A problem in Probability theory












4












$begingroup$


This is a problem in KaiLai Chung's A Course in Probability Theory.




Given a nonnegative random variable $X$ defined on $Omega$, if $mathbb{E}(X^2)=1$ and $mathbb{E}(X)geq a >0$, prove that $$mathbb{P}(Xgeq lambda a)geq (a-lambda a)^2$$
for $0leqlambda leq 1$.




Let $A={xin Omega:X(x)geq lambda a}$, we get
$$int_A (X-lambda a)geq a-int_Alambda a -int_{A^c}X$$
and $$int_A (X^2-lambda^2 a^2)=1-int_Alambda^2a^2-int_{A^c}X^2$$
I want to contrast $int_A (X-lambda a)$ and $int_A (X^2-lambda^2 a^2)$, but I don't know how to do it, could anyone gives me some hints?






probability integration lp-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    Chebyshev might be useful.
    $endgroup$
    – copper.hat
    3 hours ago
















4












$begingroup$


This is a problem in KaiLai Chung's A Course in Probability Theory.




Given a nonnegative random variable $X$ defined on $Omega$, if $mathbb{E}(X^2)=1$ and $mathbb{E}(X)geq a >0$, prove that $$mathbb{P}(Xgeq lambda a)geq (a-lambda a)^2$$
for $0leqlambda leq 1$.




Let $A={xin Omega:X(x)geq lambda a}$, we get
$$int_A (X-lambda a)geq a-int_Alambda a -int_{A^c}X$$
and $$int_A (X^2-lambda^2 a^2)=1-int_Alambda^2a^2-int_{A^c}X^2$$
I want to contrast $int_A (X-lambda a)$ and $int_A (X^2-lambda^2 a^2)$, but I don't know how to do it, could anyone gives me some hints?






probability integration lp-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    Chebyshev might be useful.
    $endgroup$
    – copper.hat
    3 hours ago














4












4








4


1



$begingroup$


This is a problem in KaiLai Chung's A Course in Probability Theory.




Given a nonnegative random variable $X$ defined on $Omega$, if $mathbb{E}(X^2)=1$ and $mathbb{E}(X)geq a >0$, prove that $$mathbb{P}(Xgeq lambda a)geq (a-lambda a)^2$$
for $0leqlambda leq 1$.




Let $A={xin Omega:X(x)geq lambda a}$, we get
$$int_A (X-lambda a)geq a-int_Alambda a -int_{A^c}X$$
and $$int_A (X^2-lambda^2 a^2)=1-int_Alambda^2a^2-int_{A^c}X^2$$
I want to contrast $int_A (X-lambda a)$ and $int_A (X^2-lambda^2 a^2)$, but I don't know how to do it, could anyone gives me some hints?






probability integration lp-spaces







share|cite|improve this question









$endgroup$




This is a problem in KaiLai Chung's A Course in Probability Theory.




Given a nonnegative random variable $X$ defined on $Omega$, if $mathbb{E}(X^2)=1$ and $mathbb{E}(X)geq a >0$, prove that $$mathbb{P}(Xgeq lambda a)geq (a-lambda a)^2$$
for $0leqlambda leq 1$.




Let $A={xin Omega:X(x)geq lambda a}$, we get
$$int_A (X-lambda a)geq a-int_Alambda a -int_{A^c}X$$
and $$int_A (X^2-lambda^2 a^2)=1-int_Alambda^2a^2-int_{A^c}X^2$$
I want to contrast $int_A (X-lambda a)$ and $int_A (X^2-lambda^2 a^2)$, but I don't know how to do it, could anyone gives me some hints?






probability integration lp-spaces




probability integration lp-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 3 hours ago









Xin FuXin Fu

1568




1568












  • $begingroup$
    Chebyshev might be useful.
    $endgroup$
    – copper.hat
    3 hours ago


















  • $begingroup$
    Chebyshev might be useful.
    $endgroup$
    – copper.hat
    3 hours ago
















$begingroup$
Chebyshev might be useful.
$endgroup$
– copper.hat
3 hours ago




$begingroup$
Chebyshev might be useful.
$endgroup$
– copper.hat
3 hours ago










1 Answer
1






active

oldest

votes


















5












$begingroup$

You have
$$
alemathbb E(X) = int_{Xlelambda a}X,dP + int_{Xgelambda a}X,dP,le,lambda a + int_{Xgelambda a}X,dP.
$$
Hence,
$$
a(1-lambda),le,int_{Xgelambda a}X,dP,le,left(int_{Xgelambda a}X^2,dPright)^{1/2}cdot P(Xgelambda a)^{1/2},le,P(Xgelambda a)^{1/2}.
$$
Square this and you're done.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much!
    $endgroup$
    – Xin Fu
    3 hours ago











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

You have
$$
alemathbb E(X) = int_{Xlelambda a}X,dP + int_{Xgelambda a}X,dP,le,lambda a + int_{Xgelambda a}X,dP.
$$
Hence,
$$
a(1-lambda),le,int_{Xgelambda a}X,dP,le,left(int_{Xgelambda a}X^2,dPright)^{1/2}cdot P(Xgelambda a)^{1/2},le,P(Xgelambda a)^{1/2}.
$$
Square this and you're done.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much!
    $endgroup$
    – Xin Fu
    3 hours ago
















5












$begingroup$

You have
$$
alemathbb E(X) = int_{Xlelambda a}X,dP + int_{Xgelambda a}X,dP,le,lambda a + int_{Xgelambda a}X,dP.
$$
Hence,
$$
a(1-lambda),le,int_{Xgelambda a}X,dP,le,left(int_{Xgelambda a}X^2,dPright)^{1/2}cdot P(Xgelambda a)^{1/2},le,P(Xgelambda a)^{1/2}.
$$
Square this and you're done.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much!
    $endgroup$
    – Xin Fu
    3 hours ago














5












5








5





$begingroup$

You have
$$
alemathbb E(X) = int_{Xlelambda a}X,dP + int_{Xgelambda a}X,dP,le,lambda a + int_{Xgelambda a}X,dP.
$$
Hence,
$$
a(1-lambda),le,int_{Xgelambda a}X,dP,le,left(int_{Xgelambda a}X^2,dPright)^{1/2}cdot P(Xgelambda a)^{1/2},le,P(Xgelambda a)^{1/2}.
$$
Square this and you're done.






share|cite|improve this answer









$endgroup$



You have
$$
alemathbb E(X) = int_{Xlelambda a}X,dP + int_{Xgelambda a}X,dP,le,lambda a + int_{Xgelambda a}X,dP.
$$
Hence,
$$
a(1-lambda),le,int_{Xgelambda a}X,dP,le,left(int_{Xgelambda a}X^2,dPright)^{1/2}cdot P(Xgelambda a)^{1/2},le,P(Xgelambda a)^{1/2}.
$$
Square this and you're done.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 3 hours ago









amsmathamsmath

3,364419




3,364419












  • $begingroup$
    Thank you very much!
    $endgroup$
    – Xin Fu
    3 hours ago


















  • $begingroup$
    Thank you very much!
    $endgroup$
    – Xin Fu
    3 hours ago
















$begingroup$
Thank you very much!
$endgroup$
– Xin Fu
3 hours ago




$begingroup$
Thank you very much!
$endgroup$
– Xin Fu
3 hours ago


















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