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466 lines
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<TITLE>Scaling Poisson variates</TITLE>
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<H2><A NAME="SECTION00626000000000000000"></A><A NAME="sec:3"></A>
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<BR>
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Scaling Poisson variates
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</H2>
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<P>
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If we have a count value <IMG
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WIDTH="23" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img128.png"
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ALT="$ C_0$">
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that follows a Poisson distribution,
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we can assume immediately that the average is <IMG
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WIDTH="23" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img128.png"
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ALT="$ C_0$">
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and the s.d. is <!-- MATH
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$\sqrt{C_0}$
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-->
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<IMG
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WIDTH="36" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
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SRC="img129.png"
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ALT="$ \sqrt{C_0}$">
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.
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I.e., repeated experiments would give values <IMG
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WIDTH="14" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img130.png"
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ALT="$ n$">
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distributed according to the normalized distribution
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<P><!-- MATH
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\begin{displaymath}
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P(n)={\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{C_0^n{\ensuremath{\mathrm{e}}}^{-C_0}
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}}}}{{\ensuremath{\displaystyle{
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n!}}}}}}}
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\end{displaymath}
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-->
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</P>
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<DIV ALIGN="CENTER">
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<IMG
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WIDTH="118" HEIGHT="58" ALIGN="MIDDLE" BORDER="0"
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SRC="img131.png"
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ALT="$\displaystyle P(n)={\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{C_0^n{\ensuremath{\mathrm{e}}}^{-C_0}
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}}}}{{\ensuremath{\displaystyle{
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n!}}}}}}}
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$">
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</DIV><P>
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</P>
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This obeys
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<P><!-- MATH
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\begin{displaymath}
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\mathop{\sum}_{n=0}^{+\infty}
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P(n)=1\ ;
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\end{displaymath}
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-->
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</P>
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<DIV ALIGN="CENTER">
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<IMG
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WIDTH="104" HEIGHT="65" ALIGN="MIDDLE" BORDER="0"
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SRC="img132.png"
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ALT="$\displaystyle \mathop{\sum}_{n=0}^{+\infty}
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P(n)=1\ ;
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$">
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</DIV><P>
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</P>
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<P><!-- MATH
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\begin{displaymath}
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\langle n\rangle=\mathop{\sum}_{n=0}^{+\infty}
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nP(n)=C_0\ ;
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\end{displaymath}
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-->
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</P>
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<DIV ALIGN="CENTER">
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<IMG
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WIDTH="168" HEIGHT="65" ALIGN="MIDDLE" BORDER="0"
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SRC="img133.png"
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ALT="$\displaystyle \langle n\rangle=\mathop{\sum}_{n=0}^{+\infty}
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nP(n)=C_0\ ;
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$">
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</DIV><P>
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</P>
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<P><!-- MATH
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\begin{displaymath}
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\langle n^2\rangle=\mathop{\sum}_{n=0}^{+\infty}
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n^2 P(n)=C_0^2+C_0\ ;
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\end{displaymath}
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-->
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</P>
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<DIV ALIGN="CENTER">
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<IMG
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WIDTH="221" HEIGHT="65" ALIGN="MIDDLE" BORDER="0"
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SRC="img134.png"
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ALT="$\displaystyle \langle n^2\rangle=\mathop{\sum}_{n=0}^{+\infty}
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n^2 P(n)=C_0^2+C_0\ ;
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$">
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</DIV><P>
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</P>
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<P><!-- MATH
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\begin{displaymath}
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\sigma_{C_0}=\sqrt{\langle n^2\rangle-\langle n\rangle^2}=\sqrt{C_0}
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\end{displaymath}
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-->
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</P>
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<DIV ALIGN="CENTER">
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<IMG
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WIDTH="200" HEIGHT="40" ALIGN="MIDDLE" BORDER="0"
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SRC="img135.png"
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ALT="$\displaystyle \sigma_{C_0}=\sqrt{\langle n^2\rangle-\langle n\rangle^2}=\sqrt{C_0}
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$">
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</DIV><P>
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</P>
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Suppose now that
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our observable is
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<P><!-- MATH
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\begin{displaymath}
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X_0=\eta_0 C_0
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\end{displaymath}
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-->
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</P>
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<DIV ALIGN="CENTER">
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<IMG
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WIDTH="80" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img212.png"
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ALT="$\displaystyle X_0=\eta_0 C_0
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$">
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</DIV><P>
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</P>
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where <IMG
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WIDTH="20" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img213.png"
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ALT="$ \eta_0$">
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is a known error-free scaling factor.
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The distribution of <IMG
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WIDTH="19" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img214.png"
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ALT="$ X$">
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is
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<P><!-- MATH
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\begin{displaymath}
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P'(X)=P(X/\eta_0)=P(n)\qquad\Biggl|\Biggr.\qquad \frac{X}{\eta_0}\equiv n\in\mathbb{Z}
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\end{displaymath}
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-->
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</P>
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<DIV ALIGN="CENTER">
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<IMG
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WIDTH="335" HEIGHT="64" ALIGN="MIDDLE" BORDER="0"
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SRC="img215.png"
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ALT="$\displaystyle P'(X)=P(X/\eta_0)=P(n)\qquad\Biggl\vert\Biggr.\qquad \frac{X}{\eta_0}\equiv n\in\mathbb{Z}
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$">
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</DIV><P>
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</P>
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and now,
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<P><!-- MATH
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\begin{displaymath}
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\mathop{\sum}_{n=0}^{+\infty}
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P(n)=1\ ;
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\end{displaymath}
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-->
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</P>
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<DIV ALIGN="CENTER">
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<IMG
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WIDTH="104" HEIGHT="65" ALIGN="MIDDLE" BORDER="0"
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SRC="img132.png"
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ALT="$\displaystyle \mathop{\sum}_{n=0}^{+\infty}
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P(n)=1\ ;
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$">
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</DIV><P>
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</P>
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<P><!-- MATH
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\begin{displaymath}
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\langle X\rangle=\mathop{\sum}_{n=0}^{+\infty}
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\eta_0 nP(n)=\eta_0 C_0=X_0\ ;
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\end{displaymath}
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-->
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</P>
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<DIV ALIGN="CENTER">
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<IMG
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WIDTH="244" HEIGHT="65" ALIGN="MIDDLE" BORDER="0"
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SRC="img216.png"
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ALT="$\displaystyle \langle X\rangle=\mathop{\sum}_{n=0}^{+\infty}
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\eta_0 nP(n)=\eta_0 C_0=X_0\ ;
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$">
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</DIV><P>
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</P>
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<P><!-- MATH
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\begin{displaymath}
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\langle X^2\rangle=\mathop{\sum}_{n=0}^{+\infty}
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\eta_0^2 n^2 P(n)=\eta_0^2(C_0^2+C_0)=X_0^2+\eta_0 X_0\ ;
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\end{displaymath}
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-->
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</P>
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<DIV ALIGN="CENTER">
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<IMG
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WIDTH="367" HEIGHT="65" ALIGN="MIDDLE" BORDER="0"
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SRC="img217.png"
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ALT="$\displaystyle \langle X^2\rangle=\mathop{\sum}_{n=0}^{+\infty}
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\eta_0^2 n^2 P(n)=\eta_0^2(C_0^2+C_0)=X_0^2+\eta_0 X_0\ ;
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$">
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</DIV><P>
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</P>
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<P><!-- MATH
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\begin{displaymath}
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\sigma_X=\sqrt{\langle X^2\rangle-\langle X\rangle^2}=\sqrt{\eta_0 X_0}=\eta_0\sqrt{C_0}=\sqrt{\eta_0}\sqrt{X_0}
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\end{displaymath}
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-->
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</P>
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<DIV ALIGN="CENTER">
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<IMG
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WIDTH="379" HEIGHT="40" ALIGN="MIDDLE" BORDER="0"
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SRC="img218.png"
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ALT="$\displaystyle \sigma_X=\sqrt{\langle X^2\rangle-\langle X\rangle^2}=\sqrt{\eta_0 X_0}=\eta_0\sqrt{C_0}=\sqrt{\eta_0}\sqrt{X_0}
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$">
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</DIV><P>
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</P>
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Now it is no more valid that <!-- MATH
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$\sigma_X=\sqrt{\langle X\rangle}=\sqrt{X_0}$
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-->
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<IMG
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WIDTH="144" HEIGHT="38" ALIGN="MIDDLE" BORDER="0"
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SRC="img219.png"
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ALT="$ \sigma_X=\sqrt{\langle X\rangle}=\sqrt{X_0}$">
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, instead
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<P><!-- MATH
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\begin{displaymath}
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\sigma_X=\sqrt{\eta_0}\sqrt{X_0}=\eta_0\sqrt{C_0}=\eta_0\sigma_{C_0}
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\end{displaymath}
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-->
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</P>
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<DIV ALIGN="CENTER">
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<IMG
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WIDTH="244" HEIGHT="40" ALIGN="MIDDLE" BORDER="0"
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SRC="img220.png"
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ALT="$\displaystyle \sigma_X=\sqrt{\eta_0}\sqrt{X_0}=\eta_0\sqrt{C_0}=\eta_0\sigma_{C_0}
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$">
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</DIV><P>
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</P>
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that is the characteristic relationship for a normal-variate distribution.
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<P>
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Moreover, assume now that the scaling factor is not exctly known
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but instead it is a normal-variate itself with average <IMG
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WIDTH="20" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img213.png"
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ALT="$ \eta_0$">
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, s.d.
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<!-- MATH
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$\sigma_{\eta_0}$
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-->
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<IMG
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WIDTH="27" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img221.png"
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ALT="$ \sigma_{\eta_0}$">
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, and distribution
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<P><!-- MATH
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\begin{displaymath}
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\widehat{P}(\eta)={\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{
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{\ensuremath{\mathrm{e}}}^{
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-\frac{1}{2}
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{\ensuremath{\left({
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\frac{\eta-\eta_0}{\sigma_{\eta_0}}
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}\right)}}^2
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}
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}}}}{{\ensuremath{\displaystyle{
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\sigma_{\eta_0}\sqrt{2\pi}
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}}}}}}}
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\end{displaymath}
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-->
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</P>
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<DIV ALIGN="CENTER">
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<IMG
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WIDTH="142" HEIGHT="77" ALIGN="MIDDLE" BORDER="0"
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SRC="img222.png"
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ALT="$\displaystyle \widehat{P}(\eta)={\ensuremath{\displaystyle{\frac{{\ensuremath{\...
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...ght)}}^2
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}
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}}}}{{\ensuremath{\displaystyle{
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\sigma_{\eta_0}\sqrt{2\pi}
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}}}}}}}
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$">
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</DIV><P>
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</P>
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Then,
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<P><!-- MATH
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\begin{displaymath}
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\int_{-\infty}^{+\infty}{\ensuremath{\mathrm{d}{\eta}\, }}\mathop{\sum}_{n=0}^{+\infty}
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P(n)\widehat{P}(\eta)=1\ ;
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\end{displaymath}
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-->
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</P>
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<DIV ALIGN="CENTER">
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<IMG
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WIDTH="202" HEIGHT="65" ALIGN="MIDDLE" BORDER="0"
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SRC="img223.png"
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ALT="$\displaystyle \int_{-\infty}^{+\infty}{\ensuremath{\mathrm{d}{\eta}\, }}\mathop{\sum}_{n=0}^{+\infty}
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P(n)\widehat{P}(\eta)=1\ ;
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$">
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</DIV><P>
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</P>
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<P><!-- MATH
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\begin{displaymath}
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\langle X\rangle=\int_{-\infty}^{+\infty}{\ensuremath{\mathrm{d}{\eta}\, }}\mathop{\sum}_{n=0}^{+\infty}
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\widehat{P}(\eta)\eta nP(n)=
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\mathop{\sum}_{n=0}^{+\infty}
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nP(n)
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\int_{-\infty}^{+\infty}{\ensuremath{\mathrm{d}{\eta}\, }} \widehat{P}(\eta)\eta
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=
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\eta_0 C_0=X_0\ ;
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\end{displaymath}
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-->
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</P>
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<DIV ALIGN="CENTER">
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<IMG
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WIDTH="533" HEIGHT="65" ALIGN="MIDDLE" BORDER="0"
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SRC="img224.png"
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ALT="$\displaystyle \langle X\rangle=\int_{-\infty}^{+\infty}{\ensuremath{\mathrm{d}{...
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...}{\ensuremath{\mathrm{d}{\eta}\, }} \widehat{P}(\eta)\eta
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=
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\eta_0 C_0=X_0\ ;
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$">
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</DIV><P>
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</P>
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<P><!-- MATH
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\begin{displaymath}
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\langle X^2\rangle=\int_{-\infty}^{+\infty}{\ensuremath{\mathrm{d}{\eta}\, }}\mathop{\sum}_{n=0}^{+\infty}
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\widehat{P}(\eta)\eta^2 n^2 P(n)=
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\int_{-\infty}^{+\infty}{\ensuremath{\mathrm{d}{\eta}\, }}\widehat{P}(\eta)\eta^2
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\mathop{\sum}_{n=0}^{+\infty}
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n^2 P(n)
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=
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(\eta_0^2+\sigma_{\eta_0}^2)(C_0^2+C_0)\ ;
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\end{displaymath}
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-->
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</P>
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<DIV ALIGN="CENTER">
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<IMG
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WIDTH="631" HEIGHT="65" ALIGN="MIDDLE" BORDER="0"
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SRC="img225.png"
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ALT="$\displaystyle \langle X^2\rangle=\int_{-\infty}^{+\infty}{\ensuremath{\mathrm{d...
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...p{\sum}_{n=0}^{+\infty}
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n^2 P(n)
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=
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(\eta_0^2+\sigma_{\eta_0}^2)(C_0^2+C_0)\ ;
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$">
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</DIV><P>
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</P>
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<P><!-- MATH
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\begin{displaymath}
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{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{\sigma_X}}}}{{\ensuremath{\displaystyle{\langle X\rangle}}}}}}}={\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{\sqrt{\langle X^2\rangle-\langle X\rangle^2}}}}}{{\ensuremath{\displaystyle{\langle X\rangle}}}}}}}
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={\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{\sqrt{
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\eta_0^2 C_0+\sigma_{\eta_0}^2C_0^2+\sigma_{\eta_0}^2 C_0
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}}}}}{{\ensuremath{\displaystyle{\eta_0C_0}}}}}}}=
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\sqrt{
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{\ensuremath{\left({{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{ \sigma_{C_0}}}}}{{\ensuremath{\displaystyle{C_0}}}}}}}}\right)}}^2
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+{\ensuremath{\left({{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{\sigma_{\eta_0}}}}}{{\ensuremath{\displaystyle{\eta_0}}}}}}}}\right)}}^2+{\ensuremath{\left({{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{\sigma_{\eta_0}}}}}{{\ensuremath{\displaystyle{\eta_0}}}}}}}{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{\sigma_{C_0}}}}}{{\ensuremath{\displaystyle{C_0}}}}}}}}\right)}}^2
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}\approx\sqrt{
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{\ensuremath{\left({{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{ \sigma_{C_0}}}}}{{\ensuremath{\displaystyle{C_0}}}}}}}}\right)}}^2
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+{\ensuremath{\left({{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{\sigma_{\eta_0}}}}}{{\ensuremath{\displaystyle{\eta_0}}}}}}}}\right)}}^2
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}
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\end{displaymath}
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-->
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</P>
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<DIV ALIGN="CENTER">
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<IMG
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WIDTH="812" HEIGHT="79" ALIGN="MIDDLE" BORDER="0"
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SRC="img226.png"
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ALT="$\displaystyle {\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{\sigm...
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...yle{\sigma_{\eta_0}}}}}{{\ensuremath{\displaystyle{\eta_0}}}}}}}}\right)}}^2
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}
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$">
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</DIV><P>
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</P>
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where in the last we discard, as usual, the 4th order in the relative errors. Both the exact and approximated forms
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are exactly the same as if both distributions were to be normal.
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<P>
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<HR>
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Thattil Dhanya
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2018-09-28
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