# When can I decompose a random variable $Y=X'-X''$?

1. Feb 16, 2012

### aspiring88

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 $Pr(m> Y'-Y'')=Pr(m>X'-X'')$ 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 constructing a measure on the difference first that satisfies the above then decomposing it. Is this possible?

Mod note: fixed LaTeX

Last edited by a moderator: Feb 16, 2012
2. Feb 16, 2012

### mathman

fix the latex.

3. Feb 25, 2012

### bpet

Firstly your Y must be symmetric. If it has a characteristic function g(t) you could check whether g(t) = h(t)h(-t) for some other c.f. h(t).

4. Feb 25, 2012

### chiro

Hey aspiring88 and welcome to the forums.

Since you are using the i.i.d property for your random variables, what you can do is use the convolution property, but you have to map your probabilities to the 'inverse' values instead of the positive values: in other words if your domain for the B RV in X = A + B is [0,infinity), then you have to change the mapping from (-infinity,0] and this can be done by just flipping the sign.

Using the convolution theorem, you can substitute your identities in and you will have a relationship that has to hold.

Because the convolution should return the CDF directly, this means that you should basically get a relationship between two integral expressions and from there you can get more specific with the properties of your density functions as you wish.

5. Apr 21, 2012

### aspiring88

Thanks so much @chiro and @bpet. I'm still a bit loss. So you're recommendation is to start with one arbitrary characteristic function that is hopefully decomposable and see if I can craft another one?

Thanks again.