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<H2><A NAME="SECTION00624000000000000000">
Poisson and normal statistics for diffraction</A>
</H2>
<P>
The normal situation for diffraction data
is that the observed signal is a photon count.
Therefore it follows a Poisson distribution.
If we have a count value <IMG
WIDTH="23" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
SRC="img128.png"
ALT="$ C_0$">
that follows a Poisson distribution,
we can assume immediately that the average is equal to <IMG
WIDTH="23" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
SRC="img128.png"
ALT="$ C_0$">
and the s.d. is <!-- MATH
$\sqrt{C_0}$
-->
<IMG
WIDTH="36" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
SRC="img129.png"
ALT="$ \sqrt{C_0}$">
.
I.e., repeated experiments would give values <IMG
WIDTH="14" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img130.png"
ALT="$ n$">
distributed according to the normalized distribution
<P><!-- MATH
\begin{displaymath}
P(n)={\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{C_0^n{\ensuremath{\mathrm{e}}}^{-C_0}
}}}}{{\ensuremath{\displaystyle{
n!}}}}}}}
\end{displaymath}
-->
</P>
<DIV ALIGN="CENTER">
<IMG
WIDTH="118" HEIGHT="58" ALIGN="MIDDLE" BORDER="0"
SRC="img131.png"
ALT="$\displaystyle P(n)={\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{C_0^n{\ensuremath{\mathrm{e}}}^{-C_0}
}}}}{{\ensuremath{\displaystyle{
n!}}}}}}}
$">
</DIV><P>
</P>
This obeys
<P><!-- MATH
\begin{displaymath}
\mathop{\sum}_{n=0}^{+\infty}
P(n)=1\ ;
\end{displaymath}
-->
</P>
<DIV ALIGN="CENTER">
<IMG
WIDTH="104" HEIGHT="65" ALIGN="MIDDLE" BORDER="0"
SRC="img132.png"
ALT="$\displaystyle \mathop{\sum}_{n=0}^{+\infty}
P(n)=1\ ;
$">
</DIV><P>
</P>
<P><!-- MATH
\begin{displaymath}
\langle n\rangle=\mathop{\sum}_{n=0}^{+\infty}
nP(n)=C_0\ ;
\end{displaymath}
-->
</P>
<DIV ALIGN="CENTER">
<IMG
WIDTH="168" HEIGHT="65" ALIGN="MIDDLE" BORDER="0"
SRC="img133.png"
ALT="$\displaystyle \langle n\rangle=\mathop{\sum}_{n=0}^{+\infty}
nP(n)=C_0\ ;
$">
</DIV><P>
</P>
<P><!-- MATH
\begin{displaymath}
\langle n^2\rangle=\mathop{\sum}_{n=0}^{+\infty}
n^2 P(n)=C_0^2+C_0\ ;
\end{displaymath}
-->
</P>
<DIV ALIGN="CENTER">
<IMG
WIDTH="221" HEIGHT="65" ALIGN="MIDDLE" BORDER="0"
SRC="img134.png"
ALT="$\displaystyle \langle n^2\rangle=\mathop{\sum}_{n=0}^{+\infty}
n^2 P(n)=C_0^2+C_0\ ;
$">
</DIV><P>
</P>
The standard deviation comes then to
<P><!-- MATH
\begin{displaymath}
\sigma_{C_0}=\sqrt{\langle n^2\rangle-\langle n\rangle^2}=\sqrt{C_0}
\end{displaymath}
-->
</P>
<DIV ALIGN="CENTER">
<IMG
WIDTH="200" HEIGHT="40" ALIGN="MIDDLE" BORDER="0"
SRC="img135.png"
ALT="$\displaystyle \sigma_{C_0}=\sqrt{\langle n^2\rangle-\langle n\rangle^2}=\sqrt{C_0}
$">
</DIV><P>
</P>
<P>
When the data have to be analyzed, one must compare observations with a model
which gives calculated values of the observations in dependence of a certain set of
parameters. The best values of the parameters (the target of investigation)
are the one that maximize the likelihood function [4,5]. The likelihood function for
Poisson variates is pretty difficult to use; furthermore, even simple data manipulations
are not straightforward with Poisson variates (see Sec.&nbsp;<A HREF="node63.html#sec:3">5.2.6</A>). The common choice is to approximate
Poisson variates with normal variates, and then use the much easier formalism
of normal distribution to a) do basic data manipulations and b) fit data with model.
To the latter task, in fact, the likelihood function is maximized simply by minimizing
the usual weighted-<IMG
WIDTH="22" HEIGHT="34" ALIGN="MIDDLE" BORDER="0"
SRC="img1.png"
ALT="$ \chi ^2$">
[4] :
<P><!-- MATH
\begin{displaymath}
\chi^2 = \mathop{\sum}_{j=1}^{N_{\mathrm{obs}}}
{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{{\ensuremath{\left({F_j-O_j}\right)}}^2
}}}}{{\ensuremath{\displaystyle{
\sigma_j^2
}}}}}}}
\end{displaymath}
-->
</P>
<DIV ALIGN="CENTER">
<IMG
WIDTH="151" HEIGHT="67" ALIGN="MIDDLE" BORDER="0"
SRC="img136.png"
ALT="$\displaystyle \chi^2 = \mathop{\sum}_{j=1}^{N_{\mathrm{obs}}}
{\ensuremath{\dis...
...\left({F_j-O_j}\right)}}^2
}}}}{{\ensuremath{\displaystyle{
\sigma_j^2
}}}}}}}
$">
</DIV><P>
</P>
where <IMG
WIDTH="23" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
SRC="img137.png"
ALT="$ O_j$">
are the experimentally observed values, <IMG
WIDTH="21" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
SRC="img138.png"
ALT="$ F_j$">
the calculated model values,
<IMG
WIDTH="20" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
SRC="img139.png"
ALT="$ \sigma_j$">
the s.d.s of the observations.
<P>
Substituting directly the counts (and derived s.d.s) for the observations in the former :
<P><!-- MATH
\begin{displaymath}
\chi_{(0)}^2 = \mathop{\sum}_{j=1}^{N_{\mathrm{obs}}}
{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{{\ensuremath{\left({F_j-C_j}\right)}}^2
}}}}{{\ensuremath{\displaystyle{
C_j
}}}}}}}
\end{displaymath}
-->
</P>
<DIV ALIGN="CENTER">
<IMG
WIDTH="160" HEIGHT="67" ALIGN="MIDDLE" BORDER="0"
SRC="img140.png"
ALT="$\displaystyle \chi_{(0)}^2 = \mathop{\sum}_{j=1}^{N_{\mathrm{obs}}}
{\ensuremat...
...remath{\left({F_j-C_j}\right)}}^2
}}}}{{\ensuremath{\displaystyle{
C_j
}}}}}}}
$">
</DIV><P>
</P>
is the most common way. It is <I>slightly</I> wrong to do so, however [6],
the error being large only when the counts are low.
There is also a divergence for zero counts.
In fact, a slightly modified form [6] exists, reading
<P><!-- MATH
\begin{displaymath}
\chi_{(1)}^2 = \mathop{\sum}_{j=1}^{N_{\mathrm{obs}}}
{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{{\ensuremath{\left({F_j-{\ensuremath{\left({C_j+\min{\ensuremath{\left({1,C_j}\right)}}}\right)}}}\right)}}^2
}}}}{{\ensuremath{\displaystyle{
C_j+1
}}}}}}}
\end{displaymath}
-->
</P>
<DIV ALIGN="CENTER">
<IMG
WIDTH="266" HEIGHT="67" ALIGN="MIDDLE" BORDER="0"
SRC="img141.png"
ALT="$\displaystyle \chi_{(1)}^2 = \mathop{\sum}_{j=1}^{N_{\mathrm{obs}}}
{\ensuremat...
...\right)}}}\right)}}}\right)}}^2
}}}}{{\ensuremath{\displaystyle{
C_j+1
}}}}}}}
$">
</DIV><P>
</P>
Minimizing this form of <IMG
WIDTH="22" HEIGHT="34" ALIGN="MIDDLE" BORDER="0"
SRC="img1.png"
ALT="$ \chi ^2$">
is equivalent - to an exceptionally good approximation [6]-
to maximizing the proper Poisson-likelihood.
<P>
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<ADDRESS>
Thattil Dhanya
2018-09-28
</ADDRESS>
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