What is the expected distance between normally distributed points on a plane? What about the distance in higher dimensions?

Question

Let $X = (x_1, x_2)$ and $Y = (y_1, y_2)$ where the random variables $x_1$, $x_2$, $y_1$, $y_2$ are independent standard normal. What is the expected distance between $X$ and $Y$, i.e. what is $$D_2=E\left(\sqrt{(x_1 - y_1)^2 + (x_2 - y_2)^2}\right)\ ?$$ Does this value increase or decrease when the number $n$ of dimensions increases, that is when $X = (x_1, \cdots, x_n)$ and $Y = (y_1, \cdots, y_n)$ for some independent standard normal random variables $x_i$, $y_i$, and $n>2$? Does it converge when $n \rightarrow \infty$?

Note: I do not know if the resulting integrals are tractable (I would suspect that they are not).

Answer 1

I do not understand why this problem reappears without a complete solution seven years later. However it is not hard. Suppose that $X,Y,Z$ are independent and $N(0,I_n)$ distributed. Since $X-Y\sim N(0,2I_n)$ then $E(\|X-Y\|)=2^aE(\|Z\|)$ with $a=n/2.$ We use the formula $$\sqrt{r}=\int_{0}^{\infty}\frac{1}{2\sqrt{\pi}s^{3/2}}(1-e^{-sr})ds$$ (derive in $r$ to check it) applied to $r=\|Z\|^2.$ We get $$E(\|Z\|)=\int_{0}^{\infty}\frac{1}{2\sqrt{\pi}s^{3/2}}\left(1-\frac{1}{(1+2s)^a}\right)ds$$$$=\int_{0}^{\infty}\frac{1}{2\sqrt{\pi}s^{3/2}}\left(\int_1^{1+2s}\frac{dx}{ax^{a+1}}\right)ds$$$$=\int_{0}^{\infty}\frac{1}{\sqrt{\pi}s^{1/2}}\left(\int_0^{1}\frac{du}{a(1+2su)^{a+1}}\right)ds$$$$=\frac{1}{a\sqrt{\pi}}\int_{0}^{1}\left(\int_0^{\infty}\frac{1}{s^{1/2}}\frac{ds}{(1+2su)^{a+1}}\right)du$$$$=\frac{1}{a\sqrt{\pi}}\int_{0}^{1}\left(\int_0^{\infty}\frac{1}{v^{1/2}}\frac{dv}{(1+v)^{a+1}}\right)\frac{du}{\sqrt{2u}}=\sqrt{2}\frac{\Gamma(\frac{n+1}{2}) }{\Gamma(\frac{n}{2})}.$$

Answer 2

If everything is independent, then this is no longer a two-point problem: the variables $x_i - y_i$ are independent $N(0,2)$. The expectation is not terribly hard to compute, see here. It is increasing (moreover, the distribution itself is increasing, in terms of stochastic order) and, as @Did commented, equivalent to $\sqrt{2n}$.

Answer 3

Here is a simplified example for $n=2$ which might help you to start out.

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Assume an archer is shooting an arrow onto an infinite target with center $(0,0)$. The arrow hits at the coordinates $(X,Y)$ where $X$ and $Y$ are independent and $X,Y\sim\mathcal{N}(0,1)$. It should be clear that

$$ f_{X,Y}(x,y)=\frac{\exp(-1/2\cdot(x^2+y^2))}{2\pi} $$

since $X$ and $Y$ are independent. Let $Z$ be a random variable describing the distance from the center - since $Z<0$ does not make sense we have $F_Z(z)=0$ for $z<0$. Therefore we can assume $z\geq 0$ now. Now with some steps we can get

\begin{align*} F_Z(z) &= \Pr[Z\leq z] = \Pr[(X,Y)\in B_z(0)] = \int_{B_z(0)}f_{X,Y}(s,t)\,\mathrm{d}s\,\mathrm{d}t\\ &= \int_0^z\int_0^{2\pi}r\cdot f_{X,Y}(r\cos\Theta,r\sin\Theta)\,\mathrm{d}\Theta\,\mathrm{d}r\\ &= \int_0^z\int_0^{2\pi}r\cdot \frac{\exp(-1/2\cdot(r^2\cos^2\Theta+r^2\sin^2\Theta))}{2\pi}\,\mathrm{d}\Theta\,\mathrm{d}r\\ &= \int_0^z\int_0^{2\pi}r\cdot \frac{\exp(-1/2\cdot r^2)}{2\pi}\,\mathrm{d}\Theta\,\mathrm{d}r\\ &= \int_0^zr\cdot \frac{\exp(-1/2\cdot r^2)}{2\pi}\int_0^{2\pi}\,\mathrm{d}\Theta\,\mathrm{d}r\\ &= \int_0^zr\cdot \frac{\exp(-1/2\cdot r^2)}{2\pi}\cdot 2\pi\,\mathrm{d}r\\ &= \int_0^z r\cdot\exp(-1/2\cdot r^2)\,\mathrm{d}r \end{align*}

which yields

\begin{align*} f_Z(z) = F_Z'(z)=\begin{cases} r\cdot\exp(-1/2\cdot r^2),&z\geq 0,\\ 0,&\text{otherwise}. \end{cases} \end{align*}

Now it follows that

\begin{align*} \mathbb{E}[Z] &= \int_{-\infty}^\infty t\cdot \left (t\cdot\exp(-1/2\cdot t^2)\right )\,\mathrm{d}t\\ &= \left[-\exp(-1/2\cdot t^2)\cdot t\right |_0^\infty-\int_0^\infty -\exp(-1/2\cdot t^2)\,\mathrm{d}t \\ &= 0 + \sqrt{2\pi}\int_0^\infty\frac{\exp(-1/2\cdot t^2)}{\sqrt{2\pi}}\,\mathrm{d}t\\ &= \sqrt{2\pi}\cdot \frac{1}{2} = \sqrt{\frac{\pi}{2}}. \end{align*}

Hence the archer will miss the center by about $\sqrt{\pi/2}$ units.