Donald F. Burrill ([EMAIL PROTECTED]) wrote:
: Haven't seen a public response to this question that I find credible,
: and am curious. Is the problem as described solvable?
: If n is finite, what meaning attaches to "P(X=n+2)" and "P(X=n+1)"?
: If n is infinite, shouldn't the description read, without reference
: to an apparently finite n, "X(omega)={1,2,3,...}"?
: Or should the description read "for every i in {1,2,3,...,n-2}"?
:
: 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.
--
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
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