$X$ is a random variable, if $Bbb E(X^2)=1$ and $Bbb E(X)geq a>0$, prove that $Bbb P(Xgeqlambda...
$begingroup$
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
edited Mar 28 at 9:19
Asaf Karagila♦
307k33440773
asked Mar 28 at 3:11
Xin FuXin Fu
34319
$endgroup$
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 $mathbb{E}(X^2)=1$ 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
edited Mar 28 at 9:19
Asaf Karagila♦
307k33440773
asked Mar 28 at 3:11
Xin FuXin Fu
34319
$endgroup$
$begingroup$
Chebyshev might be useful.
$endgroup$
– copper.hat
Mar 28 at 3:30
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 $mathbb{E}(X^2)=1$ 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
edited Mar 28 at 9:19
Asaf Karagila♦
307k33440773
asked Mar 28 at 3:11
Xin FuXin Fu
34319
$endgroup$
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
edited Mar 28 at 9:19
Asaf Karagila♦
307k33440773
asked Mar 28 at 3:11
Xin FuXin Fu
34319
edited Mar 28 at 9:19
Asaf Karagila♦
307k33440773
asked Mar 28 at 3:11
Xin FuXin Fu
34319
edited Mar 28 at 9:19
Asaf Karagila♦
307k33440773
edited Mar 28 at 9:19
Asaf Karagila♦
307k33440773
edited Mar 28 at 9:19
Asaf Karagila♦
307k33440773
307k33440773
asked Mar 28 at 3:11
Xin FuXin Fu
34319
asked Mar 28 at 3:11
Xin FuXin Fu
34319
asked Mar 28 at 3:11
Xin FuXin Fu
34319
34319
$begingroup$
Chebyshev might be useful.
$endgroup$
– copper.hat
Mar 28 at 3:30
add a comment |
$begingroup$
Chebyshev might be useful.
$endgroup$
– copper.hat
Mar 28 at 3:30
$begingroup$
Chebyshev might be useful.
$endgroup$
– copper.hat
Mar 28 at 3:30
$begingroup$
Chebyshev might be useful.
$endgroup$
– copper.hat
Mar 28 at 3:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Mar 28 at 3:37
amsmathamsmath
3,273420
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Xin Fu
Mar 28 at 3:44
add a comment |
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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
$$
alemathbb E(X) = int_{Xlelambda a}X,dP + int_{Xgelambda a}X,dP,le,lambda a + int_{Xgelambda a}X,dP.
$$
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.
answered Mar 28 at 3:37
amsmathamsmath
3,273420
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Xin Fu
Mar 28 at 3:44
add a comment |
$begingroup$
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.
answered Mar 28 at 3:37
amsmathamsmath
3,273420
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Xin Fu
Mar 28 at 3:44
add a comment |
$begingroup$
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.
answered Mar 28 at 3:37
amsmathamsmath
3,273420
$endgroup$
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.
answered Mar 28 at 3:37
amsmathamsmath
3,273420
answered Mar 28 at 3:37
amsmathamsmath
3,273420
answered Mar 28 at 3:37
amsmathamsmath
3,273420
answered Mar 28 at 3:37
amsmathamsmath
3,273420
3,273420
$begingroup$
Thank you very much!
$endgroup$
– Xin Fu
Mar 28 at 3:44
add a comment |
$begingroup$
Thank you very much!
$endgroup$
– Xin Fu
Mar 28 at 3:44
$begingroup$
Thank you very much!
$endgroup$
– Xin Fu
Mar 28 at 3:44
$begingroup$
Thank you very much!
$endgroup$
– Xin Fu
Mar 28 at 3:44
add a comment |
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$begingroup$
Chebyshev might be useful.
$endgroup$
– copper.hat
Mar 28 at 3:30