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Is there a clever way to find two density functions, $f$ and $g$, that satisfy the following conditions?

$$\begin{align*} \int_{\infty}^{m}\int_{-\infty}^{\infty}f(w)f(w+z)\,dw\,dz&=\int_{\infty}^{m}\int_{-\infty}^{\infty}g(w)g(w+z)\,dw\,dz\\ \int_{\infty}^{m+2}\int_{-\infty}^{\infty}f(w)f(w+z)\,dw\,dz&=\int_{\infty}^{m+1}\int_{-\infty}^{\infty}g(w)g(w+z)\,dw\,dz\\ \int_{-\infty}^{\infty}f(w)\,dw&=\int_{-\infty}^{\infty}g(w)\,dw=M\\ \end{align*}$$ where $f\gt 0$ and $g\gt 0$ almost everywhere?

for $m\in (-\delta,\delta)$ and $\delta$ is some small number.

My main intent is to come up with two i.i.d. random variable, $X'$ and $X''$ and $Y$ and $Y''$, such that $\operatorname{\mathbb{Pr}}(m> Y'-Y'')=\operatorname{\mathbb{Pr}}(m>X'-X'')$ for $m \in (-b,b)$ for some $b$ small enough, while $\operatorname{\mathbb{Pr}}(m+2> Y'-Y'')=\operatorname{\mathbb{Pr}}(m+1> X'-X'')$.Is this possible?

Thanks so much in advance for your much appreciated help.

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What's wrong with $f=g$? –  Robert Israel Feb 16 '12 at 20:51
    
@Robert: I am not sure that in general $f=g$ will satisfy the second condition, one with $m+2$ and the other with $m+1$ in the limits of integration. SOmething similar is in math.stackexchange.com/questions/109786/… –  Henry Feb 16 '12 at 22:56
    
@RobertIsrael Since construction of such two functions appear to be difficult, any advice on how to show existence? –  Stuck_pls_help Feb 21 '12 at 21:47
    
I'm assuming the $= M$ at the end only applies to the last equation between the total weight of the densities $f$ and $g$ and not to the above equations, right? –  josh Feb 25 '12 at 16:54
    
Yes, @josh. Thanks for the correction. –  Stuck_pls_help Feb 26 '12 at 15:26

1 Answer 1

Why don't you see if you can work with characteristic function. This is almost related to divisibility, in which case one need only to examine characteristic functions with roots.

Look into Characteristic Functions by Eugene Lukacs.

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This isn't really the answer I was looking for, but thanks for the helpful reference. –  Stuck_pls_help Apr 3 '12 at 14:34

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