Prove the following using Chebyshev and Markov inequality.












2












$begingroup$


Suppose $X$ is a random variable with mean $mu$ and variance $sigma^2$. Show that
$$P(|X-mu| geq ksigma) leq frac{1}{k^2}$$



Question: I know the exercise wants me to use Markov inequality and Chebyshev inequality, but I can't reach the same answer. If someone can help me, it will be appreciated.



Thanks!






probability probability-theory statistics







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$endgroup$








  • 1




    $begingroup$
    You have stated the standard Chebyshev Inequality. Proofs can be found in many places, including the Wikipedia article.
    $endgroup$
    – André Nicolas
    Nov 28 '15 at 22:12










  • $begingroup$
    Usually the Chebyshev inequality is either written that way or as $P(|X-mu| geq varepsilon) leq frac{sigma^2}{varepsilon^2}$. Starting from the latter way of writing it, you can identify $k$ such that $varepsilon=k sigma$, and then the result follows.
    $endgroup$
    – Ian
    Nov 28 '15 at 22:17










  • $begingroup$
    "I know the exercise want me to use Markov inequality and chebyshev inequality, but I can't reach the same answer." Which answer can you reach?
    $endgroup$
    – Did
    Nov 28 '15 at 22:35
















2












$begingroup$


Suppose $X$ is a random variable with mean $mu$ and variance $sigma^2$. Show that
$$P(|X-mu| geq ksigma) leq frac{1}{k^2}$$



Question: I know the exercise wants me to use Markov inequality and Chebyshev inequality, but I can't reach the same answer. If someone can help me, it will be appreciated.



Thanks!






probability probability-theory statistics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You have stated the standard Chebyshev Inequality. Proofs can be found in many places, including the Wikipedia article.
    $endgroup$
    – André Nicolas
    Nov 28 '15 at 22:12










  • $begingroup$
    Usually the Chebyshev inequality is either written that way or as $P(|X-mu| geq varepsilon) leq frac{sigma^2}{varepsilon^2}$. Starting from the latter way of writing it, you can identify $k$ such that $varepsilon=k sigma$, and then the result follows.
    $endgroup$
    – Ian
    Nov 28 '15 at 22:17










  • $begingroup$
    "I know the exercise want me to use Markov inequality and chebyshev inequality, but I can't reach the same answer." Which answer can you reach?
    $endgroup$
    – Did
    Nov 28 '15 at 22:35














2












2








2


1



$begingroup$


Suppose $X$ is a random variable with mean $mu$ and variance $sigma^2$. Show that
$$P(|X-mu| geq ksigma) leq frac{1}{k^2}$$



Question: I know the exercise wants me to use Markov inequality and Chebyshev inequality, but I can't reach the same answer. If someone can help me, it will be appreciated.



Thanks!






probability probability-theory statistics







share|cite|improve this question











$endgroup$




Suppose $X$ is a random variable with mean $mu$ and variance $sigma^2$. Show that
$$P(|X-mu| geq ksigma) leq frac{1}{k^2}$$



Question: I know the exercise wants me to use Markov inequality and Chebyshev inequality, but I can't reach the same answer. If someone can help me, it will be appreciated.



Thanks!






probability probability-theory statistics




probability probability-theory statistics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 10:07









Rócherz

2,7762721




2,7762721










asked Nov 28 '15 at 22:05









Electro82Electro82

536218




536218








  • 1




    $begingroup$
    You have stated the standard Chebyshev Inequality. Proofs can be found in many places, including the Wikipedia article.
    $endgroup$
    – André Nicolas
    Nov 28 '15 at 22:12










  • $begingroup$
    Usually the Chebyshev inequality is either written that way or as $P(|X-mu| geq varepsilon) leq frac{sigma^2}{varepsilon^2}$. Starting from the latter way of writing it, you can identify $k$ such that $varepsilon=k sigma$, and then the result follows.
    $endgroup$
    – Ian
    Nov 28 '15 at 22:17










  • $begingroup$
    "I know the exercise want me to use Markov inequality and chebyshev inequality, but I can't reach the same answer." Which answer can you reach?
    $endgroup$
    – Did
    Nov 28 '15 at 22:35














  • 1




    $begingroup$
    You have stated the standard Chebyshev Inequality. Proofs can be found in many places, including the Wikipedia article.
    $endgroup$
    – André Nicolas
    Nov 28 '15 at 22:12










  • $begingroup$
    Usually the Chebyshev inequality is either written that way or as $P(|X-mu| geq varepsilon) leq frac{sigma^2}{varepsilon^2}$. Starting from the latter way of writing it, you can identify $k$ such that $varepsilon=k sigma$, and then the result follows.
    $endgroup$
    – Ian
    Nov 28 '15 at 22:17










  • $begingroup$
    "I know the exercise want me to use Markov inequality and chebyshev inequality, but I can't reach the same answer." Which answer can you reach?
    $endgroup$
    – Did
    Nov 28 '15 at 22:35








1




1




$begingroup$
You have stated the standard Chebyshev Inequality. Proofs can be found in many places, including the Wikipedia article.
$endgroup$
– André Nicolas
Nov 28 '15 at 22:12




$begingroup$
You have stated the standard Chebyshev Inequality. Proofs can be found in many places, including the Wikipedia article.
$endgroup$
– André Nicolas
Nov 28 '15 at 22:12












$begingroup$
Usually the Chebyshev inequality is either written that way or as $P(|X-mu| geq varepsilon) leq frac{sigma^2}{varepsilon^2}$. Starting from the latter way of writing it, you can identify $k$ such that $varepsilon=k sigma$, and then the result follows.
$endgroup$
– Ian
Nov 28 '15 at 22:17




$begingroup$
Usually the Chebyshev inequality is either written that way or as $P(|X-mu| geq varepsilon) leq frac{sigma^2}{varepsilon^2}$. Starting from the latter way of writing it, you can identify $k$ such that $varepsilon=k sigma$, and then the result follows.
$endgroup$
– Ian
Nov 28 '15 at 22:17












$begingroup$
"I know the exercise want me to use Markov inequality and chebyshev inequality, but I can't reach the same answer." Which answer can you reach?
$endgroup$
– Did
Nov 28 '15 at 22:35




$begingroup$
"I know the exercise want me to use Markov inequality and chebyshev inequality, but I can't reach the same answer." Which answer can you reach?
$endgroup$
– Did
Nov 28 '15 at 22:35










2 Answers
2






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1












$begingroup$

The Chebyshev inequality is
$$mathbb{P}(|x - mu| geq a) leq frac{sigma^2}{a^2}$$ .Substituting $$a = ksigma $$gives the answer.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Thanks André, I found it.



    For any event $A$, let $I_A$ be the indicator random variable of $A$, i.e. $I_A$ equals $1$ if $A$ occurs and $0$ otherwise. Then $newcommand{E}{operatorname{E}}$
    begin{align*}
    Pr(|X-mu| geq ksigma) &=E(I_{|X-mu| geq ksigma}) \
    &= Eleft(I_{left(frac{X-mu}{ksigma}right)^2 geq 1}right) \
    &leq Eleft(left(frac{X-mu}{ksigma}right)^2right) \
    &=frac{1}{k^2} cdot frac{E((X-mu)^2)}{ksigma} \
    &=frac{1}{k^2}.end{align*}






    share|cite|improve this answer











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      2 Answers
      2






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      2 Answers
      2






      active

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      active

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      active

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      1












      $begingroup$

      The Chebyshev inequality is
      $$mathbb{P}(|x - mu| geq a) leq frac{sigma^2}{a^2}$$ .Substituting $$a = ksigma $$gives the answer.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        The Chebyshev inequality is
        $$mathbb{P}(|x - mu| geq a) leq frac{sigma^2}{a^2}$$ .Substituting $$a = ksigma $$gives the answer.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          The Chebyshev inequality is
          $$mathbb{P}(|x - mu| geq a) leq frac{sigma^2}{a^2}$$ .Substituting $$a = ksigma $$gives the answer.






          share|cite|improve this answer









          $endgroup$



          The Chebyshev inequality is
          $$mathbb{P}(|x - mu| geq a) leq frac{sigma^2}{a^2}$$ .Substituting $$a = ksigma $$gives the answer.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 28 '15 at 22:18









          Milen IvanovMilen Ivanov

          884311




          884311























              1












              $begingroup$

              Thanks André, I found it.



              For any event $A$, let $I_A$ be the indicator random variable of $A$, i.e. $I_A$ equals $1$ if $A$ occurs and $0$ otherwise. Then $newcommand{E}{operatorname{E}}$
              begin{align*}
              Pr(|X-mu| geq ksigma) &=E(I_{|X-mu| geq ksigma}) \
              &= Eleft(I_{left(frac{X-mu}{ksigma}right)^2 geq 1}right) \
              &leq Eleft(left(frac{X-mu}{ksigma}right)^2right) \
              &=frac{1}{k^2} cdot frac{E((X-mu)^2)}{ksigma} \
              &=frac{1}{k^2}.end{align*}






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                Thanks André, I found it.



                For any event $A$, let $I_A$ be the indicator random variable of $A$, i.e. $I_A$ equals $1$ if $A$ occurs and $0$ otherwise. Then $newcommand{E}{operatorname{E}}$
                begin{align*}
                Pr(|X-mu| geq ksigma) &=E(I_{|X-mu| geq ksigma}) \
                &= Eleft(I_{left(frac{X-mu}{ksigma}right)^2 geq 1}right) \
                &leq Eleft(left(frac{X-mu}{ksigma}right)^2right) \
                &=frac{1}{k^2} cdot frac{E((X-mu)^2)}{ksigma} \
                &=frac{1}{k^2}.end{align*}






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Thanks André, I found it.



                  For any event $A$, let $I_A$ be the indicator random variable of $A$, i.e. $I_A$ equals $1$ if $A$ occurs and $0$ otherwise. Then $newcommand{E}{operatorname{E}}$
                  begin{align*}
                  Pr(|X-mu| geq ksigma) &=E(I_{|X-mu| geq ksigma}) \
                  &= Eleft(I_{left(frac{X-mu}{ksigma}right)^2 geq 1}right) \
                  &leq Eleft(left(frac{X-mu}{ksigma}right)^2right) \
                  &=frac{1}{k^2} cdot frac{E((X-mu)^2)}{ksigma} \
                  &=frac{1}{k^2}.end{align*}






                  share|cite|improve this answer











                  $endgroup$



                  Thanks André, I found it.



                  For any event $A$, let $I_A$ be the indicator random variable of $A$, i.e. $I_A$ equals $1$ if $A$ occurs and $0$ otherwise. Then $newcommand{E}{operatorname{E}}$
                  begin{align*}
                  Pr(|X-mu| geq ksigma) &=E(I_{|X-mu| geq ksigma}) \
                  &= Eleft(I_{left(frac{X-mu}{ksigma}right)^2 geq 1}right) \
                  &leq Eleft(left(frac{X-mu}{ksigma}right)^2right) \
                  &=frac{1}{k^2} cdot frac{E((X-mu)^2)}{ksigma} \
                  &=frac{1}{k^2}.end{align*}







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 8 '18 at 10:14









                  Rócherz

                  2,7762721




                  2,7762721










                  answered Nov 28 '15 at 22:18









                  Electro82Electro82

                  536218




                  536218






























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