multidimensional Gaussian integral

Let 𝐱=[x1x2xn]T and dn𝐱i=1ndxi.

Theorem 1

Let K be a symmetricMathworldPlanetmathPlanetmath positive definitePlanetmathPlanetmath ( matrix and f:RnR, where f(x)=exp(-12xTK-1x). Then

e-12𝐱T𝐊-1𝐱dn𝐱=((2π)n|𝐊|)12 (1)

where |K|=detK.

Proof. 𝐊-1 is real and symmetric (since (𝐊-1)T=(𝐊T)-1=𝐊-1). For convenience, let 𝐀=𝐊-1. We can decompose 𝐀 into 𝐀=𝐓𝚲𝐓-1, where 𝐓 is an orthonormal (𝐓T𝐓=𝐈) matrix of the eigenvectorsMathworldPlanetmathPlanetmathPlanetmath of 𝐀 and 𝚲 is a diagonal matrixMathworldPlanetmath of the eigenvaluesMathworldPlanetmathPlanetmathPlanetmathPlanetmath of 𝐀. Then

e-12𝐱T𝐀𝐱dn𝐱=e-12𝐱T𝐓𝚲𝐓-1𝐱dn𝐱. (2)

Because 𝐓 is orthonormal, we have 𝐓-1=𝐓T. Now define a new vector variable 𝐲𝐓T𝐱, and substitute:

e-12𝐱T𝐓𝚲𝐓-1𝐱dn𝐱 =e-12𝐱T𝐓𝚲𝐓T𝐱dn𝐱 (3)
=e-12𝐲T𝚲𝐲|𝐉|dn𝐲 (4)

where |𝐉| is the determinantMathworldPlanetmath of the Jacobian matrix Jmn=xmyn. In this case, 𝐉=𝐓 and thus |𝐉|=1.

Since 𝚲 is diagonalMathworldPlanetmath, the integral may be separated into the product of n independentPlanetmathPlanetmath Gaussian distributions, each of which we can integrate separately using the well-known formula

e-12at2𝑑t=(2πa)12. (6)

Carrying out this program, we get

e-12𝐲T𝚲𝐲dn𝐲 =k=1ne-12λkyk2𝑑yk (7)
=k=1n(2πλk)12 (8)
=((2π)nk=1nλk)12 (9)
=((2π)n|𝚲|)12. (10)

Now, we have |𝐀|=|𝐓𝚲𝐓-1|=|𝐓||𝚲||𝐓-1|=|𝚲|, so this becomes

e-12𝐱T𝐀𝐱dn𝐱=((2π)n|𝐀|)12. (12)

Substituting back in for 𝐊-1, we get

e-12𝐱T𝐊-1𝐱dn𝐱=((2π)n|𝐊-1|)12=((2π)n|𝐊|)12, (13)

as promised.

Title multidimensional Gaussian integral
Canonical name MultidimensionalGaussianIntegral
Date of creation 2013-03-22 12:18:44
Last modified on 2013-03-22 12:18:44
Owner Mathprof (13753)
Last modified by Mathprof (13753)
Numerical id 22
Author Mathprof (13753)
Entry type Theorem
Classification msc 60B11
Classification msc 62H99
Classification msc 62H10
Related topic JacobiDeterminant
Related topic AreaUnderGaussianCurve
Related topic ProofOfGaussianMaximizesEntropyForGivenCovariance