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Consider the following integrals

$\int g(x,y) N(y;x,1) dy$ and $\int g(x,y) N(x;y,1) dx$

where $N(x;y,1)$ is the Normal PDF of the variable $x$ with mean $y$ and variance $1$ ($x$,$y$ both real variables). What I know is that if $g(x,y)=\mathbb 1_{\{|x|\leq |y|\}}-0.5$, i.e., the indicator over the set $\{|x|\leq |y|\}$ minus $0.5$, then

$\int g(x,y) N(y;x,1) dy>0 ~~\forall x$

and

$\int g(x,y) N(x;y,1) dx<0 ~~\forall y.$

Consider now a generic $g(x,y)$, the only assumption about $g$ is that both the following integrals makes sense, and assume that

$\int g(x,y) N(y;x,1) dy>0 ~~\forall x$

Is it true that does not exist any $g(x,y)$ such that

$\sup\limits_{m \in [-a,a]}\int g(x,y) N(x;y+m,1) dx<0 ~~\forall y\, ?$

I do not know how to prove it, neither I was able to find a counter-example. It is easy to see that if $g(x,y)=\mathbf{1}_{\{|x|\leq |y|\}}-0.5$, then the last inequality does not hold.

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    Yes, the last inequality must be verified for each y. I really do not know how to prove it?2011-08-22

0 Answers 0