Re: Maximum Likelihood Problem

*To*: mathgroup at smc.vnet.net*Subject*: [mg46456] Re: Maximum Likelihood Problem*From*: drbob at bigfoot.com (Bobby R. Treat)*Date*: Thu, 19 Feb 2004 03:02:09 -0500 (EST)*References*: <c0t2ii$2t9$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

I'm not familiar with some of your terminology, so I did this my own way. Maybe it will help you, even if I messed up somewhere. First of all, I wrote your model as you seem to really be using it: Clear[s, t, m, normal, u, dist] dist = NormalDistribution[0, 1]; u[x_] := x(1 - b t) + s(t^(1/2))*normal u[x] The important terms here (for known parameters) is a constant times x plus a constant times the normal variate. Introducing ss and bb to name those constants, if I could estimate them I could invert the transformation to get s and b: invert = First[Solve[{ss == s*Sqrt[t], bb == 1 - b*t}, {s, b}]] {s -> ss/Sqrt[t], b -> (1 - bb)/t} So I'll rewrite the model as Clear[v] v[x_]:=x bb+ss*normal That gives the next data point v[x] given the current data point x. Here's a set of generated data: data = Block[{t = 1/12, b = 10, s = 1, bb, ss, normal := Quantile[dist, Random[]]}, bb = 1 - b*t; ss = s*Sqrt[t]; NestList[v, 0.2, 100]]; ListPlot[data]; Being a little rusty, I derive the conditional density: Pr[(v - bb*x)/ss <= z] == CDF[dist, z] Pr[v - bb*x <= z] == CDF[dist, ss*z] Pr[v <= z] == CDF[dist, ss*z + bb*x] F[v] == CDF[dist, ss v + bb x] D[#, v] & /@ % logLikelihood[bb_, ss_][x_, v_] = PowerExpand[Log[ss/(E^((1/2)*(ss*v + bb*x)^2)* Sqrt[2*Pi])]] Summing over a collection of data looks like this: logLikelihood[bb_, ss_][data_List] := Plus @@ logLikelihood[bb, ss] @@@ Partition[data, 2, 1] Finally, solving for the maximum likelihood goes like this: logLikelihood[bb, ss][data] // Simplify D[%, #] == 0 & /@ {bb, ss} Solve[%] {s, b} /. invert % /. %% % /. t -> 1/12 Hope that helps. Bobby sabrinacasagrande at hotmail.com (sabbri) wrote in message news:<c0t2ii$2t9$1 at smc.vnet.net>... > I need to generate an algorithm in order to estimate the parameters of > an equation . It's probably a mistake in my codes that prevents me to > get the right results. > DATA (Sample Path generating): I generated the data using an > approximation of the differential equation dXt = - B Xt dt + s dWt: > > u[x_]:= x + m[x]*t + s[x]* > ((t)^(1/2))*(Quantile[NormalDistribution[0,1],Random[]]) > > where I state m[x_]= -10 x, and s[x_]= 1, t=1/12 Then I generate the > data with Table[NestList[u,0.2,100],{1}]]. So I have my data with the > true parameters value, B=10 and s=1. > ALGORITHM: I use an approximation (of the transition density > probability function). I constructed the approximation based on the > saddle points method (or Laplace Method). > > sigx[x_] := s > > mx[x_] := -B x > > g[x_] = Integrate[(1/sigx[u]), {u, 0, x}] > > sigy[x_] = 1 > > my[x_] = (mx[x]/sigx[x]) - (D[sigx[x], x]/2) > > fi[z_] = (1/Sqrt[2 Pi]) (Exp[-(z^2)/2]) > > lamy[x_] = (-1/2) (((my[x])^2) + D[my[x], x]) > > c[y_, x_, j_] := c[y, x, j] > = ((j)*(y - x)^(-j)) (Integrate[((w - x)^(j - 1))((lamy[w]c[w, x, j - > 1]) + ((D[c[w, x, j - 1], {w, 2}])(1/2))), {w, x,y}]) > > c[y_, x_, 0] = 1 > > py[t_, y_, x_] = (1/Sqrt[t]) (fi[(y - x)/(Sqrt[t])]) > (Exp[Integrate[my[w], {w, x,y}]]) (Sum[c[y, x, k] ((t)^k)/(k!), {k, 0, > 1}]) > > px[t_, y_, x_] = (py[t, g[y], g[x]]/sigx[g[x]]) > > And finally I get a function (the likelihood function) > > L[N_] = Sum[Log[px[t, x[n + 1], x[n]]], {n, 1, N}]/N > > where x[n_] are the values I generated above. So x[n_]:= > Extract[data,{1,n}] > Now I try to maximize it. > I tried using two methods: 1)the function Maximize > Maximize[{L[99],s>0},{B,s}] > > 2) Solve[] on the derivatives == 0 of my target function. > Solve[{D[L[99],B]==0,D[L[99],s]==0},{B,s}] > > In none of this case I can get the results. (Mind that I generated the > data attributing the values to parameters, so data were generated with > the right values: I should obtain those values while maximizing!!!) > Can you help me? Where am I wrong?