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In 2D or 3D I have a fix point y and Gaussian distribution of a random point x. I am now interested in the mean euclidean distance between x and y:

$E_x [ d(x,y) ] = \int _{-\infty }^{\infty } d(x,y) N(x; \mu, \Lambda )dx$

Many thanks in advance

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In 2D, if the components of $X$ are independent normal variables with the same variance, then look at Note the formula for the mean, and the first item in the section "Related distributions". – Shai Covo Jan 24 '11 at 18:57

You can define a new random variable $\tilde{X} = X-y$. Then the quantity you are interested in is really the expected distance of this random variable from the origin. Let $\tilde{X}$ be the column vector $[x_1 x_2 \dots x_n]^T$. Then the squared distance from the origin is $\tilde{X}^T\tilde{X}$. For the case when the components of $\tilde{X}$ are independent, this expectation is a non-central chi-square distribution with $n$ degrees of freedom. For the case when $\tilde{X}$ has a general (non-diagonal) covariance matrix, the distribution of $\tilde{X}^T \tilde{X}$ will be a generalized chi-square distribution. If you are interested in the expected value of the absolute distance (i.e., $E(\sqrt{\tilde{X}^T \tilde{X}})$), try looking at the chi distribution. The chi distribution is the square root of the chi-square distribution and hence will give you the expected absolute distance if the components of $\tilde{X}$ are independent. If they are not, there should be a corresponding generalized chi distribution that is the square root of the generalized chi-square distribution.

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Thank you! Your answer sounds very promising. Indeed the $\tilde{X}$ are dependent in my case which means I have a full covariance matrix $\tilde{\Lambda}$. However I did not find any good literatur concerning a "generalized chi-square distribution". Can you (or someone else) advise me something - a link, paper or a book? – Matthias Jan 25 '11 at 9:38
Closed form pdf etc don't exist for the generalized chi-square distribution. For simple moments like expectation and variance, you don't need the full blown pdf but only the pairwise correlation between the components of $\tilde{X}$. More than that, it really depends on what you want in your problem. Not sure if this link will help but this is what I found with some googling - – Dinesh Jan 25 '11 at 20:04

Since I could not find a formula for an expectation value of a generalized chi square distribution, I decided to ignore the dependency between the components of $\hat{X}$ and continue with a diagonal covariance matrix. This then gives me:

$E_{\tilde{X}} \left[ \sqrt{ \tilde{X}^T \tilde{X} } \right] = E_X \left[ \sqrt{ \sum_{i=1}^n \tilde{x}^2_i } \right]$

However each $\tilde{x}_i$ still has its own variance $\sigma_i^2$. Therefore I guess the expectation value of the non-central chi distribution won't help here since I can not factor out one common variance.

My approach now is the following: Since I need this expectation value as part of an upper bound, I apply the triangle inequality in order to approximate the square root of a sum:

$E_{\tilde{X}} \left[ \sqrt{ \sum_{i=1}^n \tilde{x}^2_i } \right] \le E_{\tilde{X}} \left[ \sum_{i=1}^n \sqrt{ \tilde{x}_i^2 } \right] = \sum_{i=1}^n E_{\tilde{x_i}} \left[ \| \tilde{x}_i \| \right]$

I think this is equivalent to exchanging the euclidean norm with the manhattan norm.

What do you think about this idea? Is there maybe another approximation which might be more close to the real expectation value?

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Try a change of variable to get the density of a standard multivariate normal multiplied by something else. It should make the problem clearer.

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