# Sure loss and comparison of marginal integrals. Looking for a proof or a counter-example

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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To get a negative value for a fixed $y$ is easy. Your question is about getting a negative supremum on a neighborhood of $y$ which does not depend on $y$, is that it? – Did Aug 20 '11 at 17:07
Yes, the last inequality must be verified for each y. I really do not know how to prove it? – vatna Aug 22 '11 at 8:03