Clear[q,belief,p0,su,sstarmax,Cond,Intg]

The first step is to compute the prior belief as a function of the signal, given the registration decision being in some range (see the proof of Proposition 3 in the appendix of the paper for the derivation of this expression);

belief[s2_,sl_,su_]=1/(1+((1-p0)/p0)*((1-s2)/s2)*((2-sl-su)/(sl+su)));

Next, we compute the difference in the expected belief, given that the researcher’s belief being equal to some fixed value, say q, and the threshold which is indifferent between registration decisions is sstar--this uses that early registration leads to a belief that s_{1} is in the “top part” of the interval whereas not pre-registering leads to a belief that s_{1} is in the “bottom part” of the interval (note: this calculation may take some time):

Integrate[(belief[s2,sstar,1-sbar]-belief[s2,sbar,sstar])*(q*2*s2+(1-q)*2*(1-s2)),{s2,0,1}]

We note that since sstar and p0 are both real numbers, the conditioning events are not relevant. We then consider this expression as a function of q, the researcher’s belief that the hypothesis is true:

Intg[q_]=4 (-1+p0) p0 (-1+2 sbar) ((1/(2 (-1+2 sbar)))(((-1+(sbar^2)-2 sbar sstar+(sstar^2)) (q (-1-sbar+sstar)+p0 (1+(-1+2 q) sbar+sstar-2 q sstar)) Log[-((-1+p0) (1+sbar-sstar))])/(((-1+2 p0-sbar+sstar))^3)+(((sbar^2)+2 sbar (-1+sstar)+(-2+sstar) sstar) (-p0 (sbar+sstar)+q (2-sbar-sstar+2 p0 (-1+sbar+sstar))) Log[(-1+p0) (-2+sbar+sstar)])/(((-2+2 p0+sbar+sstar))^3))+(1/2) (((1-2 q)/(2-6 p0+4 (p0^2)+sbar-(sbar^2)-3 sstar+4 p0 sstar+(sstar^2)))+(2 ((p0^2) (-2+(-4+8 q) sstar)+q (-2-sbar+(sbar^2)+3 sstar-(sstar^2))-2 p0 (-1+sbar-(sbar^2)-sstar+(sstar^2)+q (-1-2 sbar+2 (sbar^2)+4 sstar-2 (sstar^2)))))/(((2+4 (p0^2)+sbar-(sbar^2)-3 sstar+(sstar^2)+p0 (-6+4 sstar)))^2)+((-1+(sbar^2)-2 sbar sstar+(sstar^2)) (q (-1-sbar+sstar)+p0 (1+(-1+2 q) sbar+sstar-2 q sstar)) Log[p0 (1-sbar+sstar)])/((-1+2 sbar) (((1-2 p0+sbar-sstar))^3))-(((sbar^2)+2 sbar (-1+sstar)+(-2+sstar) sstar) (-p0 (sbar+sstar)+q (2-sbar-sstar+2 p0 (-1+sbar+sstar))) Log[p0 (sbar+sstar)])/((-1+2 sbar) (((-2+2 p0+sbar+sstar))^3))));

Increasing gains to early registration holds if and only if this expression is increasing. Thus, we write take the derivative and consider this as a function of sstar and p0. We then verify that it is positive in the range of parameters that are relevant for our calibration:

Intg'[q]

The last step is to find the relevant range of sstar. If (Subscript[p, 0])=.3 and sbar=.2 (the most extreme case we consider, we have sstar is at most

sstarmax[p0_,sbar_]=p0*(1-sbar)/(p0*(1-sbar)+(1-p0)*sbar); 
sstarmax[.3,.2]

Below, we verify that in fact, the increasing differences condition is satisfied for our problem. While the expression is admittedly lengthy, we note that it the derivative of the difference is a continuous and smooth function of the parameters. We therefore have that expected payoff difference is increasing in the prior, as desired. (The graph takes as an input a value for (Subscript[(UnderBar[s]), 1])but is most restrictive for the lowest possible value, which as per the proposition is .2; we note that this value is only the one we use in our calibration exercise, but one can verify that it does hold for a larger (but not unrestricted) set of parameters)

Cond[sstar_,p0_,sbar_]=Intg'[q]; 
sbar1=.2; 
Plot3D[{Cond[sstar,p0,sbar1],0},{sstar,0,sstarmax[.3,sbar1]},{p0,0,.3}]
Clear[sbar1]