Let $d$ be a large positive integer. Let $x$ be uniformly distributed on the unit-sphere in $\mathbb R^d$ and let $a$ and $b$ be perpendicular vectors in $\mathbb R^d$, i.e such that $a^\top b=0$. Let $f:\mathbb R \to \mathbb R$ be a continuous function which is twice-continuously differentiable at $0$ (you may assume more regularity if that helps).



> 
> **Question.** *How large can the absolute difference* $$
> \Delta(a,b)=|\mathbb E[f(x^\top a)f(x^\top b)] - \mathbb E[f(x^\top a)]\mathbb E[f(x^\top b)]|
> $$
> *be as a function of the dimension $d$ ?*
> 
> 
> 


**Observation 1.** In the case where $x \sim N(0,I\_d)$, or more generally, with any diagonal covariance matrix, we have $\Delta(a,b) = 0$. Indeed, in this case, the joint distribution of $(x^\top a,x^\top b)$ is $N(0, C)$, where $C$ is the $2 \times 2$ psd matrix with entries given by


* $c\_{11}=\mathbb E[(x^\top a)^2] = \mbox{trace}(aa^\top)=\|a\|^2$,
* $c\_{12}=c\_{21}=\mathbb E[(x^\top a)(x^\top b)] = \mbox{trace}(ab^\top) = a^\top b = 0$,
* $c\_{22}=\mathbb E[(x^\top b)^2] = \mbox{trace}(bb^\top)=\|b\|^2$.


Thus, $x^\top a$ and $x^\top b$ are **jointly gaussian** random variables with zero correlation, and so are independent. Thus $\Delta(a,b) = 0$.


**Observation 2.** Since $d$ is large both $x^\top a$ and $x^\top b$ are close to centered normal random variables with variance $\|a\|^2/d$ and $\|b\|^2/d$ respectively, so maybe an adaptation of the previous argument might be of use ?



Almost independence of $x^\top a$ and $x^\top b$ for $x$ uniform on the sphere in $\mathbb R^d$ and $a,b \in \mathbb R^d$ with $a^\top b = 0$
$\newcommand{\de}{\delta}\newcommand{\ep}{\varepsilon}$The covariance of two random variables (r.v.'s) does not change if one of them is shifted by a constant. So, without loss of generality $f(0)=0$. Let $n:=d$.


To compute the asymptotics, we need to assume that
\begin{equation\*}
 f(x)=Ax+Bx^2+Cx^3+Dx^4+e\_1x^5+O(x^6) \tag{1}
\end{equation\*}
for some real $A,B,C,D,e\_1$ and all real $x\in[-1,1]$.


The joint distribution of $x^\top a$ and $x^\top b$ is the same as that of $Y\_1$ and $Y\_2$, where
\begin{equation\*}
 Y\_j:=X\_j/|X|,
\end{equation\*}
$X\_1,\dots,X\_n$ are iid standard normal r.v.'s and $|X|:=\sqrt{X\_1^2+\dots+X\_n^2}$. So, we want to find the asymptotics of
\begin{equation\*}
 \de:=Ef(Y\_1)f(Y\_2)-(Ef(Y\_1))^2 \tag{2}
\end{equation\*}
(as $n\to\infty$).
Note that
\begin{equation\*}
 EY\_1^2=\dots=EY\_n^2=\frac1n, \tag{3}
\end{equation\*}
since the $Y\_i$'s are exchangeable and $Y\_1^2+\dots+Y\_n^2=1$. Also, using e.g. a Chernoff exponential concentration inequality for $|X|$, we see that $|X|$ is highly concentrated near $\sqrt n$, say in the sense that, for each real $\ep>0$,  

$P(|\,|X|-\sqrt n|>\ep\sqrt n)$ goes to $0$ faster than any negative power of $n$. It follows that for each real $p>0$
\begin{equation\*}
 E|Y\_1|^p\sim E|X\_1|^p n^{-p/2}. \tag{4}
\end{equation\*}


So, by (1) and the symmetry of (the distribution of) $Y\_1$,  

\begin{equation\*}
 Ef(Y\_1)=\frac Bn+\frac{(3+o(1))D}{n^2}. \tag{5}
\end{equation\*}


Next, by (1), the symmetry of the $Y\_i$'s, and (4)
\begin{equation\*}
 Ef(Y\_1)f(Y\_1)\\
 =B^2\,EY\_1^2Y\_2^2+2BD\,EY\_1^2Y\_2^4+o(n^{-3}). \tag{6}
\end{equation\*}
Similarly to (4),
\begin{equation\*}
 EY\_1^2Y\_2^4\sim EX\_1^2X\_2^4 n^{-3}=\frac3{n^3}. \tag{7}
\end{equation\*}


The main difficulty here is to estimate $EY\_1^2Y\_2^2$. We have
\begin{equation}
 EY\_1^2Y\_2^2=Eh(G), \tag{8}
\end{equation}
where $G^2$ has the $\chi^2$ distribution with $n-2$ degrees of freedom and
\begin{align\*}
 h(x)&:=E\frac{X\_1^2X\_2^2}{(X\_1^2+X\_2^2+x^2)^2} \\ 
 &=\frac1{2\pi}\int\_0^{2\pi}dt\int\_0^\infty r\,dr\,e^{-r^2/2}\frac{r^4\cos^2t\sin^2t}{(r^2+x^2)^2} \\ 
 &=\frac1{16}\int\_0^\infty du\,\frac{u^2e^{-u/2}}{(u+x^2)^2} \\ 
 &=\frac{x^2}{16}\int\_0^\infty dt\,\frac{t^2e^{-x^2 t/2}}{(1+t)^2}. 
\end{align\*}
Writing $1/(1+t)^2=1-2t+O(t^2)$ for $t\in(0,1)$, we now get
\begin{equation}
 h(x)=\frac1{16}\,(x^{-4}-(12+o(1))x^{-6}) \tag{9}
\end{equation}
as $x\to\infty$.
Also,
\begin{equation}
 EG^{-4}=\frac1{n^2}+\frac{10+o(1)}{n^3},\quad EG^{-6}=\frac{1+o(1)}{n^3}.
\end{equation}
Hence, in view of (9),
\begin{equation}
 EY\_1^2Y\_2^2=Eh(G)=\frac1{n^2}-\frac{2+o(1)}{n^3}. 
\end{equation}


Finally, recalling (2), (5), and (7), we get
\begin{equation}
 \de=-\frac{2B^2+o(1)}{n^3}. 
\end{equation}
So, as should be expected, we have a small negative correlation between $x^\top a$ and $x^\top b$ (on the order of $1/n^2$, in view of (3)).


