Let $X$ be random vector on the unit-sphere $S\_{n-1}$ in $\mathbb R^n$. We don't assume that the distribution of $X$ is uniform on $S\_{n-1}$



> 
> I'm interested in proving the existence of a (deterministic) direction $v \in S\_{n-1}$ such that
> $$
> \mathbb P(|\langle X,v\rangle | \ge \alpha) \ge \beta,
> \tag{1}
> $$
> where $\alpha,\beta \in (0,1]$ are absolute constants.
> 
> 
> 


Thus in a sense, the random vector $X$ "likes" the direction $v$. The larger the constants $\alpha,\beta$ the better. For example, if the support of $X$ is made up of a bounded number of atoms (i.e not dependent on the dimension $n$), then (1) holds.


**Assumption.** Suppose $X$ admits a second-moment matrix $\Sigma := \mathbb E\, X \otimes X \in \mathbb R^{n \times n}$ and largest eigenvalue $s\_1$ of $\Sigma$ is lower-bounded by an absolute constant $c>0$.


Let $v$ be a unit-vector in the corresponding eigenspace. Since $Z:=|\langle X,v\rangle|^2$ takes values in the interval $[0,1]$, it is clear that the variance $\sigma^2$ of $Z$ is at most $1/4$. On the other hand, by linearity of trace and expectation, the expectation of $Z$ is $$
\mathbb E Z = \mathbb E [\mbox{trace}((v\otimes v)(X \otimes X))] = \mbox{trace}((v \otimes v)\Sigma) = v^\top \Sigma v = v^\top (\Sigma v) = s\_1\|v\|^2 = s\_1.
$$


Therefore, by Cantelli's inequality, for any $0<a<s\_1$, we have $$
\mathbb P(Z \ge s\_1-a) \ge 1-\sigma^2/(\sigma^2+a^2)=(1+a^2/\sigma^2)^{-1} \in (0,1).
$$


That is, for any $0 < a < s\_1$, (1) holds with $\alpha = s\_1 - a$ and $\beta=(1+a^2/\sigma^2)^{-1}$.



> 
> **Question.** *Is my above reasoning correct ? Is there an alternative / more powerful way to use the above assumptions to obtain a stronger inequality (i.e large $\alpha$ and $\beta$) of the form (1) ?*
> 
> 
> 



Existence of preferred direction for a random vector with arbitrary distribution on sphere, under a condition on its covariance matrix
$\newcommand\R{\mathbb R}\newcommand{\Si}{\Sigma}\newcommand{\si}{\sigma}$The answer is no: in general (and usually) there are no positive absolute constants $a$ and $b$ such that for some unit vector $v$ one has
$$P(|X\cdot v|\ge a)\ge b.$$


Indeed, otherwise one would have $E(X\cdot v)^2\ge c:=ba^2>0$. However, if $X$ is uniformly distributed on the unit sphere in $\mathbb R^n$ and $v$ is a unit vector, then $(X\cdot v)^2$ has the beta distribution with parameters $1/2,(n-1)/2$ and hence $E(X\cdot v)^2=1/n<c$ if $n>1/c$.




---


The OP has rectified the confusion raised by the initial formulation of the their question. The changes invalidate the above answer. Here is an updated answer to the current state of the question.


Let $Y:=|X\cdot v|$, where $v$ is a unit eigenvector corresponding to the eigenvalue $s\_1$. Then, $0\le Y\le1$ and $EY^2=s\_1$. So, for all $a\in(0,1)$ we have the inequality
$$1(Y>a)\ge\frac{Y^2-a^2}{1-a^2},$$
with the equality on the event $\{Y\in\{a,1\}\}$,
and hence taking expectations gives
\begin{equation}
 P(|X\cdot v|>a)=P(Y>a)\ge\frac{\max(0,s\_1-a^2)}{1-a^2}. \tag{1}\label{1}
\end{equation}


This lower bound on $P(|X\cdot v|>a)$ is exact: It is attained if


(i) $a^2\le s\_1\le1$ and $(X\cdot v)^2$ only takes values $a^2$ and $1$ (with mean $E(X\cdot v)^2=s\_1\in[a^2,1]$) or if


(ii) $0\le s\_1<a^2$ and $(X\cdot v)^2$ only takes value $s\_1$.


**Addendum 1:** Strictly speaking, to show that the lower bound on $P(|X\cdot v|>a)$ in \eqref{1} is exact, we also need to show that


(I) for any $s\_1\in[a^2,1]$ and at least for some $n$, there exist a random unit vector $X$ in $\R^n$ and a unit vector $v\in\R^n$ such that $(X\cdot v)^2$ only takes values $a^2$ and $1$, with mean $E(X\cdot v)^2=s\_1$, and, moreover, $s\_1$ is the largest eigenvalue of the covariance matrix $\Si$ of $X$;


(II) for any $s\_1\in(0,a^2)$ and at least for some $n$, there exist a random unit vector $X$ in $\R^n$ and a unit vector $v\in\R^n$ such that $(X\cdot v)^2$ only takes value $s\_1$ and, moreover, $s\_1$ is the largest eigenvalue of the covariance matrix $\Si$ of $X$.


To prove (I), do take any $s\_1\in[a^2,1]$, and take any unit vector $v\in\R^n$, where $n\ge2$.
Then let $\mu\_X=p\mu\_V+\frac q2\,\mu\_W+\frac q2\,\mu\_{-W}$, where $\mu\_Y$ denotes the distribution of a random vector $Y$,
\begin{equation}
 p:=\frac{s\_1-a^2}{1-a^2},\quad q:=1-p=\frac{1-s\_1}{1-a^2}, 
\end{equation}
$P(V=v)=P(V=-v)=1/2$, $W:=av+\sqrt{1-a^2}\,U$, and $U$ is uniformly distributed on the unit sphere of the vector space that is the orthogonal complement of $\text{span}(\{v\})$ to $\R^n$. Then $P((X\cdot v)^2=1)=p$, $P((X\cdot v)^2=a^2)=q$, $E(X\cdot v)^2=s\_1$, and the eigenvalues of the covariance matrix $\Si$ of $X$ are $s\_1$ and $\dfrac{1-s\_1}{n-1}$ (the latter one of multiplicity $n-1$). So, if $n\ge1/s\_1$, then $s\_1$ is the largest eigenvalue of $\Si$. Thus, all the desired conditions are satisfied, and (I) is proved.


The proof of (II) is similar, and a bit simpler. Here, we let $\mu\_X=\frac12\,\mu\_T+\frac12\,\mu\_{-T}$, where $T:=\sqrt{s\_1}\,v+\sqrt{1-s\_1}\,U$.


**Addendum 2:** The lower bound in your post depends on the variance $\si^2$ of $Z=Y^2$. You did not fully specify a value of $\si^2$, just noting that $\si^2\le1/4$, since $0\le Z\le1$. The latter bound on $\si^2$ can be improved to the optimal (in terms of $EZ$) bound $EZ(1-EZ)=s\_1(1-s\_1)$, which is attained when $Z$ only takes values $0$ and $1$. It now follows that your lower bound cannot be exact -- because, as seen from above, the exact lower bound is only attained when $Z=(X\cdot v)^2$ takes values in the set $\{a^2,s\_1,1\}$, which does not contain the value $0$. Also, it can probably be shown directly that, in distinction from the lower bound on $P(|X\cdot v|>a)$ in \eqref{1}, the lower bound in your post cannot be exact.


