Know the concentration estimate expectation？












1












$begingroup$


Here is the problem:



let $X$ be a random variable such that:
$$P{X > c(m+t) }<2e^{-t^2} forall t >0$$
where $c>0,m>0$ are constant.



Then I was asked to prove that :
$$ mathbb{E}[X] leq cm$$



however I can't get rid of constant, here is my attempt:



begin{align}
mathbb{E}[X] &= int_0^infty P{X>u}mathrm{d}u\
&= int_{cm}^infty P{X>u}mathrm{d}u +int_{0}^{cm} P{X>u} mathrm{d}u \
&leq int_{cm}^infty P{X>u}mathrm{d}u+ cm
\&= int_{cm}^infty P{X>cm+cdelta}mathrm{d}u+ cm (delta>0)
\&= cint_{0}^infty P{X>cm+cdelta}mathrm{d}delta (Change of vairible)
\& leq cint_{0}^infty e^{-delta^2} mathrm{d}delta +cm (assumption)
end{align}



what I have been missing here?






probability concentration-of-measure







share|cite|improve this question











$endgroup$












  • $begingroup$
    In light of the answer, this begs the question; who asked you to prove that?
    $endgroup$
    – Clement C.
    Dec 30 '18 at 18:11










  • $begingroup$
    @ClementC. math.uci.edu/~rvershyn/papers/HDP-book/HDP-book.pdf exercise 4.4.6
    $endgroup$
    – ShaoyuPei
    Dec 30 '18 at 23:07












  • $begingroup$
    Doesn't Exercise 4.4.6 include another constant $C$? (I.e., I wonder if the author doesn't use the "notation" that $C$ is an absolute constant, which may not take the same value across results).
    $endgroup$
    – Clement C.
    Jan 1 at 17:57










  • $begingroup$
    @ClementC. you are right , C is an absolute constant ,so Constant C is free to choose ? For small c in this question, I think if constant part is $frac{1}{c}int_{0}^infty e^{-delta^2}$,we could take c to be large , then the constant term can be neglect, or I did integration wrong ...
    $endgroup$
    – ShaoyuPei
    Jan 2 at 13:19












  • $begingroup$
    Well, with what your wrote in your OP, you get $ccdot(frac{sqrt{pi}}{2}+m)$. so since $mgeq 1$ in the book ($m,n$ are integers, right, and "your" $m$ is their $sqrt{n}+sqrt{m}$), you do have $ccdot(frac{sqrt{pi}}{2}+m) leq ccdot(frac{sqrt{pi}}{2}+1)m = c'm$, setting $c'stackrel{rm def}{=} ccdot(frac{sqrt{pi}}{2}+1)$.
    $endgroup$
    – Clement C.
    Jan 2 at 13:52


















1












$begingroup$


Here is the problem:



let $X$ be a random variable such that:
$$P{X > c(m+t) }<2e^{-t^2} forall t >0$$
where $c>0,m>0$ are constant.



Then I was asked to prove that :
$$ mathbb{E}[X] leq cm$$



however I can't get rid of constant, here is my attempt:



begin{align}
mathbb{E}[X] &= int_0^infty P{X>u}mathrm{d}u\
&= int_{cm}^infty P{X>u}mathrm{d}u +int_{0}^{cm} P{X>u} mathrm{d}u \
&leq int_{cm}^infty P{X>u}mathrm{d}u+ cm
\&= int_{cm}^infty P{X>cm+cdelta}mathrm{d}u+ cm (delta>0)
\&= cint_{0}^infty P{X>cm+cdelta}mathrm{d}delta (Change of vairible)
\& leq cint_{0}^infty e^{-delta^2} mathrm{d}delta +cm (assumption)
end{align}



what I have been missing here?






probability concentration-of-measure







share|cite|improve this question











$endgroup$












  • $begingroup$
    In light of the answer, this begs the question; who asked you to prove that?
    $endgroup$
    – Clement C.
    Dec 30 '18 at 18:11










  • $begingroup$
    @ClementC. math.uci.edu/~rvershyn/papers/HDP-book/HDP-book.pdf exercise 4.4.6
    $endgroup$
    – ShaoyuPei
    Dec 30 '18 at 23:07












  • $begingroup$
    Doesn't Exercise 4.4.6 include another constant $C$? (I.e., I wonder if the author doesn't use the "notation" that $C$ is an absolute constant, which may not take the same value across results).
    $endgroup$
    – Clement C.
    Jan 1 at 17:57










  • $begingroup$
    @ClementC. you are right , C is an absolute constant ,so Constant C is free to choose ? For small c in this question, I think if constant part is $frac{1}{c}int_{0}^infty e^{-delta^2}$,we could take c to be large , then the constant term can be neglect, or I did integration wrong ...
    $endgroup$
    – ShaoyuPei
    Jan 2 at 13:19












  • $begingroup$
    Well, with what your wrote in your OP, you get $ccdot(frac{sqrt{pi}}{2}+m)$. so since $mgeq 1$ in the book ($m,n$ are integers, right, and "your" $m$ is their $sqrt{n}+sqrt{m}$), you do have $ccdot(frac{sqrt{pi}}{2}+m) leq ccdot(frac{sqrt{pi}}{2}+1)m = c'm$, setting $c'stackrel{rm def}{=} ccdot(frac{sqrt{pi}}{2}+1)$.
    $endgroup$
    – Clement C.
    Jan 2 at 13:52
















1












1








1





$begingroup$


Here is the problem:



let $X$ be a random variable such that:
$$P{X > c(m+t) }<2e^{-t^2} forall t >0$$
where $c>0,m>0$ are constant.



Then I was asked to prove that :
$$ mathbb{E}[X] leq cm$$



however I can't get rid of constant, here is my attempt:



begin{align}
mathbb{E}[X] &= int_0^infty P{X>u}mathrm{d}u\
&= int_{cm}^infty P{X>u}mathrm{d}u +int_{0}^{cm} P{X>u} mathrm{d}u \
&leq int_{cm}^infty P{X>u}mathrm{d}u+ cm
\&= int_{cm}^infty P{X>cm+cdelta}mathrm{d}u+ cm (delta>0)
\&= cint_{0}^infty P{X>cm+cdelta}mathrm{d}delta (Change of vairible)
\& leq cint_{0}^infty e^{-delta^2} mathrm{d}delta +cm (assumption)
end{align}



what I have been missing here?






probability concentration-of-measure







share|cite|improve this question











$endgroup$




Here is the problem:



let $X$ be a random variable such that:
$$P{X > c(m+t) }<2e^{-t^2} forall t >0$$
where $c>0,m>0$ are constant.



Then I was asked to prove that :
$$ mathbb{E}[X] leq cm$$



however I can't get rid of constant, here is my attempt:



begin{align}
mathbb{E}[X] &= int_0^infty P{X>u}mathrm{d}u\
&= int_{cm}^infty P{X>u}mathrm{d}u +int_{0}^{cm} P{X>u} mathrm{d}u \
&leq int_{cm}^infty P{X>u}mathrm{d}u+ cm
\&= int_{cm}^infty P{X>cm+cdelta}mathrm{d}u+ cm (delta>0)
\&= cint_{0}^infty P{X>cm+cdelta}mathrm{d}delta (Change of vairible)
\& leq cint_{0}^infty e^{-delta^2} mathrm{d}delta +cm (assumption)
end{align}



what I have been missing here?






probability concentration-of-measure




probability concentration-of-measure






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 12:18







ShaoyuPei

















asked Dec 30 '18 at 11:40









ShaoyuPeiShaoyuPei

1778




1778












  • $begingroup$
    In light of the answer, this begs the question; who asked you to prove that?
    $endgroup$
    – Clement C.
    Dec 30 '18 at 18:11










  • $begingroup$
    @ClementC. math.uci.edu/~rvershyn/papers/HDP-book/HDP-book.pdf exercise 4.4.6
    $endgroup$
    – ShaoyuPei
    Dec 30 '18 at 23:07












  • $begingroup$
    Doesn't Exercise 4.4.6 include another constant $C$? (I.e., I wonder if the author doesn't use the "notation" that $C$ is an absolute constant, which may not take the same value across results).
    $endgroup$
    – Clement C.
    Jan 1 at 17:57










  • $begingroup$
    @ClementC. you are right , C is an absolute constant ,so Constant C is free to choose ? For small c in this question, I think if constant part is $frac{1}{c}int_{0}^infty e^{-delta^2}$,we could take c to be large , then the constant term can be neglect, or I did integration wrong ...
    $endgroup$
    – ShaoyuPei
    Jan 2 at 13:19












  • $begingroup$
    Well, with what your wrote in your OP, you get $ccdot(frac{sqrt{pi}}{2}+m)$. so since $mgeq 1$ in the book ($m,n$ are integers, right, and "your" $m$ is their $sqrt{n}+sqrt{m}$), you do have $ccdot(frac{sqrt{pi}}{2}+m) leq ccdot(frac{sqrt{pi}}{2}+1)m = c'm$, setting $c'stackrel{rm def}{=} ccdot(frac{sqrt{pi}}{2}+1)$.
    $endgroup$
    – Clement C.
    Jan 2 at 13:52




















  • $begingroup$
    In light of the answer, this begs the question; who asked you to prove that?
    $endgroup$
    – Clement C.
    Dec 30 '18 at 18:11










  • $begingroup$
    @ClementC. math.uci.edu/~rvershyn/papers/HDP-book/HDP-book.pdf exercise 4.4.6
    $endgroup$
    – ShaoyuPei
    Dec 30 '18 at 23:07












  • $begingroup$
    Doesn't Exercise 4.4.6 include another constant $C$? (I.e., I wonder if the author doesn't use the "notation" that $C$ is an absolute constant, which may not take the same value across results).
    $endgroup$
    – Clement C.
    Jan 1 at 17:57










  • $begingroup$
    @ClementC. you are right , C is an absolute constant ,so Constant C is free to choose ? For small c in this question, I think if constant part is $frac{1}{c}int_{0}^infty e^{-delta^2}$,we could take c to be large , then the constant term can be neglect, or I did integration wrong ...
    $endgroup$
    – ShaoyuPei
    Jan 2 at 13:19












  • $begingroup$
    Well, with what your wrote in your OP, you get $ccdot(frac{sqrt{pi}}{2}+m)$. so since $mgeq 1$ in the book ($m,n$ are integers, right, and "your" $m$ is their $sqrt{n}+sqrt{m}$), you do have $ccdot(frac{sqrt{pi}}{2}+m) leq ccdot(frac{sqrt{pi}}{2}+1)m = c'm$, setting $c'stackrel{rm def}{=} ccdot(frac{sqrt{pi}}{2}+1)$.
    $endgroup$
    – Clement C.
    Jan 2 at 13:52


















$begingroup$
In light of the answer, this begs the question; who asked you to prove that?
$endgroup$
– Clement C.
Dec 30 '18 at 18:11




$begingroup$
In light of the answer, this begs the question; who asked you to prove that?
$endgroup$
– Clement C.
Dec 30 '18 at 18:11












$begingroup$
@ClementC. math.uci.edu/~rvershyn/papers/HDP-book/HDP-book.pdf exercise 4.4.6
$endgroup$
– ShaoyuPei
Dec 30 '18 at 23:07






$begingroup$
@ClementC. math.uci.edu/~rvershyn/papers/HDP-book/HDP-book.pdf exercise 4.4.6
$endgroup$
– ShaoyuPei
Dec 30 '18 at 23:07














$begingroup$
Doesn't Exercise 4.4.6 include another constant $C$? (I.e., I wonder if the author doesn't use the "notation" that $C$ is an absolute constant, which may not take the same value across results).
$endgroup$
– Clement C.
Jan 1 at 17:57




$begingroup$
Doesn't Exercise 4.4.6 include another constant $C$? (I.e., I wonder if the author doesn't use the "notation" that $C$ is an absolute constant, which may not take the same value across results).
$endgroup$
– Clement C.
Jan 1 at 17:57












$begingroup$
@ClementC. you are right , C is an absolute constant ,so Constant C is free to choose ? For small c in this question, I think if constant part is $frac{1}{c}int_{0}^infty e^{-delta^2}$,we could take c to be large , then the constant term can be neglect, or I did integration wrong ...
$endgroup$
– ShaoyuPei
Jan 2 at 13:19






$begingroup$
@ClementC. you are right , C is an absolute constant ,so Constant C is free to choose ? For small c in this question, I think if constant part is $frac{1}{c}int_{0}^infty e^{-delta^2}$,we could take c to be large , then the constant term can be neglect, or I did integration wrong ...
$endgroup$
– ShaoyuPei
Jan 2 at 13:19














$begingroup$
Well, with what your wrote in your OP, you get $ccdot(frac{sqrt{pi}}{2}+m)$. so since $mgeq 1$ in the book ($m,n$ are integers, right, and "your" $m$ is their $sqrt{n}+sqrt{m}$), you do have $ccdot(frac{sqrt{pi}}{2}+m) leq ccdot(frac{sqrt{pi}}{2}+1)m = c'm$, setting $c'stackrel{rm def}{=} ccdot(frac{sqrt{pi}}{2}+1)$.
$endgroup$
– Clement C.
Jan 2 at 13:52






$begingroup$
Well, with what your wrote in your OP, you get $ccdot(frac{sqrt{pi}}{2}+m)$. so since $mgeq 1$ in the book ($m,n$ are integers, right, and "your" $m$ is their $sqrt{n}+sqrt{m}$), you do have $ccdot(frac{sqrt{pi}}{2}+m) leq ccdot(frac{sqrt{pi}}{2}+1)m = c'm$, setting $c'stackrel{rm def}{=} ccdot(frac{sqrt{pi}}{2}+1)$.
$endgroup$
– Clement C.
Jan 2 at 13:52












1 Answer
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$begingroup$

The inequality is false. For example if $Z$ has uniform distribution on $(0,1)$ and $X=cZ+cm$ then the hypothesis reduces to $P(Z>t) leq 2e^{-t^{2}}$ or $1-t leq 2e^{-t^{2}}$ for $tin (0,1)$ which can be proved easily by writing the inequality as $e^{t^{2}}(1-t)-2leq 0$. [ The left side is decreasing and it has the value $-1$ at $t=0$]. Note that $EX=frac c 2 +cm>cm$ if $c>0$.






share|cite|improve this answer









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    oldest

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    $begingroup$

    The inequality is false. For example if $Z$ has uniform distribution on $(0,1)$ and $X=cZ+cm$ then the hypothesis reduces to $P(Z>t) leq 2e^{-t^{2}}$ or $1-t leq 2e^{-t^{2}}$ for $tin (0,1)$ which can be proved easily by writing the inequality as $e^{t^{2}}(1-t)-2leq 0$. [ The left side is decreasing and it has the value $-1$ at $t=0$]. Note that $EX=frac c 2 +cm>cm$ if $c>0$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The inequality is false. For example if $Z$ has uniform distribution on $(0,1)$ and $X=cZ+cm$ then the hypothesis reduces to $P(Z>t) leq 2e^{-t^{2}}$ or $1-t leq 2e^{-t^{2}}$ for $tin (0,1)$ which can be proved easily by writing the inequality as $e^{t^{2}}(1-t)-2leq 0$. [ The left side is decreasing and it has the value $-1$ at $t=0$]. Note that $EX=frac c 2 +cm>cm$ if $c>0$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The inequality is false. For example if $Z$ has uniform distribution on $(0,1)$ and $X=cZ+cm$ then the hypothesis reduces to $P(Z>t) leq 2e^{-t^{2}}$ or $1-t leq 2e^{-t^{2}}$ for $tin (0,1)$ which can be proved easily by writing the inequality as $e^{t^{2}}(1-t)-2leq 0$. [ The left side is decreasing and it has the value $-1$ at $t=0$]. Note that $EX=frac c 2 +cm>cm$ if $c>0$.






        share|cite|improve this answer









        $endgroup$



        The inequality is false. For example if $Z$ has uniform distribution on $(0,1)$ and $X=cZ+cm$ then the hypothesis reduces to $P(Z>t) leq 2e^{-t^{2}}$ or $1-t leq 2e^{-t^{2}}$ for $tin (0,1)$ which can be proved easily by writing the inequality as $e^{t^{2}}(1-t)-2leq 0$. [ The left side is decreasing and it has the value $-1$ at $t=0$]. Note that $EX=frac c 2 +cm>cm$ if $c>0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 30 '18 at 12:33









        Kavi Rama MurthyKavi Rama Murthy

        63.7k42464




        63.7k42464






























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