing n ones into r groups is n−1 r−1 
, and that of dividing m zeros into r + 1 groups is m−1 r 
. The conditional probability of getting r + 1 runs of zeros and r runs of ones is therefore m−1 r n−1 r−1 
m+n m  .
To complete the proof, note that the total number of runs is 2r + 1 if and only if there are either r + 1 runs of zeros and r runs of ones or r runs of zeros and r + 1 runs of ones.]