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<B> Next:</B> <A NAME="tex2html946"
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HREF="node62.html">Numerical comparison of averages</A>
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<B> Up:</B> <A NAME="tex2html942"
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HREF="node54.html">Average vs. weighted average</A>
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<B> Previous:</B> <A NAME="tex2html936"
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HREF="node60.html">Comparison</A>
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<H3><A NAME="SECTION00625700000000000000">
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Analytical comparison of averages</A>
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</H3>
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<P>
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First we give an analytical comparison between simple average and Mighell-Poisson weighted average
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for <!-- MATH
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$N_{\mathrm{obs}}=2$
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-->
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<IMG
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WIDTH="66" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img175.png"
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ALT="$ N_{\mathrm{obs}}=2$">
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.
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If the two events are <IMG
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WIDTH="23" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img176.png"
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ALT="$ C_1$">
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and <IMG
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WIDTH="23" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img177.png"
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ALT="$ C_2$">
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, then
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<P><!-- MATH
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\begin{displaymath}
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\langle x \rangle={\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{C_1+C_2}}}}{{\ensuremath{\displaystyle{2}}}}}}}; \qquad \sigma_x={\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{\sqrt{C_1+C_2}}}}}{{\ensuremath{\displaystyle{2}}}}}}}
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\end{displaymath}
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-->
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</P>
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<DIV ALIGN="CENTER">
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<IMG
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WIDTH="261" HEIGHT="63" ALIGN="MIDDLE" BORDER="0"
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SRC="img178.png"
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ALT="$\displaystyle \langle x \rangle={\ensuremath{\displaystyle{\frac{{\ensuremath{\...
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...nsuremath{\displaystyle{\sqrt{C_1+C_2}}}}}{{\ensuremath{\displaystyle{2}}}}}}}
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$">
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</DIV><P>
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</P>
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For the M-P weighted average,
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<P><!-- MATH
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\begin{displaymath}
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\langle x \rangle_{\mathrm{w(2)}}={\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{2(C_1+1)(C_2+1)}}}}{{\ensuremath{\displaystyle{C_1+C_2+2}}}}}}}; \qquad
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\sigma_{\langle x \rangle_{\mathrm{w(2)}}}=\sqrt{{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{(C_1+1)(C_2+1)}}}}{{\ensuremath{\displaystyle{C_1+C_2+2}}}}}}}}
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\end{displaymath}
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-->
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</P>
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<DIV ALIGN="CENTER">
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<IMG
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WIDTH="448" HEIGHT="71" ALIGN="MIDDLE" BORDER="0"
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SRC="img179.png"
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ALT="$\displaystyle \langle x \rangle_{\mathrm{w(2)}}={\ensuremath{\displaystyle{\fra...
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...{\displaystyle{(C_1+1)(C_2+1)}}}}{{\ensuremath{\displaystyle{C_1+C_2+2}}}}}}}}
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$">
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</DIV><P>
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</P>
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<P>
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Now, supposing that the common 'true' value of <IMG
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WIDTH="49" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img180.png"
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ALT="$ C_1,C_2$">
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is <IMG
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WIDTH="14" HEIGHT="15" ALIGN="BOTTOM" BORDER="0"
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SRC="img181.png"
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ALT="$ \lambda$">
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,
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we use the Poisson distribution to compare the expectation values of the two results. The expectation value of the simple average is
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<P><!-- MATH
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\begin{displaymath}
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E{\ensuremath{\left({\langle x \rangle}\right)}} = \mathop{\sum}_{m,n=0}^{+\infty}
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{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{n+m}}}}{{\ensuremath{\displaystyle{2}}}}}}}P(n)P(m)=\mathop{\sum}_{m=0}^{+\infty}
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{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{m}}}}{{\ensuremath{\displaystyle{2}}}}}}}{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{\lambda^m{\ensuremath{\mathrm{e}}}^{-\lambda}}}}}{{\ensuremath{\displaystyle{m!}}}}}}}
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+\mathop{\sum}_{n=0}^{+\infty}
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{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{n}}}}{{\ensuremath{\displaystyle{2}}}}}}}{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{\lambda^n{\ensuremath{\mathrm{e}}}^{-\lambda}}}}}{{\ensuremath{\displaystyle{n!}}}}}}}
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=\lambda
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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="491" HEIGHT="65" ALIGN="MIDDLE" BORDER="0"
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SRC="img182.png"
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ALT="$\displaystyle E{\ensuremath{\left({\langle x \rangle}\right)}} = \mathop{\sum}_...
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...ath{\mathrm{e}}}^{-\lambda}}}}}{{\ensuremath{\displaystyle{n!}}}}}}}
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=\lambda
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$">
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</DIV><P>
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</P>
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As expected, the simple average gives the true value.
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For its variance,
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<P><!-- MATH
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\begin{displaymath}
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E{\ensuremath{\left({\sigma_x^2}\right)}} = \mathop{\sum}_{m,n=0}^{+\infty}
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{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{n+m}}}}{{\ensuremath{\displaystyle{4}}}}}}}P(n)P(m)={\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{
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\lambda}}}}{{\ensuremath{\displaystyle{2}}}}}}};\qquad E{\ensuremath{\left({\sigma_x}\right)}} =\sqrt{{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{\lambda}}}}{{\ensuremath{\displaystyle{2}}}}}}}}
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\end{displaymath}
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-->
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</P>
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<DIV ALIGN="CENTER">
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<IMG
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WIDTH="397" HEIGHT="65" ALIGN="MIDDLE" BORDER="0"
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SRC="img183.png"
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ALT="$\displaystyle E{\ensuremath{\left({\sigma_x^2}\right)}} = \mathop{\sum}_{m,n=0}...
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...ac{{\ensuremath{\displaystyle{\lambda}}}}{{\ensuremath{\displaystyle{2}}}}}}}}
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$">
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</DIV><P>
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</P>
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<P>
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In order to evaluate the difference with the M-P weighted average, we rewrite the latter as
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<P><!-- MATH
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\begin{displaymath}
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\langle x \rangle_{\mathrm{w(2)}}=\langle x \rangle + 1 -{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{(C_1-C_2)^2}}}}{{\ensuremath{\displaystyle{4(\langle x \rangle+1)}}}}}}}
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\end{displaymath}
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-->
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</P>
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<DIV ALIGN="CENTER">
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<IMG
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WIDTH="222" HEIGHT="57" ALIGN="MIDDLE" BORDER="0"
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SRC="img184.png"
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ALT="$\displaystyle \langle x \rangle_{\mathrm{w(2)}}=\langle x \rangle + 1 -{\ensure...
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...style{(C_1-C_2)^2}}}}{{\ensuremath{\displaystyle{4(\langle x \rangle+1)}}}}}}}
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$">
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</DIV><P>
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</P>
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and calculate the expectation value of the last term:
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<P><!-- MATH
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\begin{displaymath}
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E{\ensuremath{\left({{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{(C_1-C_2)^2}}}}{{\ensuremath{\displaystyle{4(\langle x \rangle+1)}}}}}}}}\right)}} =
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\mathop{\sum}_{m,n=0}^{+\infty}
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{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{(n-m)^2}}}}{{\ensuremath{\displaystyle{2(n+m+2) }}}}}}}P(n)P(m)={\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{{\ensuremath{\mathrm{e}}}^{-2\lambda}}}}}{{\ensuremath{\displaystyle{2}}}}}}}
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\mathop{\sum}_{m,n=0}^{+\infty}
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{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{(n-m)^2}}}}{{\ensuremath{\displaystyle{(n+m+2) }}}}}}}{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{\lambda^{n+m}}}}}{{\ensuremath{\displaystyle{n!m!}}}}}}}
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\end{displaymath}
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-->
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</P>
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<DIV ALIGN="CENTER">
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<IMG
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WIDTH="578" HEIGHT="65" ALIGN="MIDDLE" BORDER="0"
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SRC="img185.png"
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ALT="$\displaystyle E{\ensuremath{\left({{\ensuremath{\displaystyle{\frac{{\ensuremat...
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...uremath{\displaystyle{\lambda^{n+m}}}}}{{\ensuremath{\displaystyle{n!m!}}}}}}}
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$">
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</DIV><P>
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</P>
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Rearranging the sums with <IMG
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WIDTH="76" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img186.png"
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ALT="$ s=n+m$">
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, <!-- MATH
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$s=0\ldots +\infty$
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-->
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<IMG
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WIDTH="98" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img187.png"
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ALT="$ s=0\ldots +\infty$">
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; <IMG
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WIDTH="112" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img188.png"
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ALT="$ n-m=s-2k$">
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, <!-- MATH
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$k=0\ldots s$
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-->
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<IMG
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WIDTH="74" HEIGHT="15" ALIGN="BOTTOM" BORDER="0"
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SRC="img189.png"
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ALT="$ k=0\ldots s$">
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,
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we get
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<P><!-- MATH
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\begin{displaymath}
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E{\ensuremath{\left({{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{(C_1-C_2)^2}}}}{{\ensuremath{\displaystyle{4(\langle x \rangle+1)}}}}}}}}\right)}} =
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{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{{\ensuremath{\mathrm{e}}}^{-2\lambda}}}}}{{\ensuremath{\displaystyle{2}}}}}}}
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\mathop{\sum}_{s=0}^{+\infty}
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\mathop{\sum}_{k=0}^{s}
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{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{(s-2k)^2(s+1)}}}}{{\ensuremath{\displaystyle{(s+2)! }}}}}}}{\lambda^{s}}
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\binom{s}{k}={\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{1}}}}{{\ensuremath{\displaystyle{2}}}}}}}-{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{1}}}}{{\ensuremath{\displaystyle{2\lambda}}}}}}}+{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{1-{\ensuremath{\mathrm{e}}}^{-2\lambda}}}}}{{\ensuremath{\displaystyle{4\lambda^2}}}}}}}
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%{n!m!}
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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="550" HEIGHT="65" ALIGN="MIDDLE" BORDER="0"
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SRC="img190.png"
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ALT="$\displaystyle E{\ensuremath{\left({{\ensuremath{\displaystyle{\frac{{\ensuremat...
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...hrm{e}}}^{-2\lambda}}}}}{{\ensuremath{\displaystyle{4\lambda^2}}}}}}}
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%{n!m!}
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$">
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</DIV><P>
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</P>
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So, the relative difference between averages is
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<P><!-- MATH
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\begin{displaymath}
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{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{E{\ensuremath{\left({\langle x \rangle_{\mathrm{w(2)}}-\langle x \rangle}\right)}}}}}}{{\ensuremath{\displaystyle{E{\ensuremath{\left({\langle x \rangle}\right)}}}}}}}}}=
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{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{1}}}}{{\ensuremath{\displaystyle{2\lambda}}}}}}}+{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{1}}}}{{\ensuremath{\displaystyle{2\lambda^2}}}}}}}-{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{1-{\ensuremath{\mathrm{e}}}^{-2\lambda}}}}}{{\ensuremath{\displaystyle{4\lambda^3}}}}}}}
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\end{displaymath}
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-->
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</P>
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<DIV ALIGN="CENTER">
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<IMG
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WIDTH="296" HEIGHT="59" ALIGN="MIDDLE" BORDER="0"
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SRC="img191.png"
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ALT="$\displaystyle {\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{E{\en...
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...math{\mathrm{e}}}^{-2\lambda}}}}}{{\ensuremath{\displaystyle{4\lambda^3}}}}}}}
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$">
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</DIV><P>
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</P>
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The relative error on <!-- MATH
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$\langle x \rangle$
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-->
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<IMG
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WIDTH="26" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img192.png"
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ALT="$ \langle x \rangle$">
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is
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<P><!-- MATH
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\begin{displaymath}
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\epsilon = {\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{\sigma_x}}}}{{\ensuremath{\displaystyle{\langle x \rangle}}}}}}} =
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{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{\lambda^{1/2}}}}}{{\ensuremath{\displaystyle{\sqrt{2} \lambda}}}}}}}={\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{1}}}}{{\ensuremath{\displaystyle{\sqrt{2\lambda}}}}}}}}
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\end{displaymath}
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-->
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</P>
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<DIV ALIGN="CENTER">
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<IMG
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WIDTH="169" HEIGHT="61" ALIGN="MIDDLE" BORDER="0"
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SRC="img193.png"
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ALT="$\displaystyle \epsilon = {\ensuremath{\displaystyle{\frac{{\ensuremath{\display...
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...suremath{\displaystyle{1}}}}{{\ensuremath{\displaystyle{\sqrt{2\lambda}}}}}}}}
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$">
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</DIV><P>
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</P>
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therefore
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<P><!-- MATH
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\begin{displaymath}
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{\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{E{\ensuremath{\left({\langle x \rangle_{\mathrm{w(2)}}-\langle x \rangle}\right)}}}}}}{{\ensuremath{\displaystyle{E{\ensuremath{\left({\langle x \rangle}\right)}}}}}}}}}=
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O(\epsilon^2)
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\end{displaymath}
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-->
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</P>
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<DIV ALIGN="CENTER">
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<IMG
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WIDTH="184" HEIGHT="59" ALIGN="MIDDLE" BORDER="0"
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SRC="img194.png"
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ALT="$\displaystyle {\ensuremath{\displaystyle{\frac{{\ensuremath{\displaystyle{E{\en...
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...aystyle{E{\ensuremath{\left({\langle x \rangle}\right)}}}}}}}}}=
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O(\epsilon^2)
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$">
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</DIV><P>
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</P>
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Therefore, the expectation value of the error (relative) involved in taking
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the M-P weighted average instead of the simple average is negligible.
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<ADDRESS>
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Thattil Dhanya
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2018-09-28
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