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Introduction to turbulence/Statistical analysis/Generalization to the estimator of any quantity

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Let <math>f=(x-X)^2</math> in equation 2.55 so that <math>F_{N}= var_{N}\left\{ x \right\}</math>, <math>\left\langle f \right\rangle = var \left\{ x \right\} </math> and <math>var \left\{f \right\} = var \left\{ \left( x-X \right)^{2} - var \left[ x-X \right] \right\}</math>. Then:
Let <math>f=(x-X)^2</math> in equation 2.55 so that <math>F_{N}= var_{N}\left\{ x \right\}</math>, <math>\left\langle f \right\rangle = var \left\{ x \right\} </math> and <math>var \left\{f \right\} = var \left\{ \left( x-X \right)^{2} - var \left[ x-X \right] \right\}</math>. Then:
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<table width="100%"><tr><td> 
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:<math>   
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\epsilon^{2}_{F_{N}}= \frac{1}{N} \frac{var \left\{ \left( x-X \right)^{2} - var \left[x \right] \right\} }{ \left( var \left\{ x \right\} \right)^{2}
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</math> 
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</td><td width="5%">(2)</td></tr></table>
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This is easiest to understand if we first expand only the numerator to oblain:
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<table width="100%"><tr><td> 
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:<math>   
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var \left\{ \left( x- X \right)^{2} - var\left[x \right] \right\} = \left\langle \left( x- X \right)^{4} \right\rangle  - \left[ var \left\{ x \right\} \right]^2
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</math> 
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</td><td width="5%">(2)</td></tr></table>
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Thus
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<table width="100%"><tr><td> 
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:<math>   
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\epsilon^{2}_{var_{N}} = \frac{\left\langle \left( x- X \right)^4 \right\rangle}{\left[ var \left\{ x \right\} \right]^2 } - 1
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</math> 
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</td><td width="5%">(2)</td></tr></table>

Revision as of 11:37, 10 June 2006

Similar relations can be formed for the estimator of any function of the random variable say f(x). For example, an estimator for the average of f based on N realizations is given by:

    
F_{N}\equiv\frac{1}{N}\sum^{N}_{n=1}f_{n}
(2)

where f_{n}\equiv f(x_{n}). It is straightforward to show that this estimator is unbiased, and its variability (squared) is given by:

    
\epsilon^{2}_{F_{N}}= \frac{1}{N} \frac{var \left\{f \left( x \right) \right\}}{\left\langle f \left( x \right) \right\rangle^{2} }
(2)

Example: Suppose it is desired to estimate the variability of an estimator for the variance based on a finite number of samples as:

    
var_{N} \left\{x \right\} \equiv \frac{1}{N} \sum^{N}_{n=1} \left( x_{n} - X \right)^{2}
(2)

(Note that this estimator is not really very useful since it presumes that the mean value, X, is known, whereas in fact usually only X_{N} is obtainable).

Answer

Let f=(x-X)^2 in equation 2.55 so that F_{N}= var_{N}\left\{ x \right\}, \left\langle f \right\rangle = var \left\{ x \right\} and var \left\{f \right\} = var \left\{ \left( x-X \right)^{2} - var \left[ x-X \right] \right\}. Then:

Failed to parse (syntax error): \epsilon^{2}_{F_{N}}= \frac{1}{N} \frac{var \left\{ \left( x-X \right)^{2} - var \left[x \right] \right\} }{ \left( var \left\{ x \right\} \right)^{2}
(2)

This is easiest to understand if we first expand only the numerator to oblain:

    
var \left\{ \left( x- X \right)^{2} - var\left[x \right] \right\} = \left\langle \left( x- X \right)^{4} \right\rangle  - \left[ var \left\{ x \right\} \right]^2
(2)

Thus

    
\epsilon^{2}_{var_{N}} = \frac{\left\langle \left( x- X \right)^4 \right\rangle}{\left[ var \left\{ x \right\} \right]^2 } - 1
(2)
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