Decomposition of the sum of two random variables We observe the input samples as $Z = X+Y$, where distribution of $Z$ could be estimated by histogram of samples and $X,Y$ are two independent random variables. One of the variable was known following exponential distribution (as $X \sim Exp(\lambda)$).
I want to get the parameter of $X$, $\lambda$, and probablity density of $Y$ with its parameters also. How to decompose the random variable $X+Y$?
Further more, what if the distribution of $X$ is unknown?
 A: As stated, the law of $Z$ is the convolution of the law of $X$ and the law of $Y$. Finding the density of $X$ is called density deconvolution and $Y$ is generally considered as additive noise. Usually such a problem is considered by taking the fourier transforms of each variable (provided they exist) such that $f^{Ft} _Z = f^{Ft} _X \times f^{Ft} _Y$, where $f_Y ^{Ft}$ is the charateristic function of an exponential distribution in your case. Even if you were willing to give a parametric law to $X$, it is unusual that an analytic expression of $Z$ can be recovered. 
Provided the Fourier transform of $Y$ never vanishes you may write $f^{Ft} _X = f^{Ft} _Z / f^{Ft} _Y$. Given a sample of realizations of $Z$ and replacing $f_Z ^{Ft}$ by the empirical characteristic function, the numerical integration back from the Fourier domain is a difficult task since the integral is generally not defined on $\mathbb{R}$. There are several methods that allow to do so if you have realizations of thes r.v. and a large sample.  and either truncate the integral over a reasonable domain or use an appropriate kernel function. In both cases you are stuck with the choice of something like a bandwidth parameter for which there are solutions.
I must say however that the joint estimation of $\lambda$ and $f_X$ seems very ambitious and would require to construct an optimization scheme that encompasses this parameter as well...writing this down properly might indeed turn out to be ill-posed in most cases...
A: The problem is ill-posed.  There is no unique solution if you don't specify Z.  Think about it.  Pick any Y with any distribution you choose.  X+Y will have some distribution that depends on what you chose for Y.  So if you specify Z then Y is determined but two distinct choices for Y will give two different distributions for Z.  There are infinitely many solutions to your problem.
Now given that of course your second problem is also ill-posed. Just knowing that X and Y are independent tells you almost nothing about Z.  Take X and Y normal and Z will be normal.  If X and Y are identically distributed Cauchys Z will be Cauchy.  Two independent chi-squares lead to Z being a chi-square.  So there are three solutions to problem 2 and that is only scratching the surface
In problem 1 you need to know Z's distribution to determine Y's.  In problem 2 you can't get Z without specifying the distributions of X and Y.
