$X$ is a random variable, if $Bbb E(X^2)=1$ and $Bbb E(X)geq a>0$, prove that $Bbb P(Xgeqlambda a)geq(a-lambda a)^2$ for $0leqlambdaleq 1$.How to show $P[Xgeq k] leq (fraclambda ek)^k e^-lambda$ (X is Poisson random variable)Show that $P(X > lambda) geq frac(EX - lambda)^2EX^2$An equivalent condition for a random variable to be integrableProof that absolute value of a random variable is a random variableIs PDF unique for a random variable $X$ in given probability space?Any technique to show that $PleftXgeq 2lambdarightleq (e/4)^lambda$Is this a valid identity for the expected value of a nonnegative random variable?Convergence of $y^q P(X>y)$ for $y to infty$ , $X geq 0$ random variable, $q>0$, $mathbbE[X^q]<1$Explicitly representing a random variable such as $ X(omega):=frac1lambda ln frac11-omega$, which is exponentialNotation for expectation of a random variable/general Lebesgue integral
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$X$ is a random variable, if $Bbb E(X^2)=1$ and $Bbb E(X)geq a>0$, prove that $Bbb P(Xgeqlambda a)geq(a-lambda a)^2$ for $0leqlambdaleq 1$.
How to show $P[Xgeq k] leq (fraclambda ek)^k e^-lambda$ (X is Poisson random variable)Show that $P(X > lambda) geq frac(EX - lambda)^2EX^2$An equivalent condition for a random variable to be integrableProof that absolute value of a random variable is a random variableIs PDF unique for a random variable $X$ in given probability space?Any technique to show that $PleftXgeq 2lambdarightleq (e/4)^lambda$Is this a valid identity for the expected value of a nonnegative random variable?Convergence of $y^q P(X>y)$ for $y to infty$ , $X geq 0$ random variable, $q>0$, $mathbbE[X^q]<1$Explicitly representing a random variable such as $ X(omega):=frac1lambda ln frac11-omega$, which is exponentialNotation for expectation of a random variable/general Lebesgue integral
$begingroup$
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
edited Mar 28 at 9:19
Asaf Karagila♦
308k33441774
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 $mathbbE(X^2)=1$ 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
edited Mar 28 at 9:19
Asaf Karagila♦
308k33441774
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 $mathbbE(X^2)=1$ 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
edited Mar 28 at 9:19
Asaf Karagila♦
308k33441774
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 $mathbbE(X^2)=1$ 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
edited Mar 28 at 9:19
Asaf Karagila♦
308k33441774
asked Mar 28 at 3:11
Xin FuXin Fu
34319
edited Mar 28 at 9:19
Asaf Karagila♦
308k33441774
asked Mar 28 at 3:11
Xin FuXin Fu
34319
edited Mar 28 at 9:19
Asaf Karagila♦
308k33441774
edited Mar 28 at 9:19
Asaf Karagila♦
308k33441774
edited Mar 28 at 9:19
Asaf Karagila♦
308k33441774
308k33441774
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
$$
alemathbb E(X) = int_Xlelambda aX,dP + int_Xgelambda aX,dP,le,lambda a + int_Xgelambda aX,dP.
$$
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 Mar 28 at 3:37
amsmathamsmath
3,287421
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Xin Fu
Mar 28 at 3:44
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 Mar 28 at 3:37
amsmathamsmath
3,287421
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Xin Fu
Mar 28 at 3:44
add a comment |
$begingroup$
You have
$$
alemathbb E(X) = int_Xlelambda aX,dP + int_Xgelambda aX,dP,le,lambda a + int_Xgelambda aX,dP.
$$
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 Mar 28 at 3:37
amsmathamsmath
3,287421
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Xin Fu
Mar 28 at 3:44
add a comment |
$begingroup$
You have
$$
alemathbb E(X) = int_Xlelambda aX,dP + int_Xgelambda aX,dP,le,lambda a + int_Xgelambda aX,dP.
$$
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 Mar 28 at 3:37
amsmathamsmath
3,287421
$endgroup$
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 Mar 28 at 3:37
amsmathamsmath
3,287421
answered Mar 28 at 3:37
amsmathamsmath
3,287421
answered Mar 28 at 3:37
amsmathamsmath
3,287421
answered Mar 28 at 3:37
amsmathamsmath
3,287421
3,287421
$begingroup$
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