covariate that indicates if a trial is a k/1-streak successor. Theorem 9 Let x 1 with I1k (x) If (x) = (β0k (x), (x)) is defined to be the ∈ {0, /= ∅. βk β1k n solution to the least squares problem, (x) argminβ x [1 d] where d 1 is βk β ∈ − ∈ {0, R2 2 n defined so that di := 1I1k for i = 1, . . . , n, then (x)(i) 55 (x) + (x) = Pˆ1k (x) β0k β1k Proof: (xi )(di Ln − − If (x) minimizes that sum of squares then (x) = Ln βk β1k (di First, working with the numerator, letting I1k I1k (x) we have − ≡ n )(xi = ) (xi + ) (xi − − − − − − ) (xi i=1 i∈I i∈I C C = 1 |I1k | | ) (xi n i∈I1k i∈I1k − − |I1k − I n − + = 1 |I1k | n ) xi i∈I1k ) xi (n |I1k | − |I |)x¯ i∈I C |I1k | xi i∈I1k xi = 1 |I1k | − − I1k | | Li∈I C n n |I1k | − |I1k | second, with the denominator of (x) we have β1k n |I1k | 2 |I1k | = ) 2 − − i=1 I1k n 2 i∈I1k
covariate that indicates if a trial is a k/1-streak successor. Theorem 9 Let x ∈ {0, 1 n with I1k (x) /= ∅. If βk (x) = (β0k (x), β1k (x)) is defined to be the solution to the least squares problem, βk (x) ∈ argminβ R2 x − [1 d] β 2 defined so that di := 1I1k (x)(i) for i = 1, . . . , n, then where d ∈ {0, 1 n is 55 β0k (x) + β1k (x) = Pˆ1k (x) Proof: If βk (x) minimizes that sum of squares then β1k (x) = Ln working with the numerator, letting I1k ≡ I1k (x) we have (xi − − Ln (di − First, n )(xi − − = ) (xi − − + ) (xi − − i=1 i∈I1k = 1 |I1k | n ) (xi i∈I1k i∈I C − − |I1k | ) (xi i∈I C − = 1 |I1k | ) xi i∈I1k |I1k | − n ) xi i∈I C I n + |I1k | (n − |I |)x¯ = |I1k | 1 − | I1k | i∈I1k xi Li∈I C xi − n |I1k | n − |I1k | second, with the denominator of β1k (x) we have n )(di − = ) |I1k | 2 − |I1k | 2 i=1 i∈I1k n I1k 2
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