We all know that for a geometric distribution (= Pascal of order 1), i.e.
P[X(1)=r] = p*(q^r), r = 0,1,2, ...
we have
P[X(1) >= r] = q^r, r = 0,1,2, ...
If X(n) has a Pascal distribution of order n, by which I mean
P[X(n)=r] = (r+n-1)C(r)*(p^n)*(q^r), r = 0,1,2, ...
I can show that
P[X(n) >= r] = (q^r)*[(sigma k=0 to k=n-1)(r+k-1)C(k)*(p^k)], r = 0,1,2, ...
so that the expression on the RHS has just n terms for each r, rather than
the r terms required from use of the pmf - which is nice!
Clearly this result must be "well-known" (it is hardly the Poincare
conjecture!) but I can't find it in the standard texts I have to hand. Can
anyone point me in the right direction for a reference, please? I don't want
to re-invent the wheel, again.
Also, how about the case of general non-integer n to cover the negative
binomial distribution of index n?
Any hints or pointers would be much appreciated over this bank holiday
weekend!
Many thanks
Quentin Burrell
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