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A problem in Probability theory


If $G(x)=P[Xgeq x]$ then $Xgeq c$ is equivalent to $G(X)leq G(c)$ $P$-almost surelyTrying to establish an inequality on probabilityCan some probability triple give rise to any probability distribution?Expectation of $mathbbE(X^k+1)$Is PDF unique for a random variable $X$ in given probability space?Conditional expectation on different probability measureAverage of Random variables converges in probability.Range of a random variable is measurableIn probability theory what does the notation $int_Omega X(omega) P(domega)$ mean?Probability theory: Convergence













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 $mathbbE(X^2)=1$ and $mathbbE(X)geq a >0$, prove that $$mathbbP(Xgeq lambda a)geq (a-lambda a)^2$$
for $0leqlambda leq 1$.




Let $A=xin Omega:X(x)geq lambda a$, we get
$$int_A (X-lambda a)geq a-int_Alambda a -int_A^cX$$
and $$int_A (X^2-lambda^2 a^2)=1-int_Alambda^2a^2-int_A^cX^2$$
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 $mathbbE(X^2)=1$ and $mathbbE(X)geq a >0$, prove that $$mathbbP(Xgeq lambda a)geq (a-lambda a)^2$$
for $0leqlambda leq 1$.




Let $A=xin Omega:X(x)geq lambda a$, we get
$$int_A (X-lambda a)geq a-int_Alambda a -int_A^cX$$
and $$int_A (X^2-lambda^2 a^2)=1-int_Alambda^2a^2-int_A^cX^2$$
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 $mathbbE(X^2)=1$ and $mathbbE(X)geq a >0$, prove that $$mathbbP(Xgeq lambda a)geq (a-lambda a)^2$$
for $0leqlambda leq 1$.




Let $A=xin Omega:X(x)geq lambda a$, we get
$$int_A (X-lambda a)geq a-int_Alambda a -int_A^cX$$
and $$int_A (X^2-lambda^2 a^2)=1-int_Alambda^2a^2-int_A^cX^2$$
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 $mathbbE(X^2)=1$ and $mathbbE(X)geq a >0$, prove that $$mathbbP(Xgeq lambda a)geq (a-lambda a)^2$$
for $0leqlambda leq 1$.




Let $A=xin Omega:X(x)geq lambda a$, we get
$$int_A (X-lambda a)geq a-int_Alambda a -int_A^cX$$
and $$int_A (X^2-lambda^2 a^2)=1-int_Alambda^2a^2-int_A^cX^2$$
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
$$
alemathbb E(X) = int_Xlelambda aX,dP + int_Xgelambda aX,dP,le,lambda a + int_Xgelambda aX,dP.
$$
Hence,
$$
a(1-lambda),le,int_Xgelambda aX,dP,le,left(int_Xgelambda aX^2,dPright)^1/2cdot P(Xgelambda a)^1/2,le,P(Xgelambda a)^1/2.
$$
Square this and you're done.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Thank you very much!
    $endgroup$
    – Xin Fu
    2 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
$$
alemathbb E(X) = int_Xlelambda aX,dP + int_Xgelambda aX,dP,le,lambda a + int_Xgelambda aX,dP.
$$
Hence,
$$
a(1-lambda),le,int_Xgelambda aX,dP,le,left(int_Xgelambda aX^2,dPright)^1/2cdot P(Xgelambda a)^1/2,le,P(Xgelambda a)^1/2.
$$
Square this and you're done.






share|cite|improve this answer









$endgroup$












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















5












$begingroup$

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






share|cite|improve this answer









$endgroup$












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













5












5








5





$begingroup$

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






share|cite|improve this answer









$endgroup$



You have
$$
alemathbb E(X) = int_Xlelambda aX,dP + int_Xgelambda aX,dP,le,lambda a + int_Xgelambda aX,dP.
$$
Hence,
$$
a(1-lambda),le,int_Xgelambda aX,dP,le,left(int_Xgelambda aX^2,dPright)^1/2cdot P(Xgelambda a)^1/2,le,P(Xgelambda a)^1/2.
$$
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
    2 hours ago
















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















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




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

















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