Decomposition of a random variable Having a random variable $X$ with a given density function $f_{X} : [a .. b] \rightarrow\mathcal{R}_{+}$ and a point c, $c \in [a .. b]$ I am curious if the following problem has a solution:
Find two independent random variables $X_1$ and $X_2$ such that $X = X_1 + X_2$ with the additional property that the density of $X_1$ is defined over the interval $[a .. c]$ and the density of $X_2$ is defined over the interval $[c .. b]$.
Could you also please point me out to relevant literature to read about this topic.
Thanks,
Bogdan.

Based on your observations the solution I have in mind is the following:


*

*Write $X = X_1 + X_2 - c$, where $X_1$ and $X_2$ are defined over the interval $[a ..c]$ and $[c ..b]$ respectively. 

*Assume $X_2$ to be normal with mean $\mu = \frac{b + c}{2}$ and $\sigma = 1$.

*Compute the density of $X_1$ by solving the deconvolution problem knowing the density of $X$ and the density of $X_2$.


The question I have in mind now is that for what kind of densities of X one can find the density of X1 assuming normal density for X2.
Thanks,
Bogdan.
 A: Since $f_{X_1 + X_2} = f_{X_1} * f_{X_2}$ (easily shown using MGFs and independence) you basically have to verify
$$\int^{b}_{a} f_{X}(x) dx = \int^b_a \int^{c}_{a} f_{X_1}(\xi)f_{X_2}(x-\xi) d\xi dx = 1$$
With your assumptions
$$f_{X_1}(x_1) = \chi_{[a, c]}(x_1) f_{X_1}(x_1)$$
$$f_{X_2}(x_2) = \chi_{[c, b]}(x_2) f_{X_2}(x_2)$$
The convolution integral which defines $f_X(x)$ is
$$f_X(x)=\int^{+\infty}_{-\infty} f_{X_1}(\xi)f_{X_2}(x-\xi) d\xi$$
$$=\int^{+\infty}_{-\infty} \chi_{[a, c]}(\xi) f_{X_1}(\xi) \chi_{[c, b]}(x-\xi)f_{X_2}(x-\xi) d\xi$$
$$=\int^{c}_{a} f_{X_1}(\xi) \chi_{[c, b]}(x-\xi) f_{X_2}(x-\xi) d\xi$$
Now, $\chi_{[c,b]}(x-\xi) = \chi_{[x-b, x-c]}(\xi)$ because the indicator function is nonzero for $c \leq x-\xi \leq b$ that is $x-b < \xi < x-c$ so the integral is nonzero when $[x-b, x-c] \cap [a, c] \neq \emptyset$ but since $x \in [a, b]$ by your hypotheses, then you should check the intersection of the two supports.
An example solution would be, e.g. given $X_1 \sim U([-1, 0]), X_2 \sim U([0,1])$, $$f_X(x) = 
\begin{cases}
 1-x & 0<x<1 \\
 x+1 & -1<x<0 \\
 0 & \text{otherwise} \end{cases}$$
In case you are interested, try this Mathematica code out for the previous example and another case easily derived.
A: Would you agree that this is always true? 
$a+c\le (X_1+X_2) \le c+b$
If you do, as I hope, then you'd agree that unless $f_{X}(x\in [a .. c]) =f_{X}(x\in [c .. b])=0$ it is impossible to come up with $X=X_1+X_2$. 
So, my answer is that in a general case you can't decompose $X$ into independent $X_1+X_2$ with domains $[a\dots c],[c\dots b]$. 
