- #1

- 36

- 0

## Main Question or Discussion Point

Hey all,

I have an engineering background with a pretty terrible grounding in statistics and just about all maths that I can't immediately visualize as some kind of dynamic system, so forgive me if this is an obvious question!

I am working through a set of derivations for a conditional probability problem and am a little stuck on a rigorous proof of the following...

[tex]\int_{\Re^{n}}\exp(-\frac{1}{2}A B A^T)dA = \frac{1}{|2\pi B|^{1/2}}[/tex]

I can arrive at this solution by comparing the integral with that of the multivariate normal PDF which by definition is

[tex]\int_{\Re^{n}}\frac{1}{|2\pi \Sigma|^{1/2}}\exp(-\frac{1}{2}X \Sigma^{-1} X^T)dX = 1[/tex]

(Substitute [tex]X = A[/tex] and [tex]\Sigma = B^{-1}[/tex] then cross multiply by everything outside of the exponential.)

Am I right in thinking that there is no analytical solution of this integral? It feels wrong to simply state it without proof, but if there is no solution how else can I present it?

Thanks in advance for your help!

Owen

I have an engineering background with a pretty terrible grounding in statistics and just about all maths that I can't immediately visualize as some kind of dynamic system, so forgive me if this is an obvious question!

I am working through a set of derivations for a conditional probability problem and am a little stuck on a rigorous proof of the following...

[tex]\int_{\Re^{n}}\exp(-\frac{1}{2}A B A^T)dA = \frac{1}{|2\pi B|^{1/2}}[/tex]

I can arrive at this solution by comparing the integral with that of the multivariate normal PDF which by definition is

[tex]\int_{\Re^{n}}\frac{1}{|2\pi \Sigma|^{1/2}}\exp(-\frac{1}{2}X \Sigma^{-1} X^T)dX = 1[/tex]

(Substitute [tex]X = A[/tex] and [tex]\Sigma = B^{-1}[/tex] then cross multiply by everything outside of the exponential.)

Am I right in thinking that there is no analytical solution of this integral? It feels wrong to simply state it without proof, but if there is no solution how else can I present it?

Thanks in advance for your help!

Owen