Difference between revisions of "Documentation:Monte Carlo Equilibration"

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(Synposis)
(Synposis)
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|| observables
 
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|| Python str, or Python list of Python str
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|| (list of) measurement observable(s)
 
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||  confidenceInterval
 
||  confidenceInterval
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|| <math> \mathrm{tolerance} = \frac{X^\mathrm{(fit)} (t_\mathrm{final}) - X^\mathrm{(fit)} (t_\mathrm{initial})}{\bar{X}} </math>
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|| Python float
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|| confidence interval in which steady state has not been reached
 
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|| shall we combine the checks of all observables as 1 final boolean answer?
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Revision as of 14:27, 10 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  :


\Rightarrow \left( 
\begin{array}{cc} 
N               & \sum_i x_i     \\
\sum_i x_i & \sum_i x_i^2
\end{array}
\right)
\left(
\begin{array}{c}
\beta_0 \\ 
\beta_1
\end{array}
\right) =
\left(
\begin{array}{c}
\sum_i y_i  \\ 
\sum_i x_i y_i 
\end{array}
\right)


\Rightarrow \left(
\begin{array}{c}
\beta_0 \\ 
\beta_1
\end{array}
\right) =
\frac{1}{N\sum_i x_i^2 - (\sum_i x_i)^2}
\left( 
\begin{array}{cc} 
\sum_i x_i^2 & -\sum_i x_i     \\
-\sum_i x_i     & N
\end{array}
\right)
\left(
\begin{array}{c}
\sum_i y_i  \\ 
\sum_i x_i y_i 
\end{array}
\right)


\Rightarrow \beta_1 = \frac{N \sum_i x_i y_i - \sum_i x_i \sum_i y_i}{N\sum_i x_i^2 - (\sum_i x_i)^2}


Slope of best-fitted line


\Rightarrow \beta_1 = \frac{\sum_i (x_i - \bar{x}_i)( y_i - \bar{y}_i) }{\sum_i (x_i - \bar{x}_i)^2}  \,\,\,\,\, \left( = \frac{s_{xy}}{s_{xx}} \right)


Error in slope of best-fitted line


\mathrm{Var}({\beta_1}) 
= \mathrm{Var} \left( \frac{\sum_i (x_i - \bar{x}_i)( y_i - \bar{y}_i) }{\sum_i (x_i - \bar{x}_i)^2} \right)
= \frac{\sum_i (x_i - \bar{x}_i)^2 \mathrm{Var}(y_i) }{(\sum_i (x_i - \bar{x}_i)^2)^2}


Denoting \mathrm{Var}(y_i) = \sigma_{y}^2 , we have:

 \Rightarrow \mathrm{Var}({\beta_1}) = \sigma_{\beta_1}^2 = \frac{\sigma_y^2}{\sum_i (x_i - \bar{x}_i)^2}

 \Rightarrow \mathrm{Var}({\beta_1}) = \sigma_{\beta_1}^2 = \frac{ 12 \, \sigma_y^2 }{ N(N^2 - 1) }


Hypothesis testing


 
\begin{array}{l}
H_0 : \beta_1 = 0 \\
H_1 : \beta_1 \ne 0 
\end{array}


Using the standard z-test, we reject H_0 at confidence interval \alpha if


\left| \frac{\beta_1 - 0 }{ \sigma_{\beta_1} } \right| > z_{\frac{1-\alpha}{2}}


Practice

Synposis

pyalps.checkNonSteadyState(outfile, observables, confidenceInterval=0.63, simplified=False, includeLog=False)
argument default type remark
outfile - Python str ALPS hdf5 output file(name)
observables - (list of ) Python str (list of) measurement observable(s)
confidenceInterval 0.63 Python float confidence interval in which steady state has not been reached
simplified False Python bool shall we combine the checks of all observables as 1 final boolean answer?
includeLog False Python bool shall we print the detailed log?