Processing math: 100%

We evaluate an integral having to do with vector averages over all orientations in an n-dimensional space.

Problem definition

Let ˆv be a unit vector in n-dimensions and consider the orientation average of

Jˆva1ˆva2ˆvak

where a1,,ak are some given fixed vectors. For example, if all ai are equal to ˆx, we want the orientation average of vkx.

Solution

We’ll evaluate our integral using parameter differentiation of the multivariate Gaussian integral. Let

I=1(2π)n/2e|v|22+ki=1αivaidnv=exp[12|ki=1αiai|2]

The expression in the second line follows from completing the square in the exponent in the first — for review, see our post on the normal distribution, here. Now, we consider a particular derivative of I with respect to the α parameters. From the first line of (2), we have

α1αkI|α=0=1(2π)n/2e|v|22ki=1vaidnv1(2π)n/20e|v|22vn+k1dvki=1ˆvaidΩv=2k/21πn/2Γ(n+k2)×ki=1ˆvaidΩv

The second factor above is almost our desired orientation average J — the only thing it’s missing is the normalization, which we can get by evaluating this integral without any as.

Next, we evaluate the parameter derivative considered above in a second way, using the second line of (2). This gives,

α1αkI|α=0=α1αkexp[12|ki=1αiai|2]|α=0=pairings(ai1ai2)(ai3ai4)(aik1aik)

The sum here is over all possible, unique pairings of the indices. You can see this is correct by carrying out the differentiation one parameter at a time.

To complete the calculation, we equate (3) and (4). This gives

ki=1ˆvaidΩv=πn/22k/21Γ(n+k2)pairings(ai1ai2)(ai3ai4)(aik1aik)

Again, to get the desired average, we need to divide the above by the normalization factor. This is given by the value of the integral (5) when k=0. This gives,

J=12k/2Γ(n/2)Γ(n+k2)pairings(ai1ai2)(ai3ai4)(aik1aik)

Example

Consider the case where k=2 and a1=a2=ˆx. In this case, we note that the average of ˆv2x is equal to the average along any other orientation. This means we have

ˆv2x=1nni=1ˆv2x+ˆv2y+=1n

We get this same result from our more general formula: Plugging in k=2 and a1=a2=ˆx into (6), we obtain

ˆv2x=12Γ(n/2)Γ(n2+1)=1n

The two results agree.

Like this post? Share on: Twitter | Facebook | Email


Jonathan Landy Avatar Jonathan Landy Jonathan grew up in the midwest and then went to school at Caltech and UCLA. Following this, he did two postdocs, one at UCSB and one at UC Berkeley.  His academic research focused primarily on applications of statistical mechanics, but his professional passion has always been in the mastering, development, and practical application of slick math methods/tools. He currently works as a data-scientist at Stitch Fix.

Published

Category

Theory

Contact