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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
$begingroup$
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
asked 3 hours ago
Xin FuXin Fu
1568
$endgroup$
add a comment |
$begingroup$
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
asked 3 hours ago
Xin FuXin Fu
1568
$endgroup$
$begingroup$
Chebyshev might be useful.
$endgroup$
– copper.hat
3 hours ago
add a comment |
$begingroup$
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
asked 3 hours ago
Xin FuXin Fu
1568
$endgroup$
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
asked 3 hours ago
Xin FuXin Fu
1568
asked 3 hours ago
Xin FuXin Fu
1568
asked 3 hours ago
Xin FuXin Fu
1568
asked 3 hours ago
Xin FuXin Fu
1568
asked 3 hours ago
Xin FuXin Fu
1568
1568
$begingroup$
Chebyshev might be useful.
$endgroup$
– copper.hat
3 hours ago
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered 2 hours ago
amsmathamsmath
3,364419
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Xin Fu
2 hours ago
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered 2 hours ago
amsmathamsmath
3,364419
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Xin Fu
2 hours ago
add a comment |
$begingroup$
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.
answered 2 hours ago
amsmathamsmath
3,364419
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Xin Fu
2 hours ago
add a comment |
$begingroup$
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.
answered 2 hours ago
amsmathamsmath
3,364419
$endgroup$
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.
answered 2 hours ago
amsmathamsmath
3,364419
answered 2 hours ago
amsmathamsmath
3,364419
answered 2 hours ago
amsmathamsmath
3,364419
answered 2 hours ago
amsmathamsmath
3,364419
3,364419
$begingroup$
Thank you very much!
$endgroup$
– Xin Fu
2 hours ago
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
Chebyshev might be useful.
$endgroup$
– copper.hat
3 hours ago