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I want to evaluate this multiple integral: $$ \iiint\limits_{ \sum_{i=1}^4 x_i=1,\ \ \ x_1,\, x_2,\,x_3,\, x_4\, \ge \,0 } (x_1+x_2)^{N1} (x_3+x_4)^{N2} (x_1+x_3)^{N3} (x_2+x_4)^{N4} \, dX$$

For context, this problem comes up when computing the marginal likelihood of data consisting of two dependent binary variables $(a,b)$. The conditional probability distribution for $B$ given $A$ is: \begin{align} P(B=0\mid A=0)=x_1+x_2; & & P(B=1\mid A=0)=x_3+x_4 \\ P(B=0\mid A=1)=x_1+x_3; & & P(B=0\mid A=1)=x_2+x_4 \end{align}

The marginal likelihood turns into this integral, where $N_i$ are the counts of $B\mid A$ in the data.

This looks like a dirichlet integral of type 1, but the problem is that I am unable to separate out the integral into integrals over fewer variables.

I have tried:

  1. Use dirac delta function to remove the simplex and then use laplace transform to find a closed form solution. However, it does not work here because the integrals are not separable.
  2. Using the variable substitution for beta variables as in this answer.
  3. Interpret the equation as product of two Beta functions: $B(N_1 +1, N_2 +1), B(N_3 +1, N_4 +1)$.
    However, I am stuck because $x_i$ are shared across the terms.

Any ideas?

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  • $\begingroup$ By using the substitutions $x_i=y_i^2,\; dx_i = 2y_i\,dy_i$ the integration domain becomes part of a sphere and you may switch to polar coordinates. $\endgroup$ – Jack D'Aurizio Oct 12 '16 at 19:00
  • $\begingroup$ thanks @JackD'Aurizio This works, however I get stuck again in separating the terms after substituting $y_1^2+y_2^2=r_{12}^2; y_3^2 + y_4^2=r_{34}^2$. The first two terms factor out, but then I have $(r_{12}^2 cos^2\theta_{12} +r_{34}^2 sin^2\theta_{34})^{N3} (r_{34}^2 cos^2\theta_{34} +r_{12}^2 sin^2\theta_{12})^{N4}$. How should I evaluate this? appreciate your help! $\endgroup$ – Amit Oct 12 '16 at 19:33
  • $\begingroup$ These look more like marginal probabilities than conditional probabilities. I suspect something is wrong with the problem from the beginning. $\qquad$ $\endgroup$ – Michael Hardy Oct 12 '16 at 19:42
  • $\begingroup$ Could you write $r_{12}^2\cos^2\theta_{12}$ instead of $r_{12}^2cos^2\theta_{12}$? That is standard usage. When you write $a\cos b$, coded as a\cos b, then you automatically get proper spacing to the left and right of $\cos$, and $\cos$ is not italicized, and the spacing depends on the context, as seen when you type $a\cos(b). \qquad$ $\endgroup$ – Michael Hardy Oct 12 '16 at 19:43
  • $\begingroup$ @Amit: you are taking polar coordinates as defined for the $3$-dimensional sphere, but I was actually suggesting to use polar coordinates as defined for a $4$-dimensional sphere, depending on $3$ angles. $\endgroup$ – Jack D'Aurizio Oct 12 '16 at 19:45
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First observe this simplification: \begin{align} & \int_0^1 \int_0^{1-x_1} \int_0^{1-x_1-x_2} (x_1+x_2)^{N_1}(1-x_1-x_2)^{N_2} (x_1+x_3)^{N_3} (1-x_1-x_3)^{N_4} \,dx_3 \,dx_2 \,dx_1 \\[12pt] = {} & \int_0^1 \int_0^{1-x_1}\left( (x_1+x_2)^{N_1}(1-x_1-x_2)^{N_2} \int_0^{1-x_1-x_2} (x_1+x_3)^{N_3} (1-x_1-x_3)^{N_4} \,dx_3\right) \,dx_2 \,dx_1 \end{align} In the innermost integral, $x_1$ and $x_2$ remain constant as $x_3$ goes from $0$ to $1-x_1-x_2$; hence the step above can be done.

Now look at the inner integral: $$ \int_0^{1-x_1-x_2} (x_1+x_3)^{N_3} (1-x_1-x_3)^{N_4} \,dx_3 = \int_{x_1}^{1-x_2} u^{N_3} (1-u)^{N_4} \, du. $$ This is a doubly incomplete Beta function. By expanding via the binomial theorem and integrating term by term, you can write it as a sum, but I'm not optimistic about a closed form.

This last integral is of course a polynomial function of $x_1$ and $x_2$.

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  • $\begingroup$ thanks @michael-hardy do you know of a reliable technique to approximate this integral if a closed form cannot be achieved? I'm happy using a numerical solution too. $\endgroup$ – Amit Oct 12 '16 at 20:20

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