# Expected Euclidean distance between a given point and a multivariate Gaussian variable

For simplicity, let us consider a multivariate standard Gaussian distribution $$\mathcal N_k$$ of dimension $$k$$, where each dimension follows a standard normal distribution $$\mathcal N(0, 1)$$ independently. Let $$\varphi(z) = \frac{\exp(-z^2/2)}{\sqrt{2\pi}}$$ denote the pdf of $$\mathcal N(0, 1)$$.

Now given a point $$\mathbf{s} = (s_1, s_2, \ldots, s_k) \in \mathbb R^k$$, I want to compute the expected distance between a random point $$\mathbf{x} = (x_1, x_2, \ldots, x_k) \sim \mathcal N_k$$ and $$x_0$$, which can be written as $$\int\int\cdots\int \sqrt{(s_1 - x_1)^2 + (s_2 - x_2)^2 + \cdots + (s_k - x_k)^2} \varphi(x_1)\varphi(x_2)\ldots\varphi(x_k) dx_1 dx_2 \ldots dx_k.$$

What I have tried: When $$k = 1$$, I can compute this, which reduces to $$\int \varphi(x) |x-s| dx = \frac{1}{2} e^{-\frac{s^2}{2}} \left(e^{\frac{s^2}{2}} s \text{erf}\left(\frac{s}{\sqrt{2}}\right)-e^{\frac{s^2}{2}} s \text{erfc}\left(\frac{s}{\sqrt{2}}\right)+e^{\frac{s^2}{2}} s+2 \sqrt{\frac{2}{\pi }}\right) = s \text{erf}\left(\frac{s}{\sqrt{2}}\right)+\sqrt{\frac{2}{\pi }} e^{-\frac{s^2}{2}},$$ where "erf" is the Gauss error function defined as $$\text{erf}(z) = \frac{2}{\sqrt{\pi}} \int_0^z e^{-t^2} dt$$ and "erfc" is the complementary error function defined as $$\text{erfc}(z) = 1 - \text{erf}(z).$$ But seemingly this becomes much more complicated when $$k > 1$$.

Any idea or comment would be appreciated.

Seemingly if we consider squared Euclidean distance, we have a nice solution.

• This reduces to the question of finding $\mathbb E\|X\|_2$ where $X \sim \mathcal{N}(\mu,\sigma^2)$ where $\mu \in \mathbb R^d, \sigma >0$ are arbitrary. That sounds like a problem that cannot be done without finding the actual distribution of $\|X\|_2$. Jan 20 at 5:26

Let $$\mathbf{y}=\mathbf{s}-\mathbf{x}$$, with $$\mathbf{s}$$ fixed and $$\mathbf{x}$$ with all elements i.i.d. $$\mathcal{N}(0,1)$$. We have that the elements of $$\mathbf{y}$$ are independent, following a distribution $$y_i\sim\mathcal{N}(s_i,1)$$.
Then, the distance from $$\mathbf{s}$$ to $$\mathbf{x}$$ is $$\|\mathbf{s}-\mathbf{x}\|=\|\mathbf{y}\|=\sqrt{\sum_{i=1}^k y_i^2}=\sqrt{\sum_{i=1}^k \left(\frac{y_i}{1}\right)^2}\,$$
which follows a non-central chi distribution with parameters $$k$$ (the number of dimensions) and $$\lambda=\sqrt{\sum_{i=1}^k \left(\frac{s_i}{1}\right)^2}=\sqrt{\sum_{i=1}^k s_i^2}=\|\mathbf{s}\|\,,$$ and has known expectation $$\mathbb{E}\{\|\mathbf{y}\|\}=\sqrt{\frac{\pi}{2}}L_{1/2}^{k/2-1}\left(\frac{-\|\mathbf{s}\|^2}{2}\right)\,,$$ where $$L_{1/2}^{(k/2-1)}(z)$$ is a general Laguerre function, which has involved expressions depending on $$k$$, but are possibly computable for a given value of $$k$$.