Difference between revisions of "Documentation:Monte Carlo Equilibration"

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(Theory)
(Theory)
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Suppose <math>\bar{y}_i = \beta_0 + \beta_1 x_i </math> (s.t. <math>i = 0, 1, \cdots, N-1</math>) is the least-squares best fitted line, we attempt to minimize <math> S = \sum_i (y_i - \bar{y}_i)^2 </math> w.r.t. <math>\beta_0</math> and <math> \beta_1</math>.
 
Suppose <math>\bar{y}_i = \beta_0 + \beta_1 x_i </math> (s.t. <math>i = 0, 1, \cdots, N-1</math>) is the least-squares best fitted line, we attempt to minimize <math> S = \sum_i (y_i - \bar{y}_i)^2 </math> w.r.t. <math>\beta_0</math> and <math> \beta_1</math>.
  
<math>\frac{\partial S}{\partial \beta_0 } = 0 </math> and <math>\frac{\partial S}{\partial \beta_1 } = 0 </math>
+
<math>\frac{\partial S}{\partial \beta_0 } = 0 </math> , <math>\frac{\partial S}{\partial \beta_1 } = 0 </math> :

Revision as of 11:57, 9 September 2013

Monte Carlo equilibration

Theory

We have a timeseries of N measurements obtained from a Monte Carlo simulation, i.e. y_0,y_1,\cdots,y_{N-1}.

Suppose \bar{y}_i = \beta_0 + \beta_1 x_i (s.t. i = 0, 1, \cdots, N-1) is the least-squares best fitted line, we attempt to minimize  S = \sum_i (y_i - \bar{y}_i)^2 w.r.t. \beta_0 and  \beta_1.

\frac{\partial S}{\partial \beta_0 } = 0 , \frac{\partial S}{\partial \beta_1 } = 0  :