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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?













0












$begingroup$


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







share|cite|improve this question











$endgroup$











  • $begingroup$
    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















0












$begingroup$


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







share|cite|improve this question











$endgroup$











  • $begingroup$
    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













0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday







max

















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
















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Thanks for the clear explanation.
    $endgroup$
    – max
    yesterday










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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

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.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Thanks for the clear explanation.
    $endgroup$
    – max
    yesterday















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Thanks for the clear explanation.
    $endgroup$
    – max
    yesterday













1












1








1





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered yesterday









Kavi Rama MurthyKavi Rama Murthy

66k42867




66k42867











  • $begingroup$
    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




$begingroup$
Thanks for the clear explanation.
$endgroup$
– max
yesterday

















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