2022-zelenko-paper[92250].pdf

(A11) (A12) (A13)

the number of adverse events that we expect to see in the finite sample of N patients chosen out of the general population. To calculate the confidence interval for x 0, we consider the inequality p(N, m, x) ≥ p0

(B1)

and let m1 (N, x, p0 ) be the minimum natural number and m2 (N, x, p0 ) be the maximum natural number for the parameter m that satisfies Eq. (B1). We note that both of these numbers are dependent on the sample size N, the probability x of an adverse outcome without treatment, and the p-value threshold p0 for establishing statistical confidence. Let S(N, x, p0 ) be the set of all natural numbers between m1 (N, x, p0 ) and m2 (N, x, p0 ), also including m1 (N, x, p0 ) and m2 (N, x, p0 ). Let S0 (N, x, p0 ) be the set of all m that satisfy Eq. (B1). Because p(N, m, x) is not monotonic with respect to x, the solution set S0 (N, x, p0 ) could be punctuated with empty gaps, and therefore we expect that S0 (N, x, p0 ) ⊆ S(N, x, p0 ). We also note that Eq. (B1) defines the indicator function I (N, n, x, p0 ) for the Sterne interval solution [71] of the binomial confidence interval problem, which is given by   1, if p(N, n, x) ≥ p0 I (N, n, x, p0 ) =   0, if p(N, n, x) < p 0 

(B2)

and proceed with the assumption that Sterne interval [71] has conservative coverage probability. From the above, we conclude that the probability p(N, p0 |x) for seeing an outcome of n adverse events with m1 (N, x, p0 ) ≤ n ≤ m2 (N, x, p0 ), in the absence of treatment, for a cohort of N patients, that are equivalent in every respect to the selected N patients that did receive treatment in our treatment group case series, satisfies, X p(N, p0 |x) = pr(N, n|x) (B3) n∈S(N,x, p0 )

23 X

≥

pr(N, n|x)

(B4)

I (N, n, x, p0 )pr(N, n|x)

(B5)

n∈S0 (N, x, p0 )

=

N X n=0

= c(N, p0 |x) ≥ 1 − p0

(B6)

The first inequality step follows from S0 (N, x, p0 ) ⊆ S(N, x, p0 ). The next step follows from Eq. (B2), then we apply the definition of the coverage probability, and the last inequality is based on the assumption that the Sterne interval has conservative coverage. We conclude, therefore, that there is at least 1 − p0 probability that the number n of adverse events without treatment would have been in the interval m1 (N, x, p0 ) ≤ n ≤ m2 (N, x, p0 ) for an equivalent cohort of N patients, chosen randomly out of the general population. Now let us consider the selection bias threshold, which is calculated as the minimum number x 1 that satisfies the implication x > x 1 (N, x 0, p0 ) =⇒ p(N, dx 0 Ne, x) < p0

(B7)

with dx 0 Ne the natural number obtained, if we round up the number x 0 N. To show that this threshold works, we need to show that x > x 1 (N, x 0, p0 ) implies that x 0 < m1 (N, x, p0 )/N, which is easily done with the following argument: First, we see that Eq. (B7) implies that the number dx 0 Ne is outside of the set S(N, x, p0 ), which in turn means that either dx 0 Ne < m1 (N, x, p0 ) or dx 0 Ne > m2 (N, x, p0 ). To rule out the second possibility, we observe that if x exceeds the selection bias threshold x 1 (N, x 0, p0 ), then it also exceeds the efficacy threshold x 0 and thus x > x 0 . We also observe that the population level probability x of an adverse outcome for untreated high risk patients has to be inside its own confidence interval, i.e. m1 (N, x, p0 )/N ≤ x ≤ m2 (N, x, p0 )/N. Using these inequalities, it follows that dx 0 Ne ≤ dx Ne ≤ d(m2 (N, x, p0 )/N )Ne = dm2 (N, x, p0 )e = m2 (N, x, p0 )

(B8) (B9)

and therefore we can rule out dx 0 Ne > m2 (N, x, p0 ). We conclude that dx 0 Ne < m1 (N, x, p0 ), and therefore x 0 = x 0 N/N ≤ dx 0 Ne/N < m1 (N, x, p0 )/N

(B10)

which gives us x 0 < m1 (N, x, p0 )/N. Since the true rate x 0 of an adverse event for an equivalent cohort of N patients can be bound between m1 (N, x, p0 )/N ≤ x 0 ≤ m2 (N, x, p0 )/N with 1 − p0 confidence, we can be assured that in spite of any random selection bias, x 0 exceeds the efficacy threshold x 0 with 1 − p0 confidence. This concludes the argument.

Appendix C: Monotonicity of the Bayesian factor

We prove that the function b0 (x 0, p2, t) is initially increasing and then decreasing with respect to t with a maximum in

the interval [a/N, 1]. We recall that Z t   x a (1 − x) N −a dx    p2 − x 0 0 Z p2 b0 (x 0, p2, t) = log   , t  a N −a  x (1 − x) dx   x0

(C1)

consequently maximizing the function b0 (x 0, p2, t) is equivalent to maximizing Z 1 t a g(t) = x (1 − x) N −a dx, (C2) t 0 since all other factors are independent of t. For our argument, it is simpler to work with the more abstract definition Z 1 t g(t) = f (x) dx, (C3) t 0 and assume that the function f (x) is increasing in the interval [0, a/N], decreasing in the interval [a/N, 1], and also satisfies f (1) = 0 and f (x) > 0 for all x ∈ (0, 1). These are all general assumptions that are indeed satisfied by the binomial distribution f (x) = x a (1−x) N −a . Differentiating with respect to t gives Z −1 t f (t) g 0 (t) = 2 f (x) dx + . (C4) t t 0 From the assumptions f (1) = 0 and f (x) > 0 for all x ∈ (0, 1), it immediately follows that Z 1 f (x) dx < 0. (C5) g 0 (1) = − 0

Next, we apply the integral mean-value theorem on the interval [0, a/N] which requires the assumption that f (x) > 0 for all x ∈ (0, a/N] and it follows that there exists ξ ∈ [0, a/N] such that Z a/N 1 f (ξ) = f (x) dx. (C6) a/N 0 We use this equation to show that Z a/N f (a/N ) −1 g 0 (a/N ) = f (x) dx + 2 a/N (a/N ) 0 − f (ξ) f (a/N ) = + a/N a/N ( f (a/N ) − f (ξ))N = > 0. a

(C7) (C8) (C9)

Here, the inequality step is justified by the assumption that the function f (x) is increasing in the interval [0, a/N]. It follows via the Bolzano theorem that there is at least one t 0 ∈ [a/N, 1] such that g 0 (t 0 ) = 0, making all such t 0 critical points that are the possible local minimum or maximum points of g(t). From Eq. (C4), it follows that all such critical points t 0 also satisfy the equation Z t0 1 f (t 0 ) = f (x) dx. (C10) t0 0

24 We shall now use the second derivative test to show that any such critical points have to be local maxima, which in turn implies the uniqueness of only one such local maximum point in the interval [a/N, 1]. The second derivative of the function g(t) is given by " # Z d −1 t f (t) (C11) g 00 (t) = f (x) dx + dt t 2 0 t Z t 2 f (t) f (t) f 0 (t) = 3 f (x) dx − 2 − 2 + (C12) t t 0 t t Z t 2 2 f (t) f 0 (t) = 3 f (x) dx − 2 + , (C13) t t 0 t

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and for t = t 0 , it follows that g 00 (t 0 ) =

2 f (t 0 ) f 0 (t 0 ) f 0 (t 0 ) 2 t 0 f (t 0 ) − + = < 0. (C14) 3 2 t0 t0 t0 t0

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