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\input macro.tex
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\line{\sevenrm a06.tex[162,phy] \today\hfill}

\font\rmn=cmr9
\bigskip
\line{\bf Factorial Moments\hfill}

The $n↑{\rm th}$ moment $M↓n(d)$, of a distribution~$d$, is
$\sum↓{i=0}↑∞i↑nd(i)$, a~weighted sum of~$d$ with weights that are
monomials in~$i$. Analogously, using the factorial polynomials
$i↑{\overline n}=i(i+1)(i+2)\ldots (i-n+1)$ and $i↑{\underline n}=
i(i-1)(i-2)\ldots (i-n+1)$ as weights, we get the corresponding
{\it factorial moments\/} of a distribution:
$$\eqalign{M↓{\overline n}(d)&=\sum↓{i=0}↑∞i↑{\overline n}d(i)\cr
\noalign{\medskip}
M↓{\underline n}(d)&=\sum↓{i=0}↑∞i↑{\underline n}d(i)\,.\cr}$$
The linear relations among the several kinds of polynomials carry over to the
several kinds of moments:
$$\vcenter{\halign{%
$\hfil#\;$&$#\hfil$\qquad&%
$\hfil#\;$&$#\hfil$\qquad&%
$\hfil#\;$&$#\hfil$\cr
i↑{\overline 0}&=1=i↑0&M↓{\overline 0}&=M↓0=1&M↓0&=M↓{\overline 0}=1\cr
\noalign{\smallskip}
i↑{\overline 1}&=i↑1&M↓{\overline 1}&=M↓1=E(d)&M↓1&=M↓{\overline 1}\cr
\noalign{\smallskip}
i↑{\overline 2}&=i↑2+i↑1&M↓{\overline 2}&=M↓2+M↓1&M↓2&=M↓{\overline 2}-M↓{\overline 1}\cr
\noalign{\smallskip}
i↑{\overline 3}&=i↑3+3i↑2+2i&M↓{\overline 3}&=M↓3+3M↓2+2M↓1&M↓3&=M↓{\overline 3}-3M↓{\overline 2}%
+M↓{\overline 1}\cr
\noalign{\bigskip\bigskip}
i↑{\underline 0}&=1=i↑0&M↓{\underline 0}&=M↓0=1&M↓0&=M↓{\underline 0}=1\cr
\noalign{\smallskip}
i↑{\underline 1}&=i↑1&M↓{\underline 1}&=M↓1=E(d)&M↓1&=M↓{\underline 1}\cr
\noalign{\smallskip}
i↑{\underline 2}&=i↑2-i↑1&M↓{\underline 2}&=M↓2-M↓1&M↓2&=M↓{\underline 2}+
M↓{\underline 1}\cr
\noalign{\smallskip}
i↑{\underline 3}&=i↑3-3i↑2+2i&M↓{\underline 3}&=M↓3-3M↓2+2M↓1%
&M↓3&=M↓{\underline 3}+3M↓{\underline 2}+M↓{\underline 1}\cr}}$$

$$V(d)=M↓2-M↓1↑2=M↓{\overline 2}-M↓{\overline 1}-M↓{\overline 1}↑2=M↓{\underline 2}%
+M↓{\underline 1}-M↓{\underline 1}↑2$$

\vfill\eject
%\medskip

{\rmn
{\narrower\smallskip\noindent
{\bf Example.}

Binomial distribution, $d(i)={n\choose i}p↑iq↑{n-i}$, $q=1-p$. 
$$\eqalign{M↓{\underline j}(d)&=\sum↓ii↑{\underline j}\,
{n!\over i!\,(n-i)!}\,p↑iq↑{n-i}\cr
\noalign{\medskip}
&=\sum↓in↑{\underline j}{n-j\choose i-j}p↑jp↑{i-j}q↑{(n-j)-(i-j)}\cr
\noalign{\medskip}
&=n↑{\underline j}p↑j\sum↓i{n-j\choose i-j}p↑{i-j}q↑{(n-j)-(i-j)}\cr
\noalign{\medskip}
&=n↑{\underline j}p↑j\sum↓id'(i)\cr
\noalign{\medskip}
&=n↑{\underline j}p↑j\cr}$$
where $d'$ is also a binomial distribution.
$$V(d)=M↓{\underline 2}+M↓{\underline 1}
-M↓{\underline 1}↑2=n(n-1)p↑2+np-n↑2p↑2=npq\,.$$
For such distributions, one easily finds all desired factorial moments, then
uses Stirling numbers to get monomial moments.
\smallskip}

\medskip
{\narrower\smallskip\noindent
{\bf Example.}

For the mean and variance of the hypergeometric distribution:

Take $w$ white balls and $b$ black ones in an urn. Draw $k$ of them.
The chance of drawing $i$~white ones is
${w\choose i}{b\choose k-i}/{w+b\choose k}$, so
$$\eqalign{M↓{\underline j}(d)&=\sum↓i{i↑{\underline j}{w\choose i}
{b\choose k-i}\over {w+b\choose k}}\cr
\noalign{\medskip}
&=\sum↓i{w↑{\underline j}{w-j\choose i-j}{b\choose k-i}\over
{w+b\choose k}}\cr
\noalign{\medskip}
&={w↑{\underline{j}}{w+b-j\choose k-j}\over {w+b\choose k}}\cr
%&=\sum↓i{w↑{\underline j}{w-j\choose i-j}{b\choose (k-j)-(i-j)}\over
%{(w+b)↑{\underline j}\over k↑{\underline j}}{w-j+b\choose k-j}}\cr
%\noalign{\medskip}
%&={w↑{\underline j}k↑{\underline j}\over (w+b)↑{\underline j}}\sum↓id'(i)\cr
\noalign{\medskip}
&={w↑{\underline j}k↑{\underline j}\over (w+b)↑{\underline j}}\cr}$$
%where $d'$ is also a hypergeometric distribution.

\vfill\eject

$$\eqalign{M↓1(d)&=M↓{\underline 1}(d)={wk\over w+b}\cr
\noalign{\medskip}
M↓2(d)&=M↓{\underline 2}(d)+M↓{\underline 1}(d)={wk\over w+b}
\left({(w-1)(k-1)\over (w+b-1)}+1\right)\cr
\noalign{\medskip}
V(d)&=M↓2(d)-M↓1(d)↑2\cr
\noalign{\medskip}
&={w\,b\,k(w+b-k)\over (w+b)↑2(w+b-1)}\cr}$$


\medskip\noindent
{\bf Example.}

Let an urn contain $a+1$ balls, of which $b+1$ are black, $b-a$ white.
There are ${a+1\choose b+1}$ orders in which the balls can be drawn.
Let $d(j)$ be the probability that the first black ball appears in
position~$j$; $d(j)=\left.{a+1-j\choose b}\right/{a+1\choose b+1}$.
Then
$$\eqalign{M↓{\underline p}(d)&=\biggl(\,\sum↓{j=1}↑{a+1}j↑{\underline p}
{a+1-j\choose b}\biggr)\left/{a+1\choose b+1}\right.\cr
\noalign{\medskip}
&=p!\sum↓{j=1}↑{a+1}{j\choose p}{a+1-j\choose b}\left/{a+1\choose b+1}\right.\cr}$$
%The identity (proof in appendix)
%$$\sum↓{q=1}↑L(q-1)↑{\underline{k-1}}{L-q\choose A-k}=(k-1)!\,{L\choose A}$$
%can be applied with
%$$\eqalign{q-1&=j\cr
%k-1&=p\cr
%L-q&=a+1-j\cr
%A-k&=b\cr}$$
%or
%$$\eqalign{q&=j+1\cr
%k&=p+1\cr
%L&=a+1-j+q=a+2\cr
%A&=b+k=b+p+1\cr}$$
getting for $p>0$
$$\eqalign{M↓{\underline p}(d)&=p!\,\left.{a+2\choose b+p+1}\right/
{a+1\choose b+1}\cr
\noalign{\medskip}
M↓{\underline 1}&={a+2\choose b+2}\left/{a+1\choose b+1}\right.={a+2\over b+2}\cr
\noalign{\medskip}
M↓{\underline 2}&=\left.2{a+2\choose b+3}\right/{a+1\choose b+1}=
{2(a+2)(a-b)\over (b+2)(b+3)}\cr
\noalign{\medskip}
M↓2&={(a+2)(2a-b+3)\over (b+2)(b+3)}\cr
V&=M↓2-M↓1↑2={(a+2)(a-b)(b+1)\over (b+2)↑2(b+3)}\cr}$$
\smallskip}
}

\bigskip
\parindent0pt
\copyright 1984 Robert W. Floyd

First draft (not published)
September 19, 1985.
%revised: date;
%subsequently revised.

\bye