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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 $f=g$?2012-02-16
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    @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 http://math.stackexchange.com/questions/109786/when-can-i-decompose-a-random-variable-y-x-x2012-02-16
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    @RobertIsrael Since construction of such two functions appear to be difficult, any advice on how to show existence?2012-02-21
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    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?2012-02-25
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    Yes, @josh. Thanks for the correction.2012-02-26

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