Expectation and direct integration give different results

Question

I have an integral I want to compute:

$\qquad \int_{\mathbb R^4} e^{-(x_1+x_2+x_3+x_4)} \left( 1-x_1-x_3 \right) dx$

To me, this should be equivalent (modulo some scaling factor) to computing the expectation of $(1-x_1^2-x_3^2+ 0 x_2 + 0 x_4)$ when $x$ is a zero-mean Gaussian variable with variance $2I_4$. See How to deal with complicated Gaussian integrals in Mathematica?).

So I tried

Sqrt[(2 Pi)^4] Sqrt[2^4] 
  Expectation[
    (1 - x1^2 - x3^2), 
    {x1, x2, x3, x4} \[Distributed] MultinormalDistribution[2 IdentityMatrix[4]]]

which gives $-48 \pi^2$. However, when I compute

Integrate[Exp[-x1^2 - x2^2 - x3^2 - x4^2] (1 - x1^2 - x3^2), 
  {x1, -Infinity, Infinity}, {x2,-Infinity,Infinity}, 
  {x3, -Infinity,Infinity}, {x4, -Infinity, Infinity}]

The result is $0$.

I am not sure what is going on here. Which one is correct?.

I think the variance should be 1/2 IdentityMatrix[4]. – b.gates.you.know.what Dec 16 '19 at 13:45 — b.gates.you.know.what, Dec 16 '19 at 13:45
True, which actually gives 0. – NoobNoob Dec 16 '19 at 13:51 — NoobNoob, Dec 16 '19 at 13:51

score 2 · Accepted Answer · answered Dec 16 '19 at 14:08

If you only consider the part x1,x3 of your integral Integrate[ Exp[-x1^2 - x3^2 ] (1 - x1^2 - x3^2), {x1, -Infinity, Infinity}, {x3, -Infinity, Infinity} ] it`s possible to transform in polarcoordinates (pointsymmetrical integrand!) which gives

Integrate[Exp[-r^2] (1 -r^2) 2Pi r, {r, -Infinity,Infinity}  ] 
(*0*)

That means your Integrate-result is ok!

Expectation and direct integration give different results

1 Answers1