Originally posted by humy
just noticed that if x=1 in;
x ∈ [0, 1], x ∈ β
a, b ∈ β{0}
∫[0, x] t^a (1 – t)^b dt = b! ∑[r=0, b] (-1)^r x^(a + 1 + r) / ( (a + 1 + r) r! (b – r)! )
then that very neatly simplifies to just;
∫[0, 1] t^a (1 – t)^b dt = a! b! / (a + b + 1)!
-yet another useful equation for my research.
And I just used it to find the mean average of my new probabi ...[text shortened]... s;
mean = (S + 1) / (S + F + 2)
Haven't yet checked that with a computer program though.
The full beta function can be written in terms of the gamma function, the formula is:
B(p, q) = gamma(p)gamma(q)/gamma(p + q)
Where:
gamma(x) = ∫[0, infinity] t^(x - 1) exp(-t) dt
and for integer values:
gamma(n) = (n - 1)!
for integer values of p and q the beta function is then:
B(p, q) = (p - 1)!(q - 1)! / (p + q - 1)!
Your a = p - 1, and b = q - 1
So:
B(a + 1, b + 1) = a! b! / (a + b + 1)!
Which is the expression you wrote down. Note though that you have a difference in definition between your expression (I'll call your function A(a, b)) and the standard beta function:
A(a, b) = ∫[0, 1] t^a (1 – t)^b dt
B(p, q) = ∫[0, 1] t^(p - 1) (1 – t)^(q - 1) dt
So A(a, b) = B(a + 1, b + 1). The binomial distribution is: C(N, m) = N! / [(N - m)! m!], this gives us:
C(N, m) = 1/ (N - 1) A(N - m, m)
and
C(N, m) = 1 / (N + 1)B(N - m + 1, m + 1)