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How could the following integration of a 6-dimensional Gaussian be achieved? Are there some techniques? (MMA code below)

$$\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }a \cdot e^{-\frac{1}{2} \left(z_1^2+z_2^2+z_3^2+z_4^2+z_5^2+z_6^2\right)}dx_6dx_5dx_4dx_3dx_2dx_1$$ with

$a =\sqrt{(z_1 z_5 - z_2 z_4)^2 + (z_3 z_4 - z_1 z_6)^2 + (z_2 z_6 - z_3 z_5)^2} \\ z1 = p1 + x1 \\ z2 = p2 + x2 \\ z3 = p3 + x3 \\ z4 = p4 + x4 \\ z5 = p5 + x5 \\ z6 = p6 + x6 $

where $p_1$,...,$p_6$ are free variables.

MMA code:

z1 = p1 + x1; 
z2 = p2 + x2; 
z3 = p3 + x3; 
z4 = p4 + x4; 
z5 = p5 + x5; 
z6 = p6 + x6; 
a = Sqrt[(z1*z5 - z2*z4)^2 + (z3*z4 - z1*z6)^2 + (z2*z6 - z3*z5)^2]; 
Integrate[a*E^((-1/2)*(z1^2 + z2^2 + z3^2 + z4^2 + z5^2 + z6^2)), {x1, -Infinity, Infinity}, {x2, -Infinity, Infinity}, {x3, -Infinity, Infinity}, 
  {x4, -Infinity, Infinity}, {x5, -Infinity, Infinity}, {x6, -Infinity, Infinity}]
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  • $\begingroup$ How about if you try the two dimensional version... then the three... see where it fails? $\endgroup$
    – bill s
    Commented Jan 27, 2020 at 0:29
  • $\begingroup$ Even integration in 1 dimension is not easy. $\endgroup$ Commented Jan 27, 2020 at 0:35
  • $\begingroup$ This is a duplicate of stats.stackexchange.com/questions/445185/… also asked by the OP. It's a good question but just a duplicate. $\endgroup$
    – JimB
    Commented Jan 27, 2020 at 3:51
  • $\begingroup$ Use Monte Carlo. $\endgroup$
    – Dom
    Commented Apr 14, 2021 at 9:29

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