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Re: Large control loops
*To*: mathgroup at smc.vnet.net
*Subject*: [mg122355] Re: Large control loops
*From*: Ray Koopman <koopman at sfu.ca>
*Date*: Wed, 26 Oct 2011 17:38:58 -0400 (EDT)
*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com
*References*: <j80qff$ael$1@smc.vnet.net> <j83aon$k7c$1@smc.vnet.net> <j862ol$5tl$1@smc.vnet.net>
On Oct 25, 3:24 am, Arthur <shuk... at gmail.com> wrote:
> The vector in the first column is the common vector. It is the
> independent variable in the subsequent regressions - I am
> regressing first to find the correlation between it and each of
> the other vectors. It's all numbers.
>
> OLS regressions.
>
> I am estimating the parameters through a secondary regression on
> the residuals of the first and then some basic algebra. It's an
> Ornstein-Uhlenbeck process, I am using the first procedure outlined
> here
> http://www.sitmo.com/article/calibrating-the-ornstein-uhlenbeck-model/
>
> I am not so much worried about the time (although that too is a
> concern) but the actual structure. Should it just be a long while
> loop?
par[r] expects r to be a vector of residuals. It uses van den Berg's
first procedure and returns {a,b,sd,lambda,mu,sigma}. Since the
residuals sum to zero, the formulas simplify a little.
par[r_] := Block[{n = Length@r - 1, delta = 1 (* ?? *),
Sx,Sy, Sxx,Syy, Sxy, a,b,sd},
{Sx , Sy } = -r[[{-1,1}]];
{Sxx, Syy} = n(#.#&@Take[r,{2,-2}] + {Sy,Sx}^2) - {Sx,Sy}^2
Sxy = n Most at r.Rest@r - Sx*Sy;
{ a = Sxy/Sxx,
b = (Sy - a*Sx)/n,
sd = Sqrt[(Syy - a*Sxy)/(n(n-2))],
(* lambda = *) -Log[a]/delta,
(* mu = *) b/(1-a),
(* sigma = *) sd*Sqrt[-2 Log[a]/(delta(1-a^2))]} ]
I would get all the residuals simultaneously,
and then map par over them.
{x,y} = {First@#,Last@#}&[Transpose@A - Mean@A];
p = par /@ (y + Transpose[{y.x/-x.x}].{x}])
Those two lines would be followed by a loop that scans p looking for
the best parameters, and all that would be nested within a loop that
puts different values into A. (Note: rows 1,2,... of p correspond to
column 2,3,... of A.)
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