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Introduction

use new method for measuring w pola

w pola

The theoretical equation:

$\frac{1}{N}\frac{dN}{d\cos\Psi}=\frac{3}{2}\left\lbrack F_0 \left(\frac{\sin \Psi}{\sqrt{2}} \right)^2 +F_L \left(\frac{1-\cos \Psi}{2} \right)^2 +F_R \left(\frac{1+\cos \Psi}{2} \right)^2\right\rbrack$

let $x=\cos \Psi$

$ \frac{1}{N}\frac{dN}{dx}=\frac{3}{2}\left\lbrack F_0 \frac{1-x^2}{2}} +F_L\frac{(1-x)^2}{4} +(1-F_0-F_L) \frac{(1+x)^2}{4}\right\rbrack=\frac{3}{8} \left\lbrack F_0 (1-2x-3x^2) +F_L (-4x) + (1+x)^2 \right\rbrack$

measured distribution and acceptance function

Let $ f(x) $ be measued distribution $\Im(x)$ acceptance function.

Got: $f(x)=\frac{3}{8} \left\lbrack F_0 (1-2x-3x^2) +F_L (-4x) + (1+x)^2 \right\rbrack \Im(x)$

nomalize $f(x)$

assume $\int_{-1}^{1} c_1 \Im(x)dx =1$ Got the integral:

$ \frac{3}{8} \int_{-1}^{1} \left\lbrack F_0 (1-2x-3x^2) +F_L (-4x) + (1+x)^2 \right\rbrack \Im(x) dx = \frac{3}{8c_1} \left\lbrack F_0 (1-2\overline{x}-3\overline{x^2}) - F_L (4\overline{x}) + (1+2\overline{x}+\overline{x^2} ) \right\rbrack $

so:

$f(x)= c_1 \frac{ F_0 (1-2x-3x^2) + F_L (-4x) + (1+2x+x^2) }{ F_0 (1-2\overline{x}-3\overline{x^2}) + F_L ( -4\overline{x}) + (1+2\overline{x}+\overline{x^2} ) } \Im(x) $

construct the likelyhood function using $f(x)$

Assume N times of measurements $x_i$

$L=\prod f(x_i) = \prod c_1 \frac{ F_0 (1-2x_i-3x_i^2) + F_L (-4x_i) + (1+2x_i+x_i^2) }{ F_0 (1-2\overline{x}-3\overline{x^2}) + F_L ( -4\overline{x}) + (1+2\overline{x}+\overline{x^2} ) } \Im(x_i) $

$ln L= \sum ln [ F_0 (1-2x_i-3x_i^2) + F_L (-4x_i) + (1+2x_i+x_i^2) ] - N ln [ F_0 (1-2\overline{x}-3\overline{x^2}) + F_L ( -4\overline{x}) + (1+2\overline{x}+\overline{x^2} ) ] + \sum ln \Im(x_i) + \sum ln c_1 $

get maximum likelyhood for parameter $F_0, F_L$

$ 0= \frac{\partial ln L}{\partial F_0} = \sum_{i=0}^N \frac{1-2x_i-3x_i^2} {F_0 (1-2x_i-3x_i^2) + F_L (-4x_i) + (1+2x_i+x_i^2)} - N \frac{ 1-2\overline{x}-3\overline{x^2} }{F_0 (1-2\overline{x}-3\overline{x^2}) + F_L ( -4\overline{x}) + (1+2\overline{x}+\overline{x^2} ) } $

$ 0= \frac{\partial ln L}{\partial F_L} = \sum_{i=0}^N \frac{-4x_i} {F_0 (1-2x_i-3x_i^2) + F_L (-4x_i) + (1+2x_i+x_i^2)} - N \frac{-4\overline{x}}{F_0 (1-2\overline{x}-3\overline{x^2}) + F_L ( -4\overline{x}) + (1+2\overline{x}+\overline{x^2} ) } $

It's function of $\overline x , \overline {x^2}, F_0, F_L$

calculate $\overline x, \overline{x^2}$

$ \Im(x) =\frac{1}{c_1} \frac{ F_0 (1-2\overline{x}-3\overline{x^2}) + F_L (-4\overline{x}) + (1+2\overline{x}+\overline{x^2} ) } { F_0 (1-2x-3x^2) + F_L (-4x) + (1+2x+x^2) } f(x) $

so

$\overline x =\int_{-1}^{1} c_1 x \Im(x) dx = [ F_0 (1-2\overline{x}-3\overline{x^2}) + F_L (-4\overline{x}) + (1+2\overline{x}+\overline{x^2} ) ] c_2 $. where $c_2=\int_{-1}^{1} \frac{ x f(x) dx } { F_0 (1-2x-3x^2) + F_L (-4x) + (1+2x+x^2) }$

$\overline {x^2} =\int_{-1}^{1} c_1 x^2 \Im(x) dx = [ F_0 (1-2\overline{x}-3\overline{x^2}) + F_L (-4\overline{x}) + (1+2\overline{x}+\overline{x^2} ) ] c_3$ where $c_3=\int_{-1}^{1} \frac{ x^2 f(x) dx } { F_0 (1-2x-3x^2) + F_L (-4x) + (1+2x+x^2) } $

$c_2, c_3$ is calcaulated with data using numerical method, for a combination of $F_0, F_L$

$\overline {x^2} = \frac{c_3 \overline x}{c_2} $

$\overline x = \frac{c_2 (F_0+1) }{ 1 +2c_2F_0+4c_2F_L-2c_2+3c_3F_0- c_3 }$

Procedures

  • Scan $\hat F_0,\hat F_L$ from 0 to 1 with constraint $\hat F_0+\hat F_L<=1$
  • For each pair of $\hat F_0,\hat F_L$, calculate $c_2, c_3$,
  • Calculate $\overline x, \overline {x^2} $
  • Calculate $-2ln\frac{L(F_0,F_L)}{L(\hat F_0,\hat F_L)}$, which is $-2 \lbrace \sum ln [ F_0 (1-2x_i-3x_i^2) + F_L (-4x_i) ] - N ln [ F_0 (1-2\overline{x}-3\overline{x^2}) + F_L ( -4\overline{x}) ) ] - \sum ln [ \hat F_0 (1-2x_i-3x_i^2) + \hat F_L (-4x_i) ] + N ln [ \hat F_0 (1-2\overline{x}-3\overline{x^2}) + \hat F_L ( -4\overline{x}) ] \rbrace$ where $F_0, F_L$ are theoretical value
  • Find where the least $-2ln\frac{L(F_0,F_L)}{L(\hat F_0,\hat F_L)}$ locates

-- ChengguangZhu - 2009-10-04

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Topic revision: r4 - 2009-10-06 - ChengguangZhu
 
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