# Distribution and moments of $\frac{X_iX_j}{\sum_{i=1}^n X_i^2}$ when $X_i$'s are i.i.d $N(0,\sigma^2)$

Suppose $$X_1,X_2,\ldots,X_n$$ are independent $$N(0,\sigma^2)$$ random variables.

For $$i,j\in \{1,2,\ldots,n\}$$, consider $$U=\frac{X_iX_j}{\sum_{i=1}^n X_i^2}$$

Provided $$n>1$$, we know that $$U$$ has a Beta distribution when $$i=j$$ :

$$U=\frac{X_i^2/\sigma^2}{\sum_{i=1}^n X_i^2/\sigma^2} \sim \text{Beta}\left(\frac12,\frac{n-1}2\right) \quad,\,i=1,2,\ldots,n$$

What can we say regarding the distribution of $$U$$ when $$i\ne j$$? What are the moments of $$U$$ in this case?

For $$n=2$$, if we transform $$(X_1,X_2)$$ to polar coordinates $$(R,\Theta)$$, then

$$U=\frac{X_1X_2}{X_1^2+X_2^2}=\frac{R^2\cos\Theta\sin\Theta}{R^2}=\frac{\sin(2\Theta)}{2}$$

Since $$\Theta$$ is uniformly distributed on $$(0,2\pi)$$, it seems $$\sin(2\Theta)$$ has an $$\text{Arcsine}(-1,1)$$ distribution with pdf

$$f(x)=\frac1{\pi \sqrt{1-x^2}}\mathbf 1_{(-1,1)}(x)$$

So that $$U$$ has pdf

$$f_U(u)=2 f(2u)=\frac2{\pi\sqrt{1-4u^2}}\mathbf1_{\left(-\frac12,\frac12\right)}(u)$$

If $$\boldsymbol X=(X_1,X_2,\ldots,X_n)^T$$, we can think of $$U$$ as the product of $$i$$th and $$j$$th components of the vector $$\frac{\boldsymbol X}{\lVert \boldsymbol X \rVert}$$. And we know that $$\frac{\boldsymbol X}{\lVert \boldsymbol X \rVert}$$ is uniformly distributed on the surface of a unit sphere. I am not sure if this helps in any way.

• Why do you need this distribution, if I may ask? Commented Nov 10, 2021 at 20:38
• Well, a student asked a related question in class, and we have been curious about it since, @StubbornAtom. Commented Nov 10, 2021 at 21:39
• I mean I simulated it and for $n=2$ I got a variance of $0.1249932$ close to $\frac18$, for $n=3$ of $0.06666795$ close to $\frac1{15}$, for $n=4$ of $0.04166444$ close to $\frac1{24}$ and similarly for larger $n$. It is well worth simulating as the distribution when $n=2$ is bimodal at the ends, while the distribution for larger $n$ is increasingly sharply peaked at $0$ Commented Nov 10, 2021 at 23:07
• It may be possible to find the distribution, but I don't think it is worth it. If I'm not mistaken it involves $_2F_1$ and $_3F_2$. I would stick with the mean and variance. This should be doable in closed form. Commented Nov 10, 2021 at 23:17
• Again empirically, for $n=2$ it would not surprise me if the distribution was an arcsine or $\mathrm{Beta}(\frac12,\frac12)$ distribution shifted down by $\frac12$ Commented Nov 10, 2021 at 23:39

Note that we may assume $$X_i$$'s are standard normal. Then

\begin{align*} \mu_k &:= \mathbf{E}\left[ \biggl( \frac{ X_i X_j }{X_1^2 + \cdots + X_n^2} \biggr)^k \right] \\ &= \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n} \biggl( \frac{ x_i x_j }{x_1^2 + \cdots + x_n^2} \biggr)^k e^{-(x_1^2 + \cdots + x_n^2)/2} \, \mathrm{d}x_1 \cdots \mathrm{d}x_n \\ &= \frac{1}{(2\pi)^{n/2}} \left( \int_{0}^{\infty} r^{n-1}e^{-r^2/2} \, \mathrm{d}r \right)\left( \int_{\mathbb{S}^{n-1}} \omega_i^k \omega_j^k \, \sigma(\mathrm{d}\omega) \right), \end{align*}

where $$\mathbb{S}^{n-1} = \{ \mathrm{x} \in \mathbb{R}^n : \|\mathrm{x}\| = 1\}$$ is the unit sphere in $$\mathbb{R}^n$$ and $$\sigma$$ is the surface measure on $$\mathbb{S}^{n-1}$$. Then by mimicking the proof of the beta function identity, such as in this article, it is not hard to prove that

$$\int_{\mathbb{S}^{n-1}} \omega_1^{\alpha_1} \cdots \omega_n^{\alpha_n} \, \sigma(\mathrm{d}\omega) = \begin{cases} 0, & \text{if some \alpha_i is odd,} \\[0.25em] \dfrac{2\Gamma(\beta_1)\cdots\Gamma(\beta_n)}{\Gamma(\beta_1 + \cdots + \beta_n)}, & \text{if all \alpha_i are even and \beta_i=\frac{1}{2}(\alpha_i + 1)}. \end{cases}$$

Using this and simplifying the formula, we get

$$\mu_k = \begin{cases} 0, & \text{if k is odd,} \\[0.25em] \dfrac{(1 \cdot 3 \cdot 5 \cdots (k-1))^2}{\prod_{l=0}^{k-1} (n+2l)}, & \text{if k is even.} \end{cases}$$

For instance, we get

$$\mu_2 = \frac{1}{n(n+2)} \qquad\text{and}\qquad \mu_4 = \frac{3^2}{n(n+2)(n+4)(n+6)}.$$

• for $n=2$ that $\mu_2 = \frac18$ and $\mu_4=\frac{3}{128}$ (and I suspect the higher moments too) are the same as an arcsine or Beta$(\frac12,\frac12)$ distribution shifted down by $\frac12$ would have Commented Nov 11, 2021 at 12:34

You can use the independence of $$\color{blue}{U=\frac{X_iX_j}{\sum_{i=1}^n X_i^2}}$$ and $$V=\sum\limits_{i=1}^n X_i^2$$ to find at least the mean and variance of $$U$$ for $$i\ne j$$. To argue somewhat statistically, $$U$$ and $$V$$ are independent by Basu's theorem because $$U$$ is an ancillary statistic for $$\sigma^2$$ (i.e., its distribution is free of $$\sigma^2$$), and $$V$$ is a complete sufficient statistic for $$\sigma^2$$.

So,

$$0=E\left[X_iX_j\right]=E\left[UV\right]=E\left[U\right]E\left[V\right]\,,$$

As $$E\left[V\right]\ne 0$$, you must have $$\color{blue}{E\left[U\right]=0}$$

Again,

$$\operatorname{Var}(X_iX_j)=\operatorname{Var}(UV)=E\left[U^2\right]E\left[V^2\right]$$

But $$\operatorname{Var}(X_iX_j)=E\left[X_i^2X_j^2\right]=\sigma^4$$

So, $$E\left[U^2\right]=\frac{\sigma^4}{E\left[V^2\right]}$$

As $$\frac{V}{\sigma^2} \sim \chi^2_n$$, you have

$$E\left[\left(\frac{V}{\sigma^2}\right)^2\right]=\operatorname{Var}\left(\frac V{\sigma^2}\right)+\left(E\left(\frac{V}{\sigma^2}\right)\right)^2=2n+n^2$$

so that $$E\left[V^2\right]=\sigma^4 n(n+2)$$ and

$$\color{blue}{\operatorname{Var}(U)=E\left[U^2\right]=\frac1{n(n+2)}}$$

Due to the symmetry $$\sigma_1=...=\sigma_n=\sigma$$, it is enough to consider $$i=1$$ and $$j=2$$.

While my first thought was n-dimensional spherical coordinates also, you can immediately do the integrals under consideration with a trick. We start with $$I(t)=\int_{-\infty}^\infty \cdots \int_{-\infty}^\infty {\rm d}x_1 \cdots {\rm d}x_n \, \frac{e^{-{t(x_1^2+...+x_n^2)}}}{(2\pi \sigma^2)^{n/2}} \, \frac{x_1^2x_2^2}{(x_1^2+...+x_n^2)^2}$$ $$I''(t)=\frac{1}{(2\pi \sigma^2)^{n/2}}\int_{-\infty}^\infty \cdots \int_{-\infty}^\infty {\rm d}x_1 \cdots {\rm d}x_n \, e^{-t(x_1^2+...+x_n^2)} \, {x_1^2x_2^2} \\ = \frac{\left(\pi/t\right)^{n/2-1}}{(2\pi \sigma^2)^{n/2}} \cdot \frac{\pi}{4t^3} = \frac{t^{-n/2-2}}{4(2 \sigma^2)^{n/2}} \, .$$ Then by integrating we get $$I(t)= \frac{t^{-n/2}}{(2 \sigma^2)^{n/2}} \cdot \frac{1}{n(n+2)} + c_1t + c_2$$ and since $$I(\infty)=0$$ it follows $$c_1=c_2=0$$ and the result is then obtained by setting $$t=\frac{1}{2\sigma^2}$$.

Evidently this can be generalized to k-th momenta with $$k$$ being even, while for odd $$k$$ the integrand is odd and the integral thus zero.

Let us denote this variable by $$Y$$, then \begin{align} Y &:=\frac{X_i X_j}{\sum_{s=1}^n X_s^2} \\ &=\left(\frac{X_i X_j}{\sum_{s=1}^n X_s^2}+\frac{1}{2}\right)-\frac{1}{2} \\ &= \frac{1}{2}\left(\frac{\sum_{i=1}^nX_i}{\sqrt{\sum_{i=1}^nX_i^2}} \right)^2 -\frac{1}{2} \\ &= \frac{1}{2}\left(\frac{n\bar{X}}{\sqrt{n-1}\cdot S} \right)^2 -\frac{1}{2} \\ &= \frac{1}{2}\frac{n}{n-1}\left(\frac{\bar{X}}{S / \sqrt{n}} \right)^2 -\frac{1}{2} \tag{1} \\ \end{align}

with $$\bar{X} :=\frac{1}{n}\left(\sum_{i=1}^nX_i \right)$$ and $$S:=\frac{1}{n-1}\left(\sum_{i=1}^nX_i^2 \right)$$ as defined in the Student's t-distribution.

From $$(1)$$ and the definition of the t-distribution, we deduce that $$Y {\buildrel d \over =}\frac{n}{2(n-1)}Z^2 -\frac{1}{2}$$

with $$Z$$ follows the standard t-distribution of $$n-1$$ degrees of freedom.

• How does the variable become your $Y$? Commented Nov 10, 2021 at 20:06
• Just letting you know that your answer (which you deleted) to a currently deleted question has been undeleted after the deleted post was merged with this question of mine. Commented Nov 17, 2021 at 8:12
• @StubbornAtom my answer is the answer for another question :), not the question of the initial author (I thought there was a sum before $X_iX_j$). That’s why I deleted it.
– NN2
Commented Nov 17, 2021 at 9:35