# 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
The text passage is a discussion about a mathematical problem involving the Markov inequality and the proof of a specific inequality related to a random variable and a function. The passage contains a question about how to prove a part of the inequality and includes some steps of the proof.
single_question_with_answer

the passage contains a specific question about proving a part of the inequality and provides some steps and reasoning that serve as an answer or attempt to solve the problem.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# 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
user
The text passage is a discussion about a mathematical problem involving the Markov inequality and the proof of a specific inequality related to a random variable and a function. The passage contains a question about how to prove a part of the inequality and includes some steps of the proof.

TYPE: SINGLE_QUESTION_WITH_ANSWER

The passage contains a specific question about proving a part of the inequality and provides some steps and reasoning that serve as an answer or attempt to solve the problem.