1
$\begingroup$

I'm trying to solve a Baysian inference problem (although the details are not all that important for this question)

I define two probability distributions P and Q

pdf[N_, n_, p_][x_] := (1 + N) (1 - x)^(-n + N) x^n Binomial[N, n]
P = ProbabilityDistribution[pdf[N, n, p][x], {x, 0, 1}]
Q = ProbabilityDistribution[pdf[M, m, q][x], {x, 0, 1}]

Now Mathermatica is very happy to compute the $Pr_{q\sim Q, p\sim P}[q>p]$

Probability[q > p, {q \[Distributed] Q, p \[Distributed] P}]
(* (1 + M) (1 + N) Binomial[M, m] Binomial[N, n] Gamma[1 - m + M] Gamma[
  1 + n] Gamma[
  2 + m + n] HypergeometricPFQRegularized[{1 + n, 2 + m + n, 
   n - N}, {2 + n, 3 + M + n}, 1] *)

Side note I'm quite impressed it takes only .2s for this)

However if I try to compute this from a first principles standpoint i.e $$Pr_{q\sim Q, p\sim P}[q>p] = \int_{y>x} f_Q(y) f_P(x) dx dy = \int_{0}^1\int_{x}^1 f_Q(y) f_P(x) dy dx $$ with

Integrate[PDF[Q, y] PDF[P, x], {x, 0, 1}, {y, x, 1}]

I just get stuck waiting. I've also tried adding all assumptions but no luck

$Assumptions = {{m, n} \[Element] 
   NonNegativeIntegers, {M, N} \[Element] PositiveIntegers, 
  0 < \[Alpha] < 1, M >= m, N >= n, 0 <= p <= 1, 0 <= q <= 1}

I'd like to understand where this big hypergeometric solution comes from as it's th result of a theorem I want to prove in my research so any hints would be helpful. I figured at least getting the integral result to be equal would give me a start in figuring out how to tackle that integral.

Thanks

$\endgroup$

1 Answer 1

2
$\begingroup$

Got it figured out, Mathematica just wants some assumptions about the integration variables and needs GenerateConditions->False:

$Assumptions = {{m, n} \[Element] 
   NonNegativeIntegers, {M, N} \[Element] PositiveIntegers, 
  0 < \[Alpha] < 1, M >= m, N >= n, 0 <= p <= 1, 0 <= q <= 1, 
  0 < x < 1, 0 < y < 1}
Integrate[PDF[Q, y] PDF[P, x], {x, 0, 1}, {y, x, 1}, 
 GenerateConditions -> False]

(* (Gamma[1 - m + M] Gamma[1 + n] - 
 Gamma[2 + M] Gamma[2 + m + n] Gamma[
   2 + N] HypergeometricPFQRegularized[{1 + m, m - M, 
    2 + m + n}, {2 + m, 3 + m + N}, 1])/(
Gamma[1 - m + M] Gamma[1 + n]) *)
$\endgroup$
1
  • $\begingroup$ You don't need the assumptions, just the GenerateConditions->False. $\endgroup$
    – mef
    Jun 3 at 12:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.