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?


1 Answer 1


Rewrite the expression in your $\mathbb{P}$ operator:

\begin{align} ||(x-\mu)||^2 &> ||x-a||^2 \Longleftrightarrow \\ (x-\mu)^T(x-\mu)&>(x-a)^T(x-a) \Longleftrightarrow \\ x^Tx - 2\mu x + \mu^T\mu &> x^Tx - 2a^Tx + a^Ta \Longleftrightarrow \\ 2(a^T - \mu^T)x &> a^Ta - \mu^T\mu \end{align}

And note that since $X$ is normal and $a,\mu$ are constants, $2(a^T - \mu^T)X$ will also be normal. In particular, as

$\mathbb{E}(2(a^T - \mu^T)X) = 2(a^T - \mu^T)\mu$,

$Var(2(a^T - \mu^T)X) = 2(a^T - \mu^T)^TVar(X)2(a^T - \mu^T) = 4(a - \mu)I\sigma^2(a^T - \mu^T)$,

it will hold that $\mathbb{P}(||(x-\mu)||^2 > ||x-a||^2) = \mathbb{P}(2(a^T - \mu^T)x > a^Ta - \mu^T\mu) = \mathbb{P}(R > a^Ta - \mu^T\mu)$ with

$R \sim N(2(a^T - \mu^T)\mu, 4(a - \mu)I\sigma^2(a^T - \mu^T))$.

If you know $a, \mu$ you can now find an expression for the probabilty fairly easily using the normal distribution defined by $R$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.