Compare $E[|X| cdot 1{|X| < M}]$ and $P(|X| < M)$ with $E|X|=1$












1












$begingroup$


$X$ is a random variable and $E|X|=1$. Do we know any relation between $P(|X|leq M)$ and $E[|X| cdot 1{|X| leq M }]$? Do we only have one direction, i.e., either $P(|X|leq M) geq E[|X| cdot 1{|X| leq M }]$ or $P(|X|leq M) leq E[|X| cdot 1{|X| leq M }$? Or in fact, both cases can happen? EDIT: Especially when $M to infty$, what would be the asymptotic comparison?






probability-theory inequality random-variables expected-value







share|cite|improve this question











$endgroup$












  • $begingroup$
    For every nonnegative $M$, the function $u:tmapsto t$ is nondecreasing on $tgeqslant0$ and the function $v:tmapstomathbf 1_{tleqslant M}$ is nonincreasing on $tgeqslant0$, hence, by a rearrangement inequality, $$E(u(|X|)v(|X|))leqslant E(u(|X|))E(v(|X|))$$ If, in addition, $E|X|=1$, this yields $$E(|X|mathbf 1_{|X|leqslant M})leqslant P(|X|leqslant M)$$
    $endgroup$
    – Did
    Feb 1 at 7:01












  • $begingroup$
    @Did Thank you. I found that one can also use Chebyshev's association inequality to prove this.
    $endgroup$
    – jwyao
    Feb 1 at 17:53










  • $begingroup$
    You could show how, for example by posting an answer below.
    $endgroup$
    – Did
    Feb 1 at 17:54










  • $begingroup$
    @Did Just post. Thanks for reminding me of the inequality.
    $endgroup$
    – jwyao
    Feb 1 at 18:02
















1












$begingroup$


$X$ is a random variable and $E|X|=1$. Do we know any relation between $P(|X|leq M)$ and $E[|X| cdot 1{|X| leq M }]$? Do we only have one direction, i.e., either $P(|X|leq M) geq E[|X| cdot 1{|X| leq M }]$ or $P(|X|leq M) leq E[|X| cdot 1{|X| leq M }$? Or in fact, both cases can happen? EDIT: Especially when $M to infty$, what would be the asymptotic comparison?






probability-theory inequality random-variables expected-value







share|cite|improve this question











$endgroup$












  • $begingroup$
    For every nonnegative $M$, the function $u:tmapsto t$ is nondecreasing on $tgeqslant0$ and the function $v:tmapstomathbf 1_{tleqslant M}$ is nonincreasing on $tgeqslant0$, hence, by a rearrangement inequality, $$E(u(|X|)v(|X|))leqslant E(u(|X|))E(v(|X|))$$ If, in addition, $E|X|=1$, this yields $$E(|X|mathbf 1_{|X|leqslant M})leqslant P(|X|leqslant M)$$
    $endgroup$
    – Did
    Feb 1 at 7:01












  • $begingroup$
    @Did Thank you. I found that one can also use Chebyshev's association inequality to prove this.
    $endgroup$
    – jwyao
    Feb 1 at 17:53










  • $begingroup$
    You could show how, for example by posting an answer below.
    $endgroup$
    – Did
    Feb 1 at 17:54










  • $begingroup$
    @Did Just post. Thanks for reminding me of the inequality.
    $endgroup$
    – jwyao
    Feb 1 at 18:02














1












1








1





$begingroup$


$X$ is a random variable and $E|X|=1$. Do we know any relation between $P(|X|leq M)$ and $E[|X| cdot 1{|X| leq M }]$? Do we only have one direction, i.e., either $P(|X|leq M) geq E[|X| cdot 1{|X| leq M }]$ or $P(|X|leq M) leq E[|X| cdot 1{|X| leq M }$? Or in fact, both cases can happen? EDIT: Especially when $M to infty$, what would be the asymptotic comparison?






probability-theory inequality random-variables expected-value







share|cite|improve this question











$endgroup$




$X$ is a random variable and $E|X|=1$. Do we know any relation between $P(|X|leq M)$ and $E[|X| cdot 1{|X| leq M }]$? Do we only have one direction, i.e., either $P(|X|leq M) geq E[|X| cdot 1{|X| leq M }]$ or $P(|X|leq M) leq E[|X| cdot 1{|X| leq M }$? Or in fact, both cases can happen? EDIT: Especially when $M to infty$, what would be the asymptotic comparison?






probability-theory inequality random-variables expected-value




probability-theory inequality random-variables expected-value






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 1 at 7:01









Did

249k23228466




249k23228466










asked Feb 1 at 6:25









jwyaojwyao

16311




16311












  • $begingroup$
    For every nonnegative $M$, the function $u:tmapsto t$ is nondecreasing on $tgeqslant0$ and the function $v:tmapstomathbf 1_{tleqslant M}$ is nonincreasing on $tgeqslant0$, hence, by a rearrangement inequality, $$E(u(|X|)v(|X|))leqslant E(u(|X|))E(v(|X|))$$ If, in addition, $E|X|=1$, this yields $$E(|X|mathbf 1_{|X|leqslant M})leqslant P(|X|leqslant M)$$
    $endgroup$
    – Did
    Feb 1 at 7:01












  • $begingroup$
    @Did Thank you. I found that one can also use Chebyshev's association inequality to prove this.
    $endgroup$
    – jwyao
    Feb 1 at 17:53










  • $begingroup$
    You could show how, for example by posting an answer below.
    $endgroup$
    – Did
    Feb 1 at 17:54










  • $begingroup$
    @Did Just post. Thanks for reminding me of the inequality.
    $endgroup$
    – jwyao
    Feb 1 at 18:02


















  • $begingroup$
    For every nonnegative $M$, the function $u:tmapsto t$ is nondecreasing on $tgeqslant0$ and the function $v:tmapstomathbf 1_{tleqslant M}$ is nonincreasing on $tgeqslant0$, hence, by a rearrangement inequality, $$E(u(|X|)v(|X|))leqslant E(u(|X|))E(v(|X|))$$ If, in addition, $E|X|=1$, this yields $$E(|X|mathbf 1_{|X|leqslant M})leqslant P(|X|leqslant M)$$
    $endgroup$
    – Did
    Feb 1 at 7:01












  • $begingroup$
    @Did Thank you. I found that one can also use Chebyshev's association inequality to prove this.
    $endgroup$
    – jwyao
    Feb 1 at 17:53










  • $begingroup$
    You could show how, for example by posting an answer below.
    $endgroup$
    – Did
    Feb 1 at 17:54










  • $begingroup$
    @Did Just post. Thanks for reminding me of the inequality.
    $endgroup$
    – jwyao
    Feb 1 at 18:02
















$begingroup$
For every nonnegative $M$, the function $u:tmapsto t$ is nondecreasing on $tgeqslant0$ and the function $v:tmapstomathbf 1_{tleqslant M}$ is nonincreasing on $tgeqslant0$, hence, by a rearrangement inequality, $$E(u(|X|)v(|X|))leqslant E(u(|X|))E(v(|X|))$$ If, in addition, $E|X|=1$, this yields $$E(|X|mathbf 1_{|X|leqslant M})leqslant P(|X|leqslant M)$$
$endgroup$
– Did
Feb 1 at 7:01






$begingroup$
For every nonnegative $M$, the function $u:tmapsto t$ is nondecreasing on $tgeqslant0$ and the function $v:tmapstomathbf 1_{tleqslant M}$ is nonincreasing on $tgeqslant0$, hence, by a rearrangement inequality, $$E(u(|X|)v(|X|))leqslant E(u(|X|))E(v(|X|))$$ If, in addition, $E|X|=1$, this yields $$E(|X|mathbf 1_{|X|leqslant M})leqslant P(|X|leqslant M)$$
$endgroup$
– Did
Feb 1 at 7:01














$begingroup$
@Did Thank you. I found that one can also use Chebyshev's association inequality to prove this.
$endgroup$
– jwyao
Feb 1 at 17:53




$begingroup$
@Did Thank you. I found that one can also use Chebyshev's association inequality to prove this.
$endgroup$
– jwyao
Feb 1 at 17:53












$begingroup$
You could show how, for example by posting an answer below.
$endgroup$
– Did
Feb 1 at 17:54




$begingroup$
You could show how, for example by posting an answer below.
$endgroup$
– Did
Feb 1 at 17:54












$begingroup$
@Did Just post. Thanks for reminding me of the inequality.
$endgroup$
– jwyao
Feb 1 at 18:02




$begingroup$
@Did Just post. Thanks for reminding me of the inequality.
$endgroup$
– jwyao
Feb 1 at 18:02










1 Answer
1






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oldest

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$begingroup$

(Chebyshev's Association Inequality) $f$ is nonincreasing and $g$ is nondecreasing. $X$ is a real-valued random variable and $Y$ is a nonnegative random variable. Then
$$ E[Y] E[Yf(X)g(X)] leq E[Yf(X)] E[Yg(X)] .$$
Take $Y equiv 1$, $f(cdot) = 1{|X|leq cdot}$ and $g(cdot) = |cdot|$. The result would be similar to Did's comment.






share|cite|improve this answer









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    $begingroup$

    (Chebyshev's Association Inequality) $f$ is nonincreasing and $g$ is nondecreasing. $X$ is a real-valued random variable and $Y$ is a nonnegative random variable. Then
    $$ E[Y] E[Yf(X)g(X)] leq E[Yf(X)] E[Yg(X)] .$$
    Take $Y equiv 1$, $f(cdot) = 1{|X|leq cdot}$ and $g(cdot) = |cdot|$. The result would be similar to Did's comment.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      (Chebyshev's Association Inequality) $f$ is nonincreasing and $g$ is nondecreasing. $X$ is a real-valued random variable and $Y$ is a nonnegative random variable. Then
      $$ E[Y] E[Yf(X)g(X)] leq E[Yf(X)] E[Yg(X)] .$$
      Take $Y equiv 1$, $f(cdot) = 1{|X|leq cdot}$ and $g(cdot) = |cdot|$. The result would be similar to Did's comment.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        (Chebyshev's Association Inequality) $f$ is nonincreasing and $g$ is nondecreasing. $X$ is a real-valued random variable and $Y$ is a nonnegative random variable. Then
        $$ E[Y] E[Yf(X)g(X)] leq E[Yf(X)] E[Yg(X)] .$$
        Take $Y equiv 1$, $f(cdot) = 1{|X|leq cdot}$ and $g(cdot) = |cdot|$. The result would be similar to Did's comment.






        share|cite|improve this answer









        $endgroup$



        (Chebyshev's Association Inequality) $f$ is nonincreasing and $g$ is nondecreasing. $X$ is a real-valued random variable and $Y$ is a nonnegative random variable. Then
        $$ E[Y] E[Yf(X)g(X)] leq E[Yf(X)] E[Yg(X)] .$$
        Take $Y equiv 1$, $f(cdot) = 1{|X|leq cdot}$ and $g(cdot) = |cdot|$. The result would be similar to Did's comment.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 1 at 18:01









        jwyaojwyao

        16311




        16311






























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