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I am wondering if I can find a decomposition of $Y$ that is absolutely continuous nto two i.i.d. random variables $X'$ and $X''$ such that $Y=X'-X''$, where $X'$ is also Lebesgue measure with an almost everywhere positive density w.r.t to the Lebesge mesure.

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 $\operatorname{\mathbb{Pr}}(m+2> Y'-Y'')=\operatorname{\mathbb{Pr}}(m+1> X'-X'')$. I figured starting first by constructing a measure on the difference first that satisfies the above then decomposing it. Is this possible?

Thanks so much in advance for your much appreciated help.

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    See wiki page about [indecomposable distributions](http://en.wikipedia.org/wiki/Indecomposable_distribution). $f(x) \sim x^2 \exp(-x^2/2)$ is an example of an absolutely continuous indecomposable distribution.2012-02-15
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    Thanks a bunch Sasha for these counter example. I undersand that there are indecomposable distributions, but what I am wondering is if I can generate one that is decomposable. My main intent is to come up with two iid random variable, X' and X'' and Y and Y'', such that Pr(m> Y'-Y'')=Pr(m>X'-X'') for m \in (-b,b) for some b small enough, while Pr(m+2> Y'-Y'')=Pr(m+1> X'-X''). I figured starting first by working with the difference in coming up with a measure satisfyinig this then decomposing it. Is this possible?2012-02-15

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