text: # Markov inequality
  
  • Jan 26th 2011, 01:03 AM
  jovial
  Markov inequality
  Dear All
  I have problem with the following inequality
  
  Let $f = f(x)$ be a nonnegative even function that is strictly increasing for positive $x$. Then for a random variable $\xi$ with $|\xi(\omega)|\leq C$.
  
  $\frac{{Ef(\xi ) - f(\varepsilon )}}{{f(C)}} \le P\left\{ {\left| {\xi - E\xi } \right| \ge \varepsilon } \right\} \le \frac{{Ef(\xi - E\xi )}}{{f(\varepsilon )}}.$
  
  By Markov inequality
  
  $P\{\xi \ge \epsilon\} \le \frac{E\xi}{\epsilon}$
  
  I consider $\eta = |\xi - E\xi|$. Now if we have a function $f: X \rightarrow R^{+}$ then
  
  $P\{f(\eta) \ge f(\epsilon)\} \le \frac{Ef(\eta)}{f(\epsilon)}$
  (Is it correct and why)
  
  Now if $f$ is strictly increasing function then
  
  $P\{\eta \ge \epsilon\}=P\{ f(\eta) \ge f(\epsilon)\} \le \frac{Ef(\eta)}{f(\epsilon)}$
  
  So I obtained:
  
  $P\{|\xi - E\xi| \ge \epsilon\} \le \frac{Ef(|\xi - E\xi|)}{f(\epsilon)}$
  
  How can be proved this part.
  
  $\frac{{Ef(\xi ) - f(\varepsilon )}}{{f(C)}} \le P\left\{ {\left| {\xi - E\xi } \right| \ge \varepsilon } \right\}$
  
  Thanx
  • Jan 31st 2011, 12:38 PM
  Moo
  Hello,
  
  A few remarks, while I'm trying to solve it.
  Quote:
  
  Originally Posted by jovial
  Dear All
  I have problem with the following inequality
  
  Let $f = f(x)$ be a nonnegative even function that is strictly increasing for positive $x$. Then for a random variable $\xi$ with $|\xi(\omega)|\leq C$.
  
  $\frac{{Ef(\xi ) - f(\varepsilon )}}{{f(C)}} \le P\left\{ {\left| {\xi - E\xi } \right| \ge \varepsilon } \right\} \le \frac{{Ef(\xi - E\xi )}}{{f(\varepsilon )}}.$
  
  By Markov inequality
  
  $P\{\xi \ge \epsilon\} \le \frac{E\xi}{\epsilon}$
  
  False. It's $P\{|\xi |\ge\epsilon\}\le\frac{E|\xi|}{\epsilon}$
  
  Quote:
  
  I consider $\eta = |\xi - E\xi|$. Now if we have a function $f: X \rightarrow R^{+}$ then
  
  $P\{f(\eta) \ge f(\epsilon)\} \le \frac{Ef(\eta)}{f(\epsilon)}$
  (Is it correct and why)
  It's correct :
  since f is increasing, $P\{\eta\ge\epsilon\}=P\{f(\eta)\ge f(\epsilon)\}$ and by Markov's inequality, this is $\le \frac{Ef(\eta)}{f(\epsilon)}$
  
  Quote:
  
  Now if $f$ is strictly increasing function then
  
  $P\{\eta \ge \epsilon\}=P\{ f(\eta) \ge f(\epsilon)\} \le \frac{Ef(\eta)}{f(\epsilon)}$
  
  So I obtained:
  
  $P\{|\xi - E\xi| \ge \epsilon\} \le \frac{Ef(|\xi - E\xi|)}{f(\epsilon)}$
  But you're asked for $Ef(\xi-E\xi)$ in the RHS, there is no absolute value.
  Although since f is even, $f(|\xi-E\xi|)=f(\xi-E\xi)$.
  
  P.S. : Are you French ? :p
explanation: The provided text is a forum post from a mathematics discussion board. The user "jovial" presents a problem related to the Markov inequality and asks for help in proving a specific part of an inequality involving a nonnegative even function \( f \) and a random variable \( \xi \). The post includes a detailed attempt at solving the problem, and another user "Moo" provides feedback and corrections.
  
  The text contains a math question and an attempt at an answer, but it is not a complete answer to the question. The user is seeking further assistance to prove a specific part of the inequality.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Markov inequality
  
  • Jan 26th 2011, 01:03 AM
  jovial
  Markov inequality
  Dear All
  I have problem with the following inequality
  
  Let $f = f(x)$ be a nonnegative even function that is strictly increasing for positive $x$. Then for a random variable $\xi$ with $|\xi(\omega)|\leq C$.
  
  $\frac{{Ef(\xi ) - f(\varepsilon )}}{{f(C)}} \le P\left\{ {\left| {\xi - E\xi } \right| \ge \varepsilon } \right\} \le \frac{{Ef(\xi - E\xi )}}{{f(\varepsilon )}}.$
  
  By Markov inequality
  
  $P\{\xi \ge \epsilon\} \le \frac{E\xi}{\epsilon}$
  
  I consider $\eta = |\xi - E\xi|$. Now if we have a function $f: X \rightarrow R^{+}$ then
  
  $P\{f(\eta) \ge f(\epsilon)\} \le \frac{Ef(\eta)}{f(\epsilon)}$
  (Is it correct and why)
  
  Now if $f$ is strictly increasing function then
  
  $P\{\eta \ge \epsilon\}=P\{ f(\eta) \ge f(\epsilon)\} \le \frac{Ef(\eta)}{f(\epsilon)}$
  
  So I obtained:
  
  $P\{|\xi - E\xi| \ge \epsilon\} \le \frac{Ef(|\xi - E\xi|)}{f(\epsilon)}$
  
  How can be proved this part.
  
  $\frac{{Ef(\xi ) - f(\varepsilon )}}{{f(C)}} \le P\left\{ {\left| {\xi - E\xi } \right| \ge \varepsilon } \right\}$
  
  Thanx
  • Jan 31st 2011, 12:38 PM
  Moo
  Hello,
  
  A few remarks, while I'm trying to solve it.
  Quote:
  
  Originally Posted by jovial
  Dear All
  I have problem with the following inequality
  
  Let $f = f(x)$ be a nonnegative even function that is strictly increasing for positive $x$. Then for a random variable $\xi$ with $|\xi(\omega)|\leq C$.
  
  $\frac{{Ef(\xi ) - f(\varepsilon )}}{{f(C)}} \le P\left\{ {\left| {\xi - E\xi } \right| \ge \varepsilon } \right\} \le \frac{{Ef(\xi - E\xi )}}{{f(\varepsilon )}}.$
  
  By Markov inequality
  
  $P\{\xi \ge \epsilon\} \le \frac{E\xi}{\epsilon}$
  
  False. It's $P\{|\xi |\ge\epsilon\}\le\frac{E|\xi|}{\epsilon}$
  
  Quote:
  
  I consider $\eta = |\xi - E\xi|$. Now if we have a function $f: X \rightarrow R^{+}$ then
  
  $P\{f(\eta) \ge f(\epsilon)\} \le \frac{Ef(\eta)}{f(\epsilon)}$
  (Is it correct and why)
  It's correct :
  since f is increasing, $P\{\eta\ge\epsilon\}=P\{f(\eta)\ge f(\epsilon)\}$ and by Markov's inequality, this is $\le \frac{Ef(\eta)}{f(\epsilon)}$
  
  Quote:
  
  Now if $f$ is strictly increasing function then
  
  $P\{\eta \ge \epsilon\}=P\{ f(\eta) \ge f(\epsilon)\} \le \frac{Ef(\eta)}{f(\epsilon)}$
  
  So I obtained:
  
  $P\{|\xi - E\xi| \ge \epsilon\} \le \frac{Ef(|\xi - E\xi|)}{f(\epsilon)}$
  But you're asked for $Ef(\xi-E\xi)$ in the RHS, there is no absolute value.
  Although since f is even, $f(|\xi-E\xi|)=f(\xi-E\xi)$.
  
  P.S. : Are you French ? :p
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The provided text is a forum post from a mathematics discussion board. The user "jovial" presents a problem related to the Markov inequality and asks for help in proving a specific part of an inequality involving a nonnegative even function \( f \) and a random variable \( \xi \). The post includes a detailed attempt at solving the problem, and another user "Moo" provides feedback and corrections.
  
  The text contains a math question and an attempt at an answer, but it is not a complete answer to the question. The user is seeking further assistance to prove a specific part of the inequality.
  
  Label: question-and-answer