Re: NIntegrate within FixedPoint (HoldAll)

*To*: mathgroup at smc.vnet.net*Subject*: [mg24582] Re: NIntegrate within FixedPoint (HoldAll)*From*: Albert Maydeu-Olivares <amaydeu at tinet.fut.es>*Date*: Tue, 25 Jul 2000 00:56:14 -0400 (EDT)*Organization*: Universitat de Barcelona*References*: <8lgpvk$1r3@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Hello everyone, I have been able to locate the source of the problem I posted in the group, but I still do not know how to solve it. In any case, you may skip my previous message. Let, pdf2[h_, k_, x_] := N[E^((-(k*(k/(1 - x^2) - (h*x)/(1 - x^2))) - h*(h/(1 - x^2) - (k*x)/(1 - x^2)))/2)/(2*Pi*Sqrt[1 - x^2]), 16]; cdf1[h_] := Chop[N[(1 + Erf[h/Sqrt[2]])/2, 16]]; cdf2[h_, k_, r_] := Chop[Module[{x}, NIntegrate[E^((-(k*(k/(1 - x^2) - (h*x)/(1 - x^2))) - h*(h/(1 - x^2) - (k*x)/(1 - x^2)))/2)/(2*Pi* Sqrt[1 - x^2]), {x, 0, r}, PrecisionGoal -> 10] + cdf1[h] cdf1[k]]]; where cdf1 and cdf2 are the uni and bivariate standard normal CDF functions. For some values of t1, t2, and r I generate some "data" In[2] = t1 = 1.; t2 = -1.; r = .3; obs = cdf2[t1, t2, r] Clear[r] Out[2]= 0.148338 Now, I recover the value r from obs using Newton's method (for fixed t1 and t2) In[41]:= step[r_] := r - (cdf2[t1, t2, r] - obs)/pdf2[t1, t2, r]; FixedPointList[step, .2, SameTest -> (Abs[#1 - #2] < 10.^(-6) &)] Out[42]= {0.2, 0.293054, 0.299959, 0.3, 0.3} However, when I divide the step in FixedPoint in parts a := cdf2[t1, t2, r]; b := obs; c := pdf2[t1, t2, r]; step[r_] := r - (a - b)/c; FixedPointList[step, .2, SameTest -> (Abs[#1 - #2] < 10.^(-6) &)] I obtain the following error NIntegrate::"nlim": "\!\(x$50\) = \!\(r\) is not a valid limit of \ integration." I would like to know how can I get this piece of code to work. It's not working because of the attribute HoldAll. Any help would be sincerely appreciated. Albert