78
VON NEUMANN
relative levels of excitation of the two input bundles and of the output
bundle, respectively, of the network under consideration. The question is
then: "What is the distribution of the (stochastic) variable £ in terms
of the (given) g, t) ?
Let W be the complementary set of Z . Let p, q, r be the
numbers of elements of X, Y, W, respectively, so that p = |N, q = tjN,
r = (i-£)N. Then the problem is to determine the distribution of the
(stochastic) variable r in terms of the (given) p, q — i.e., the prob
ability of any given r in combination with any given p, q.
W is clearly the intersection of the sets X, Y: W = X-Y. Let
U, V be the (relative) complements of W in X, Y, respectively:
U = X - W, V = Y - W, and let S be the (absolute, i.e., in the set
(1, •••, N)) complement of the sum of X and Y: S = - (X + Y). Then
W, U, V, S are pairwise disjoint sets making up together precisely the
entire set (1, ..., N), with r, p-r, q-r, N-p-q+r elements,
respectively. Apart from this they are unrestricted. Thus they offer to
gether N!/[r!(p - r)I(q - r)!(N - p - q + r )I] possible choices. Since
there are a priori N!/[pI(N - p)IJ possible choices of an X with p
elements and a priori N!/[q!(N — q )I] possible choices of a Y with q
elements, this means that the looked for probability of W having r
elements is
_/ N! / N! N!
p r F("p^rTT(q-rTi"(N"-'p-"q+r)! / p I W p T T qT(K-q7 ! /
= pi (N-p)Iql (N-q)i
r ! (p-r) 1 (q-r) f (N-p-q+r) INI
Note, that this formula also shows that p = 0 when r < 0 or
p-r<0 or q-r<0 or N-p-q+r<0, i.e., when r violates
the conditions
Max(o, p+q-N)<r< Min(p, q).
This is clear combinatorially, in view of the meaning of X, Y and W. In
terms of I, tj , (; the above conditions become
(17) 1 - Max(0, | + n — 1 ) > ? > 1 - Min(|, t] ).
Returning to the expression for p, substituting the |, t], 5
expressions for p, q, r and using Stirling1s formula for the factorials
involved, gives
( 18) p ~ - 1 = ^ e"eN ,
v 2 jtN
where
ed-e)Ti(i-n)
„ =
_
_____
SU-UTiU-ri )
_______
(5+I- 1 )(5+i)-i )(i-5)(2-5-ti-5)