CRGreathouse
May23-07, 07:21 PM
I'm not sure if I should be posting here or in General, but here goes.
I have a probability problem and I'm trying to get a closed form, or something resembling it, for a tricky recursive formulation. This problem is the 'simple' base case for a much more complicated problem, but I think I can solve it (in recursive terms) if I can somehow simplify this special case.
The variables (a, b, c, d) represent a partition of a whole, thus all must be nonnegative.
Base cases:
\mathfrak{f}(0, b, c, d) = 0
\mathfrak{f}(a, 0, c, d) = 0
\mathfrak{f}(a, b, c, d) = 0 if any of {a, b, c, d} are negative
Recursive case:
\binom{a+b+c+d}{2}\mathfrak{f}(a,b,c,d)=
ab+ac\mathfrak{f}(a-1,b,c-1,d)+bc\mathfrak{f}(a, b-1,c-1,d)
+\binom{c}{2}\mathfrak{f}(a,b,c-2,d)+cd\mathfrak{f}(a,b,c-1,d-1)
Any ideas for me? Does this resemble some famous (hopefully solved) problem? Should I just give up and use a computer algebra system for the whole messy expression (there are several layers of recursion to go for the general solution!)?
I have a probability problem and I'm trying to get a closed form, or something resembling it, for a tricky recursive formulation. This problem is the 'simple' base case for a much more complicated problem, but I think I can solve it (in recursive terms) if I can somehow simplify this special case.
The variables (a, b, c, d) represent a partition of a whole, thus all must be nonnegative.
Base cases:
\mathfrak{f}(0, b, c, d) = 0
\mathfrak{f}(a, 0, c, d) = 0
\mathfrak{f}(a, b, c, d) = 0 if any of {a, b, c, d} are negative
Recursive case:
\binom{a+b+c+d}{2}\mathfrak{f}(a,b,c,d)=
ab+ac\mathfrak{f}(a-1,b,c-1,d)+bc\mathfrak{f}(a, b-1,c-1,d)
+\binom{c}{2}\mathfrak{f}(a,b,c-2,d)+cd\mathfrak{f}(a,b,c-1,d-1)
Any ideas for me? Does this resemble some famous (hopefully solved) problem? Should I just give up and use a computer algebra system for the whole messy expression (there are several layers of recursion to go for the general solution!)?