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):
$$\mathbb{E}_{\mathbf{w}\sim\mathcal{S}^{p-1}}\!\left[{w_{1}}^{\!4}\,{w_{2}}^{\!4}\right]\,.$$$$\mathbb{E}_{\mathbf{w}\sim\mathcal{S}^{p-1}}\!\left[{w_1}^{\!4}\,{w_2}^{\!4}\right]\,.$$
Any help would be greatly appreciated!
A related expectation we have already derived:
Using an answer 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]$$\mathbb{E}_{\mathbf{w}\sim\mathcal{S}^{p-1}}\!\left[{w_1}^{\!2}\,{w_2}^{\!2}\right]$ as follows:
\begin{align}1
&=\mathbb{E} \left(\sum_{i=1}^p w_i^2\right)^2=
p(p-1)\cdot
{\mathbb{E}_{\mathbf{w}\sim\mathcal{S}^{p-1}}\!\left[{w_{1}}^{\!2}\,{w_{2}}^{\!2}\right]}+p\cdot \mathbb{E} w_1^4
\\
{\mathbb{E}_{\mathbf{w}\sim\mathcal{S}^{p-1}}\!\left[{w_{1}}^{\!2}\,{w_{2}}^{\!2}\right]}
&
=\frac{1-p\cdot \mathbb{E} w_1^4}{p(p-1)}\stackrel{(*)}{=}
\frac{1}{p\left(p+2\right)}\,,\end{align}\begin{align}1
&=\mathbb{E} \left(\sum_{i=1}^p w_i^2\right)^2 \\[10pt]
& =
p(p-1)\cdot
{\mathbb{E}_{\mathbf{w}\sim\mathcal{S}^{p-1}}\!\left[{w_{1}}^{\!2}\,{w_{2}}^{\!2}\right]}+p\cdot \mathbb{E} w_1^4
\\[10pt]
{\mathbb{E}_{\mathbf{w}\sim\mathcal{S}^{p-1}}\!\left[{w_{1}}^{\!2}\,{w_{2}}^{\!2}\right]}
&
=\frac{1-p\cdot \mathbb{E} w_1^4}{p(p-1)}\stackrel{(*)}{=}
\frac{1}{p\left(p+2\right)}\,,\end{align}
where to show $(*)$ we notice that
$w_{1}^{2}=\frac{X}{X+Y}$ where $X\sim\chi^{2}\left(1\right)=\Gamma\left(\frac{1}{2},\frac{1}{2}\right)$ and $Y\sim\chi^{2}\left(p-1\right)=\Gamma\left(\frac{p-1}{2},\frac{1}{2}\right)$, and conclude that $w_{1}^{2}=\frac{X}{X+Y}\sim B\left(\frac{1}{2},\frac{p-1}{2}\right)$ and
$\mathbb{E}_{w_{1}}\left[w_{1}^{4}\right]
=\text{Var}\left[w_{1}^{2}\right]+\left(\mathbb{E}_{w_{1}}\left[w_{1}^{2}\right]\right)^{2}=\dots=\frac{3}{p\left(p+2\right)}$$\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)}$.
