Re: Fourier Transfer and a game?!?!

*To*: mathgroup at smc.vnet.net*Subject*: [mg54207] Re: Fourier Transfer and a game?!?!*From*: Maxim <ab_def at prontomail.com>*Date*: Sun, 13 Feb 2005 00:21:36 -0500 (EST)*Organization*: MTU-Intel ISP*References*: <cuf5kg$god$1@smc.vnet.net> <cukb23$lmi$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

On Sat, 12 Feb 2005 07:25:55 +0000 (UTC), Maxim <ab_def at prontomail.com> wrote: > On Thu, 10 Feb 2005 08:22:40 +0000 (UTC), elparedblanco > <cire1611 at gmail.com> wrote: > >> let's say we have a probabability of picking an amount of money from a >> vat. I can go into the vat either 0,1,2,3 times. The probability with >> which I pick from the vat is described by the vector {.4, .35,.15,.1}. >> I call this frequency. >> >> I can pull only four amounts of money from the vat. The amounts are >> {$2500,$5000,$7500,$10000}. The probability of picking each amount is >> described by the vector {.35,.35,.15,.15}. I call this severity1. >> >> Process is this. First I choose how many times I can go into the vat. >> Then I go in that many times. I always replace what I pick out. so it >> is possible to win $30000. >> >> The Question is what's the probabililty of winning certain amounts of >> money, such as 15,000 or 7,500 or any number, given the fact that I can >> pick multiple times? >> > >> >> ANSWER = {.4000, .1225, .1409, .0935, .0995, .0499, .040, .0257, .016, >> .0074 .0034, .0010, .0003} >> > > Suppose we have a model with three vats which open independently with > probabilities p1, p2, p3 respectively. Then the generating function (GF) > for the amount of money taken from the first vat is ((1 - p1) > + p1*(.35*z^2500 + .35*z^5000 +.15*z^7500 + .15*z^10000)), and similarly > for the other vats. Of course then the solution would be simple, because > to find the GF for the sum taken from the three vats we just have to > multiply the three GFs. So if we could find p1, p2, p3 such that the > probability of all three vats opening, which is p1*p2*p3, equals .1 and > the equations for other combinations hold as well, we would be done. > > In[1]:= > sol = Solve[ > {(1 - p1)*(1 - p2)*(1 - p3) == .4, > p1*(1 - p2)*(1 - p3) + (1 - p1)*p2*(1 - p3) + > (1 - p1)*(1 - p2)*p3 == .35, > p1*p2*(1 - p3) + p1*(1 - p2)*p3 + (1 - p1)*p2*p3 == .15}, > {p1, p2, p3}][[1]] > > Out[1]= > {p1 -> 0.253124 - 0.401596 I, p2 -> 0.253124 + 0.401596 I, > p3 -> 0.443751} > > Other solutions are just the permutations of the first one. The solution > is complex-valued, but what if we substitute it in the GF? > > In[2]:= > Module[ > {gf = .35*z^2500 + .35*z^5000 +.15*z^7500 + .15*z^10000}, > ((1 - p1) + p1*gf)*((1 - p2) + p2*gf)*((1 - p3) + p3*gf) /. > sol // Expand // Chop // > Replace[List @@ #, > {k_.*z^p_. -> {p, k}, k_ -> {0, k}}, > {1}]& > ] > > Out[2]= > {{0, 0.4}, {2500, 0.1225}, {5000, 0.140875}, {7500, 0.0935375}, {10000, > 0.0994875}, {12500, 0.049875}, {15000, 0.03995}, {17500, 0.02565}, > {20000, > 0.015975}, {22500, 0.007425}, {25000, 0.003375}, {27500, 0.0010125}, > {30000, 0.0003375}} > > It would be interesting to give a rigorous justification for this trick. > > Maxim Rytin > m.r at inbox.ru > Alternatively, we can simply use the formula for the total probability P(amount = k) = Sum(P(amount = k | openvats = i)*P(openvats = i), i = 0..3). From this we can infer that the GF for amount is .4 + .35*gf + .15*gf^2 +.1*gf^3, arriving at the same expression for GF as above without the intermediate step of finding p1, p2, p3. In[1]:= Module[ {gf = .35*z^2500 + .35*z^5000 +.15*z^7500 + .15*z^10000}, .4 + .35*gf + .15*gf^2 +.1*gf^3 // Expand // Replace[List @@ #, {k_.*z^p_. -> {p, k}, k_ -> {0, k}}, {1}]& ] Out[1]= {{0, 0.4}, {2500, 0.1225}, {5000, 0.140875}, {7500, 0.0935375}, {10000, 0.0994875}, {12500, 0.049875}, {15000, 0.03995}, {17500, 0.02565}, {20000, 0.015975}, {22500, 0.007425}, {25000, 0.003375}, {27500, 0.0010125}, {30000, 0.0003375}} Here's how we can apply Fourier to this problem: In[2]:= F = Fourier; IF = InverseFourier; SetOptions[#, FourierParameters -> {1, 1}]& /@ {F, IF}; Module[{L}, L = PadRight[{0., .35, .35, .15, .15}, 13]; IF[.4*F[L]^0 + .35*F[L]^1 + .15*F[L]^2 + .1*F[L]^3] // Chop ] Out[5]= {0.4, 0.1225, 0.140875, 0.0935375, 0.0994875, 0.049875, 0.03995, 0.02565, 0.015975, 0.007425, 0.003375, 0.0010125, 0.0003375} With the default setting of FourierParameters F[Convolution[f, g]] == Sqrt[Length[f]]*F[f]*F[g]. Maxim Rytin m.r at inbox.ru

**Follow-Ups**:**Re: Re: Fourier Transfer and a game?!?!***From:*DrBob <drbob@bigfoot.com>