**Problem:**  

Given a random isotropic unit vector in $\mathbb{R}^p$ for $p\ge2$, we are trying to compute (preferably exactly, otherwise to upper bound):
$$\mathbb{E}\_{\mathbf{w}\sim\mathcal{S}^{p-1}}\!\left[{w\_1}^{\!4}\,{w\_2}^{\!4}\right]\,.$$


Any help would be greatly appreciated!




---


**A related expectation we have already derived:**  

Using an [answer](https://mathoverflow.net/a/449362/100796) to another question, we were able to compute
$\mathbb{E}\_{\mathbf{w}\sim\mathcal{S}^{p-1}}\!\left[{w\_1}^{\!2}\,{w\_2}^{\!2}\right]$ as follows:
\begin{align}1
&=\mathbb{E} \left(\sum\_{i=1}^p w\_i^2\right)^2 \\[10pt]
& =
p(p-1)\cdot
{\mathbb{E}\_{\mathbf{w}\sim\mathcal{S}^{p-1}}\!\left[{w\_{1}}^{\!2}\,{w\_{2}}^{\!2}\right]}+p\cdot \mathbb{E} w\_1^4
\\[10pt]
{\mathbb{E}\_{\mathbf{w}\sim\mathcal{S}^{p-1}}\!\left[{w\_{1}}^{\!2}\,{w\_{2}}^{\!2}\right]}
&
=\frac{1-p\cdot \mathbb{E} w\_1^4}{p(p-1)}\stackrel{(\*)}{=}
\frac{1}{p\left(p+2\right)}\,,\end{align}
where to show $(\*)$ we notice that
$w\_{1}^{2}=\frac{X}{X+Y}$ where $X\sim\chi^{2}\left(1\right)=\Gamma\left(\frac{1}{2},\frac{1}{2}\right)$ and $Y\sim\chi^{2}\left(p-1\right)=\Gamma\left(\frac{p-1}{2},\frac{1}{2}\right)$, and conclude that $w\_{1}^{2}=\frac{X}{X+Y}\sim B\left(\frac{1}{2},\frac{p-1}{2}\right)$ and
$\mathbb{E}\_{w\_1}\left[w\_1^4\right]
=\operatorname{Var}\left[w\_1^2 \right] + \left(\mathbb{E}\_{w\_1} \left[w\_1^2 \right] \right)^2= \dots =\frac{3}{p(p+2)}$.



Expectation of a function of two entries of an isotropic unit vector $\mathbb{E}_{\mathbf{w}\sim\mathcal{S}^{p-1}}\![{w_{1}}^{\!4}\,{w_{2}}^{\!4}]$
As noted in the [previous answer](https://mathoverflow.net/a/449401/36721),
the joint distribution of $w\_1^2$ and $w\_2^2$ is the [Dirichlet distribution](https://en.wikipedia.org/wiki/Dirichlet_distribution#Definitions) with parameters $1/2,1/2,p/2-1$. [Therefore](https://en.wikipedia.org/wiki/Dirichlet_distribution#Moments), after some simple calculations we get
$$Ew\_1^4w\_2^4=\frac9{p(p+2)(p+4)(p+6)}. $$


