Skip to main content
deleted 644 characters in body
Source Link
Elwood
  • 562
  • 2
  • 12

Let $p \ge 1$ be an integer. Does there exist a constant $C_p$ such that for every random variable $X \ge 0$, $$ \mathbb{E} \left[ \left(X - \mathbb{E} \left[ X \right] \right)^{2p} \right] \le C_p \mathbb{E} \left[ \left(X^p - \mathbb{E} \left[ X^p \right] \right)^{2} \right] \ \ ? $$

EDIT: the answer is yes. Let $Y$ be an independent random variable distributed as $X$. We have $$ \|X - \mathbb{E}[X]\|_{L^{2p}} = \|\mathbb{E}\left[X - Y \, | \, X\right]\|_{L^{2p}} \le \|X - Y\|_{L^{2p}}. $$ Moreover, there exists a constant $C_p$ such that for every $x,y \ge 0$, $$ |x-y|^p \le C_p |x^p - y^p|. $$ Indeed, it suffices to verify this for $x = 1$ and $y \in [0,1]$ by homogeneity and symmetry. This is then a simple exercise. As a consequence, we deduce $$ \|X - \mathbb{E}[X]\|_{L^{2p}} \le C_p \||X^p - Y^p|^{1/p}\|_{L^{2p}} = C_p \|X^p - Y^p\|_{L^{2}}^{1/p}. $$ We conclude by the triangle inequality.

Let $p \ge 1$ be an integer. Does there exist a constant $C_p$ such that for every random variable $X \ge 0$, $$ \mathbb{E} \left[ \left(X - \mathbb{E} \left[ X \right] \right)^{2p} \right] \le C_p \mathbb{E} \left[ \left(X^p - \mathbb{E} \left[ X^p \right] \right)^{2} \right] \ \ ? $$

EDIT: the answer is yes. Let $Y$ be an independent random variable distributed as $X$. We have $$ \|X - \mathbb{E}[X]\|_{L^{2p}} = \|\mathbb{E}\left[X - Y \, | \, X\right]\|_{L^{2p}} \le \|X - Y\|_{L^{2p}}. $$ Moreover, there exists a constant $C_p$ such that for every $x,y \ge 0$, $$ |x-y|^p \le C_p |x^p - y^p|. $$ Indeed, it suffices to verify this for $x = 1$ and $y \in [0,1]$ by homogeneity and symmetry. This is then a simple exercise. As a consequence, we deduce $$ \|X - \mathbb{E}[X]\|_{L^{2p}} \le C_p \||X^p - Y^p|^{1/p}\|_{L^{2p}} = C_p \|X^p - Y^p\|_{L^{2}}^{1/p}. $$ We conclude by the triangle inequality.

Let $p \ge 1$ be an integer. Does there exist a constant $C_p$ such that for every random variable $X \ge 0$, $$ \mathbb{E} \left[ \left(X - \mathbb{E} \left[ X \right] \right)^{2p} \right] \le C_p \mathbb{E} \left[ \left(X^p - \mathbb{E} \left[ X^p \right] \right)^{2} \right] \ \ ? $$

added 644 characters in body
Source Link
Elwood
  • 562
  • 2
  • 12

Let $p \ge 1$ be an integer. Does there exist a constant $C_p$ such that for every random variable $X \ge 0$, $$ \mathbb{E} \left[ \left(X - \mathbb{E} \left[ X \right] \right)^{2p} \right] \le C_p \mathbb{E} \left[ \left(X^p - \mathbb{E} \left[ X^p \right] \right)^{2} \right] \ \ ? $$

EDIT: the answer is yes. Let $Y$ be an independent random variable distributed as $X$. We have $$ \|X - \mathbb{E}[X]\|_{L^{2p}} = \|\mathbb{E}\left[X - Y \, | \, X\right]\|_{L^{2p}} \le \|X - Y\|_{L^{2p}}. $$ Moreover, there exists a constant $C_p$ such that for every $x,y \ge 0$, $$ |x-y|^p \le C_p |x^p - y^p|. $$ Indeed, it suffices to verify this for $x = 1$ and $y \in [0,1]$ by homogeneity and symmetry. This is then a simple exercise. As a consequence, we deduce $$ \|X - \mathbb{E}[X]\|_{L^{2p}} \le C_p \||X^p - Y^p|^{1/p}\|_{L^{2p}} = C_p \|X^p - Y^p\|_{L^{2}}^{1/p}. $$ We conclude by the triangle inequality.

Let $p \ge 1$ be an integer. Does there exist a constant $C_p$ such that for every random variable $X \ge 0$, $$ \mathbb{E} \left[ \left(X - \mathbb{E} \left[ X \right] \right)^{2p} \right] \le C_p \mathbb{E} \left[ \left(X^p - \mathbb{E} \left[ X^p \right] \right)^{2} \right] \ \ ? $$

Let $p \ge 1$ be an integer. Does there exist a constant $C_p$ such that for every random variable $X \ge 0$, $$ \mathbb{E} \left[ \left(X - \mathbb{E} \left[ X \right] \right)^{2p} \right] \le C_p \mathbb{E} \left[ \left(X^p - \mathbb{E} \left[ X^p \right] \right)^{2} \right] \ \ ? $$

EDIT: the answer is yes. Let $Y$ be an independent random variable distributed as $X$. We have $$ \|X - \mathbb{E}[X]\|_{L^{2p}} = \|\mathbb{E}\left[X - Y \, | \, X\right]\|_{L^{2p}} \le \|X - Y\|_{L^{2p}}. $$ Moreover, there exists a constant $C_p$ such that for every $x,y \ge 0$, $$ |x-y|^p \le C_p |x^p - y^p|. $$ Indeed, it suffices to verify this for $x = 1$ and $y \in [0,1]$ by homogeneity and symmetry. This is then a simple exercise. As a consequence, we deduce $$ \|X - \mathbb{E}[X]\|_{L^{2p}} \le C_p \||X^p - Y^p|^{1/p}\|_{L^{2p}} = C_p \|X^p - Y^p\|_{L^{2}}^{1/p}. $$ We conclude by the triangle inequality.

Source Link
Elwood
  • 562
  • 2
  • 12

Is this simple-looking moment inequality true?

Let $p \ge 1$ be an integer. Does there exist a constant $C_p$ such that for every random variable $X \ge 0$, $$ \mathbb{E} \left[ \left(X - \mathbb{E} \left[ X \right] \right)^{2p} \right] \le C_p \mathbb{E} \left[ \left(X^p - \mathbb{E} \left[ X^p \right] \right)^{2} \right] \ \ ? $$