Let there be
ways for a successful and
ways for an unsuccessful trial out of a total of
possibilities.
Take
samples and let
equal 1 if selection
is successful and 0 if it is not. Let
be the total number of
successful selections,
![\begin{displaymath}
x \equiv \sum_{i=1}^N x_i.
\end{displaymath}](h_2227.gif) |
(1) |
The probability of
successful selections is then
The
th selection has an equal likelihood of being in any trial, so the fraction of acceptable selections
is
![\begin{displaymath}
p \equiv {n\over n+m}
\end{displaymath}](h_2231.gif) |
(3) |
![\begin{displaymath}
P(x_i = 1) = {n\over n+m}\equiv p.
\end{displaymath}](h_2232.gif) |
(4) |
The expectation value of
is
The Variance is
![\begin{displaymath}
\mathop{\rm var}\nolimits (x) \equiv \sum_{i=1}^N \mathop{\r...
...p \scriptstyle j\not = i} \mathop{\rm cov}\nolimits (x_i,x_j).
\end{displaymath}](h_2235.gif) |
(6) |
Since
is a Bernoulli variable,
so
![\begin{displaymath}
\sum_{i=1}^N \mathop{\rm var}\nolimits (x_i) = {Nnm\over (n+m)^2}.
\end{displaymath}](h_2240.gif) |
(8) |
For
, the Covariance is
![\begin{displaymath}
\mathop{\rm cov}\nolimits (x_i,x_j) = \langle x_ix_j\rangle -\langle x_i\rangle \langle x_j\rangle.
\end{displaymath}](h_2242.gif) |
(9) |
The probability that both
and
are successful for
is
But since
and
are random Bernoulli variables (each 0 or 1), their product is
also a Bernoulli variable. In order for
to be 1, both
and
must be
1,
Combining (11) with
![\begin{displaymath}
\left\langle{x_i}\right\rangle{}\left\langle{x_j}\right\rangle{} = {n\over n+m}{n\over n+m} = {n^2\over (n+m)^2},
\end{displaymath}](h_2252.gif) |
(12) |
gives
There are a total of
terms in a double summation over
. However,
for
of these, so there are a total
of
terms in the Covariance summation
![\begin{displaymath}
\sum_{i=1}^N \sum^N_{\scriptstyle j=1\atop\scriptstyle j\not...
...rm cov}\nolimits (x_i,x_j) = - {N(N-1)mn\over (n+m)^2(n+m-1)}.
\end{displaymath}](h_2259.gif) |
(14) |
Combining equations (6), (8), (11), and (14) gives the Variance
so the final result is
![\begin{displaymath}
\left\langle{x}\right\rangle{} = Np
\end{displaymath}](h_2265.gif) |
(16) |
and, since
![\begin{displaymath}
1-p={m\over n+m}
\end{displaymath}](h_2266.gif) |
(17) |
and
![\begin{displaymath}
np(1-p)={mn\over (n+m)^2},
\end{displaymath}](h_2267.gif) |
(18) |
we have
The Skewness is
and the Kurtosis
![\begin{displaymath}
\gamma_2= {F(m, n, N) \over mnN( -3 + m + n) ( -2 + m + n)( -m - n + N)},
\end{displaymath}](h_2272.gif) |
(21) |
where
The Generating Function is
![\begin{displaymath}
\phi(t)={{m\choose N}\over{n+m\choose N}}{}_2F_1(-N,-n; m-N+1; e^{it}),
\end{displaymath}](h_2281.gif) |
(23) |
where
is the Hypergeometric Function.
If the hypergeometric distribution is written
![\begin{displaymath}
h_n(x,s)={{np\choose x}{nq\choose s-x}\over{n\choose s}},
\end{displaymath}](h_2283.gif) |
(24) |
then
![\begin{displaymath}
\sum_{x=0}^s h_n(x,s)u^x = A\,{}_2F_1(-s,-np;nq-s+1;u).
\end{displaymath}](h_2284.gif) |
(25) |
References
Beyer, W. H. CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC Press, pp. 532-533, 1987.
Spiegel, M. R. Theory and Problems of Probability and Statistics.
New York: McGraw-Hill, pp. 113-114, 1992.
© 1996-9 Eric W. Weisstein
1999-05-25