Michael Cohen ([EMAIL PROTECTED]) wrote:
: : On Wed, 1 Dec 1999, Yonah Russ wrote:
: :
: : > how do you solve a problem like this one?
: : > thanks in advance
: : > -------
: : > X is a chance variable such that X(omega)={1,2,3...,n}
: : > and for every i in {1,2,3...n}, 4P(X=i+2)=5P(X=i+1)-P(X=i)
: : >
: : > find the breakdown of X.
: :
:
: Let's take n finite first. Let ^ denote "raised to power". Using difference
: equations, the solutions are constant and constant times (1/4)^i or linear
: combinations thereof: P(X=i) = a (1/4)^i + b where a and b are constants. The
: requirement that the P(X=i) sum to 1 forces a=3(1-bn)4^n/(4^n-1) and b is at most
: 1/n.
:
: If n is infinite, then b must be 0 (otherwise the sum of the P(X=i) is infinite).
: Then a=3. So P(X=i)=3 (1/4)^i.
:
One revision: At the end of the second to last paragraph, I should not have
required b be at most 1/n (a can be negative). We just need to be sure P(X=1)
and P(X=n) are non-negative --- the other values are in-between.
--
Michael P. Cohen home phone 202-232-4651
1615 Q Street NW #T-1 office phone 202-219-1917
Washington, DC 20009-6310 office fax 202-219-1736
[EMAIL PROTECTED]
