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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 of $Z$ is at most $1/4$. On the other hand, by linearity of trace and expectation, the expectation of $Z$ is $$ \mathbb E Z = \mathbb E [\mbox{trace}((v\otimes v)(X \otimes X))] = \mbox{trace}(\Sigma (v \otimes v)) = v^\top \Sigma v = v^\top (\Sigma v) = s_1\|v\|^2 = s_1. $$

Therefore, by Chebychev's inequality, for any $a>0$, we have $\mathbb P(|Z-s_1| \le a) \ge 1-1/(2a)^2$. Taking $a = ts_1$ with $1/(2s_1)< t < 1$, then gives $\mathbb P(Z > (1-t)s_1) \ge 1-2/(2ts_1)^2$.

That is, (1) holds with $\alpha = (1-t)s_1$ and $\beta=1-1/(2ts_1)^2$.

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) ?

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) ?

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.

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 of $Z$ is at most $1/4$. On the other hand, by linearity of trace and expectation, the expectation of $Z$ is $$ \mathbb E Z = \mathbb E [\mbox{trace}((v\otimes v)(X \otimes X))] = \mbox{trace}(\Sigma (v \otimes v)) = v^\top \Sigma v = v^\top (\Sigma v) = s_1\|v\|^2 = s_1. $$

Therefore, by Chebychev's inequality, for any $a>0$, we have $\mathbb P(|Z-s_1| \le a) \ge 1-1/(2a)^2$. Taking $a = ts_1$ with $1/(2s_1)< t < 1$, then gives $\mathbb P(Z > (1-t)s_1) \ge 1-2/(2ts_1)^2$.

That is, (1) holds with $\alpha = (1-t)s_1$ and $\beta=1-1/(2ts_1)^2$.

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) ?

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 of $Z$ is at most $1/4$. On the other hand, by linearity of trace and expectation, the expectation of $Z$ is $$ \mathbb E Z = \mathbb E [\mbox{trace}((v\otimes v)(X \otimes X))] = \mbox{trace}(\Sigma (v \otimes v)) = v^\top \Sigma v = v^\top (\Sigma v) = s_1\|v\|^2 = s_1. $$

Therefore, by Chebychev's inequality, for any $a>0$, we have $\mathbb P(|Z-s_1| \le a) \ge 1-1/(2a)^2$. Taking $a = ts_1$ with $1/(2s_1)< t < 1$, then gives $\mathbb P(Z > (1-t)s_1) \ge 1-2/(2ts_1)^2$.

That is, (1) holds with $\alpha = (1-t)s_1$ and $\beta=1-1/(2ts_1)^2$.

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) ?

Let $X$ be random vector with an arbitrary distribution on the unit-sphere $S_{d-1}$ in $\mathbb R^n$.

I'm interested in proving the existence of a (deterministic) direction $v \in S_{d-1}$ 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 $X$ has standard gaussian iid components, then (1) holds.

Assumption. Suppose $X$ admits a second-moment matrix $\Sigma := \mathbb E\, X \otimes X \in \mathbb R^{n \times n}$ and $s_1$ 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 of $Z$ is at most $1/4$. On the other hand, by linearity of trace and expectation, the expectation of $Z$ is $$ \mathbb E Z = \mathbb E [\mbox{trace}((v\otimes v)(X \otimes X))] = \mbox{trace}(\Sigma (v \otimes v)) = v^\top \Sigma v = v^\top (\Sigma v) = s_1\|v\|^2 = s_1. $$

Therefore, by Chebychev's inequality, for any $a>0$, we have $\mathbb P(|Z-s_1| \le a) \ge 1-1/(2a)^2$. Taking $a = ts_1$ with $1/(2s_1)< t < 1$, then gives $\mathbb P(Z > (1-t)s_1) \ge 1-2/(2ts_1)^2$.

That is, (1) holds with $\alpha = (1-t)s_1$ and $\beta=1-1/(2ts_1)^2$.

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) ?

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.

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 of $Z$ is at most $1/4$. On the other hand, by linearity of trace and expectation, the expectation of $Z$ is $$ \mathbb E Z = \mathbb E [\mbox{trace}((v\otimes v)(X \otimes X))] = \mbox{trace}(\Sigma (v \otimes v)) = v^\top \Sigma v = v^\top (\Sigma v) = s_1\|v\|^2 = s_1. $$

Therefore, by Chebychev's inequality, for any $a>0$, we have $\mathbb P(|Z-s_1| \le a) \ge 1-1/(2a)^2$. Taking $a = ts_1$ with $1/(2s_1)< t < 1$, then gives $\mathbb P(Z > (1-t)s_1) \ge 1-2/(2ts_1)^2$.

That is, (1) holds with $\alpha = (1-t)s_1$ and $\beta=1-1/(2ts_1)^2$.

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) ?