Correction for finite samples.

The quantity $\left\{I\left(\vec{x}\right)\right\}$ should ideally be evaluated by averaging infinite images. We obtain a good evaluation of it by averaging a great number $N$ of images $I_n\left(\vec{x}\right)$, typically one hundred:

$$\bar{I}\left(\vec{x}\right) = \frac{1}{N} \sum{I_n\left(\vec{x}\right)} \approx \left\{I\left(\vec{x}\right)\right\}$$ (6.8)

From this evaluation, we obtain $i\left(\vec{x}\right)$:

$$i_n\left(\vec{x}\right) = \frac{I_n\left(\vec{x}\right) - \bar{I}\left(\vec{x}\right)} {\left<\bar{I}\right> }$$ (6.9)

The average value $\bar{I}\left(\vec{x}\right)$, evaluated over a given number of images, is sistematically different from the true mean value, in the direction that reduces the evaluation of the root mean square displacement from the mean. This problem is analogous to the one that leads to the so called Bessel correction for the evaluation of the variance $\sigma$ of a stochastic variable, from the knowledge of a finite number of stochastic values.

We evaluate the correlation function of $i_n\left(\vec{x}\right)$ for each $n$, then we average them, thus obtaining $C_i\left(\Delta \vec{x}\right)$. Now we want to evaluate $\left\{C_i\left(\Delta \vec{x}\right)\right\}$, that is the mean value over infinite samples, in order to correct systematic errors:

$$\left\{C_i\left(\Delta \vec{x}\right)\right\} = \frac{1}{\left<\b... ...}\sum_{m=0}^N {I_m\left(\vec{x}+\Delta \vec{x}\right)}\right] \right>} \right\}$$ (6.10)

The symbol $\left<\cdot\right>$ means the average over $\vec{x}$. We can write $\bar{I}\left(\vec{x}\right) + \delta I\left(\vec{x}\right)$ instead of $I_n\left(\vec{x}\right)$:

$$\left\{C_i\left(\Delta \vec{x}\right)\right\} = \frac{1}{\left<\b... ...m=0}^N {\delta I_m\left(\vec{x}+\Delta \vec{x}\right)}\right] \right>} \right\}$$ (6.11)

Evaluating the products:

$$\left\{C_i\left(\Delta \vec{x}\right)\right\} = \frac{1}{\left<\bar{I}\right>^2} \frac{1}{N} \sum_{n=0}^N {\left\... ...vec{x}\right) \delta I_n\left(\vec{x}+\Delta \vec{x}\right) \right>\right\} } -$$ (6.12)

$$\frac{1}{\left<\bar{I}\right>^2} \frac{1}{N^2} \sum_{n,m=0}^N {\l... ...vec{x}\right) \delta I_n\left(\vec{x}+\Delta \vec{x}\right) \right>\right\} } +$$ (6.13)

$$\frac{1}{\left<\bar{I}\right>^2} \frac{1}{N^3} \sum_{n,m,l=0}^N {... ...(\vec{x}\right) \delta I_l\left(\vec{x}+\Delta \vec{x}\right) \right>\right\} }$$ (6.14)

Since $\bar{\delta I}=0$:

$$\left\{C_i\left(\Delta \vec{x}\right)\right\} = \frac{N-1}{N} \fr... ...left(\vec{x}\right) \delta I\left(\vec{x}+\Delta \vec{x}\right) \right>\right\}$$ (6.15)

Now we can use Eq. (6.5):

$$\left\{C_i\left(\Delta \vec{x}\right)\right\} = \frac{N-1}{N}C_E\left(\Delta \vec{x}\right)$$ (6.16)

The correlation function evaluated on $N$ samples is proportional to the correlation function evaluated for $N\to \infty$. The poportionality constant is the same of the well known Bessel correction.

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