Calculate probability of distance for d-dimensional normal

Question

Is there any simple way to calculate the probability of distance in the following form for d-dimensional normal distribution?

$P(||\mathbf{x}-\mathbf{\mu}||^2>||\mathbf{x}-\mathbf{a}||^2)$, where $f(\mathbf{x})=N(\mathbf{\mu},\sigma^2\mathbf{I})$, $\mathbf{I}$ is the identity matrix, $||\cdot||$ is a Euclid distance.

How can I get a formula for this? Especially, can I convert this to one-dimensional normal distribution to get the value of probability?

Answer 1

One thing you can do is write down the probability:

$$\mathbb{P}( || \mathbf{x} - \mathbf{\mu} ||^2 > || \mathbf{x} - \mathbf{a}||^2) = \mathbb{P}\left( \sum_1^n(x_i - \mu_i)^2 > \sum_1^n (x_i - a_i)^2 \right) = \mathbb{P}\left( \sum_1^n \mu_i^2 - a_i^2 +2(a_i - \mu_i)x_i > 0 \right) = \mathbb{P}\left( Z > 0 \right)$$ Where $Z \sim \mathcal{N}(\sum \mu_i^2 + 2\mu_i(\mu_i - a_i) - a_i^2, 4 \sigma^2(\sum (a_i - \mu_i)^2) ) = \mathcal{N}(\sum -(a_i - \mu_i)^2, 4 \sigma^2(\sum (a_i - \mu_i)^2) )$

So you have reduced yourself to the onedimensional case, as you were trying to do.