Is $f(x)equiv 0$ necessary condition for $mathbbE[Xf(X)mathbb1_[0,infty)(X)]=0$?convolution of random variablesApplication of Strong Law of Large Numbers and Fubini's TheoremWhen does $mathbb E_mathbb P[X]=0$ imply $mathbb E_mathbb P[Xmidmathcal E]= 0$ $mathbb P$-a.s.$f$ convex, is $f(X)$ quasi-integrable if $X$ is?Necessary and Sufficient Conditions for Convergence to Standard NormalConditional ExpectationProblem on conditional expected value with to a random variableRelation between uncountably infinite probability space and continuous random variablesDoes there exist some probability space $(Omega,mathcal F,mathbb P)$ that admits random variables with all possible laws on $mathbb R^n$?What is the necessary and sufficient condition of Markov chain sample average converging to the expectation wrt the stationary distribution?
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Is $f(x)equiv 0$ necessary condition for $mathbbE[Xf(X)mathbb1_[0,infty)(X)]=0$?
convolution of random variablesApplication of Strong Law of Large Numbers and Fubini's TheoremWhen does $mathbb E_mathbb P[X]=0$ imply $mathbb E_mathbb P[Xmidmathcal E]= 0$ $mathbb P$-a.s.$f$ convex, is $f(X)$ quasi-integrable if $X$ is?Necessary and Sufficient Conditions for Convergence to Standard NormalConditional ExpectationProblem on conditional expected value with to a random variableRelation between uncountably infinite probability space and continuous random variablesDoes there exist some probability space $(Omega,mathcal F,mathbb P)$ that admits random variables with all possible laws on $mathbb R^n$?What is the necessary and sufficient condition of Markov chain sample average converging to the expectation wrt the stationary distribution?
$begingroup$
I have the following question:
Let $XsimmathcalN(0,1)$ and $mathbbE[Xf(X)mathbb1_[0,infty)(X)]=0$. Clearly it is sufficient that $f(x)=0$ for all values of the domain. Is it also a necessary condition?
So I separated the function $f$ into its positive and negative parts $f=f^+-f^-$. It follows that $mathbbE[Xf^+(X)mathbb1_A]>0$ for $A:=omega: g(X(omega))>0$. Now I need to show that $mathbbP(A)=0$ but I'm not sure how how to proceed. Do I need to use some sort of 0-1 law? The help would be much appreciated.
probability probability-theory probability-distributions
asked yesterday
maxmax
886
$endgroup$
add a comment |
$begingroup$
I have the following question:
Let $XsimmathcalN(0,1)$ and $mathbbE[Xf(X)mathbb1_[0,infty)(X)]=0$. Clearly it is sufficient that $f(x)=0$ for all values of the domain. Is it also a necessary condition?
So I separated the function $f$ into its positive and negative parts $f=f^+-f^-$. It follows that $mathbbE[Xf^+(X)mathbb1_A]>0$ for $A:=omega: g(X(omega))>0$. Now I need to show that $mathbbP(A)=0$ but I'm not sure how how to proceed. Do I need to use some sort of 0-1 law? The help would be much appreciated.
probability probability-theory probability-distributions
asked yesterday
maxmax
886
$endgroup$
$begingroup$
What is the meaning of $1_[0,infty)$. You probably mean $1_[0,infty)(X)$.
$endgroup$
– Kavi Rama Murthy
yesterday
$begingroup$
yes, that's what I mean
$endgroup$
– max
yesterday
add a comment |
$begingroup$
I have the following question:
Let $XsimmathcalN(0,1)$ and $mathbbE[Xf(X)mathbb1_[0,infty)(X)]=0$. Clearly it is sufficient that $f(x)=0$ for all values of the domain. Is it also a necessary condition?
So I separated the function $f$ into its positive and negative parts $f=f^+-f^-$. It follows that $mathbbE[Xf^+(X)mathbb1_A]>0$ for $A:=omega: g(X(omega))>0$. Now I need to show that $mathbbP(A)=0$ but I'm not sure how how to proceed. Do I need to use some sort of 0-1 law? The help would be much appreciated.
probability probability-theory probability-distributions
asked yesterday
maxmax
886
$endgroup$
I have the following question:
Let $XsimmathcalN(0,1)$ and $mathbbE[Xf(X)mathbb1_[0,infty)(X)]=0$. Clearly it is sufficient that $f(x)=0$ for all values of the domain. Is it also a necessary condition?
So I separated the function $f$ into its positive and negative parts $f=f^+-f^-$. It follows that $mathbbE[Xf^+(X)mathbb1_A]>0$ for $A:=omega: g(X(omega))>0$. Now I need to show that $mathbbP(A)=0$ but I'm not sure how how to proceed. Do I need to use some sort of 0-1 law? The help would be much appreciated.
probability probability-theory probability-distributions
probability probability-theory probability-distributions
asked yesterday
maxmax
886
asked yesterday
maxmax
886
edited yesterday
max
asked yesterday
maxmax
886
asked yesterday
maxmax
886
asked yesterday
maxmax
886
886
$begingroup$
What is the meaning of $1_[0,infty)$. You probably mean $1_[0,infty)(X)$.
$endgroup$
– Kavi Rama Murthy
yesterday
$begingroup$
yes, that's what I mean
$endgroup$
– max
yesterday
add a comment |
$begingroup$
What is the meaning of $1_[0,infty)$. You probably mean $1_[0,infty)(X)$.
$endgroup$
– Kavi Rama Murthy
yesterday
$begingroup$
yes, that's what I mean
$endgroup$
– max
yesterday
$begingroup$
What is the meaning of $1_[0,infty)$. You probably mean $1_[0,infty)(X)$.
$endgroup$
– Kavi Rama Murthy
yesterday
$begingroup$
What is the meaning of $1_[0,infty)$. You probably mean $1_[0,infty)(X)$.
$endgroup$
– Kavi Rama Murthy
yesterday
$begingroup$
yes, that's what I mean
$endgroup$
– max
yesterday
$begingroup$
yes, that's what I mean
$endgroup$
– max
yesterday
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It does not follow that $f$ is $0$. Example: take $f(x)=a$ for $0<x<1$ and $f(x)=b$ for $1 <x<infty$. Then the hypothesis becomes $aint_0^1xphi(x)dx+bint_1^infty x phi(x)dx=0$ where $phi$ is the standard normal density function. It is clear that you can choose non-zero $a$ and $b$ satisfying this equation.
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
66k42867
$endgroup$
$begingroup$
Thanks for the clear explanation.
$endgroup$
– max
yesterday
add a comment |
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$begingroup$
It does not follow that $f$ is $0$. Example: take $f(x)=a$ for $0<x<1$ and $f(x)=b$ for $1 <x<infty$. Then the hypothesis becomes $aint_0^1xphi(x)dx+bint_1^infty x phi(x)dx=0$ where $phi$ is the standard normal density function. It is clear that you can choose non-zero $a$ and $b$ satisfying this equation.
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
66k42867
$endgroup$
$begingroup$
Thanks for the clear explanation.
$endgroup$
– max
yesterday
add a comment |
$begingroup$
It does not follow that $f$ is $0$. Example: take $f(x)=a$ for $0<x<1$ and $f(x)=b$ for $1 <x<infty$. Then the hypothesis becomes $aint_0^1xphi(x)dx+bint_1^infty x phi(x)dx=0$ where $phi$ is the standard normal density function. It is clear that you can choose non-zero $a$ and $b$ satisfying this equation.
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
66k42867
$endgroup$
$begingroup$
Thanks for the clear explanation.
$endgroup$
– max
yesterday
add a comment |
$begingroup$
It does not follow that $f$ is $0$. Example: take $f(x)=a$ for $0<x<1$ and $f(x)=b$ for $1 <x<infty$. Then the hypothesis becomes $aint_0^1xphi(x)dx+bint_1^infty x phi(x)dx=0$ where $phi$ is the standard normal density function. It is clear that you can choose non-zero $a$ and $b$ satisfying this equation.
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
66k42867
$endgroup$
It does not follow that $f$ is $0$. Example: take $f(x)=a$ for $0<x<1$ and $f(x)=b$ for $1 <x<infty$. Then the hypothesis becomes $aint_0^1xphi(x)dx+bint_1^infty x phi(x)dx=0$ where $phi$ is the standard normal density function. It is clear that you can choose non-zero $a$ and $b$ satisfying this equation.
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
66k42867
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
66k42867
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
66k42867
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
66k42867
66k42867
$begingroup$
Thanks for the clear explanation.
$endgroup$
– max
yesterday
add a comment |
$begingroup$
Thanks for the clear explanation.
$endgroup$
– max
yesterday
$begingroup$
Thanks for the clear explanation.
$endgroup$
– max
yesterday
$begingroup$
Thanks for the clear explanation.
$endgroup$
– max
yesterday
add a comment |
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$begingroup$
What is the meaning of $1_[0,infty)$. You probably mean $1_[0,infty)(X)$.
$endgroup$
– Kavi Rama Murthy
yesterday
$begingroup$
yes, that's what I mean
$endgroup$
– max
yesterday