Deconvolution of gamma distributions If the sum of two independent random variables is gamma distributed does this imply that the individual random variables are also gamma distributed. I suspect that the answer is no, but I do not know how to construct a counter-example. [The convolution of two gamma distributions with the same scale parameter is also a gamma distribution, but I am dealing with the converse.]
If this can be answered assuming that the convolution is the exponential distribution f(x) = exp(-x) that would be sufficient for my purposes.
Any help with this would be much appreciated.   
 A: You can decompose the exponential distribution into a sum of two terms, which are not both gamma distributed.
Let A,B,ε be independent where A,B are exponentially distributed and ε takes the values 0,1 each with probability 1/2, and set X=A/2, Y=εB. You can calculate the moment generating functions of X and Y,
$$
E\left[\exp(-\lambda X)\right] = E\left[\exp(-(\lambda/2)A)\right]=1/(1+\lambda/2).
$$
$$
E\left[\exp(-\lambda Y)\right]=(1/2)E\left[\exp(-\lambda B)\right]+1/2=(2+\lambda)/(2+2\lambda).
$$
Then you can check the moment generating function function of X+Y, E[exp(-λ(X+Y)]=E[exp(-λX)]E[exp(-λY)]=1/(1+λ) to see that X+Y has the exponential distribution.
Edit:
After reading at Michael Lugo's response below, it might be more satisfying to have an answer where neither of X or Y are Gamma distributed. In fact, by iterating my argument above you can get the following example. If A1,A2,... have the exponential distribution and ε1,ε2,... take values 0,1 each with probability 1/2 (and all these rvs are independent), then X=∑n21-nεnAn has the exponential distribution (just check the moment generating function). By splitting this sum up into two smaller sums you can generate a whole load of counterexamples where neither term is gamma distributed.
Edit 2: Apologies for keeping coming back to this one, but it seems interesting and my examples above are a special case of the following.
For any k>0 and measurable subset A of the interval (0,1], you can define a random variable XA with moment generating function E[exp(-λXA)]=exp(-λk∫Adx/(1+λx)). If you partition (0,1] into two measurable sets A,B and XA,XB are independent, then XA+XB has the Gamma(k) distribution. If A and B are unions of finitely many intervals then the moment generating functions will be kth powers of rational functions of λ and its easy to make sure that XA,XB are not gamma distributed. My first example above is using k=1 and the partition (0,1/2],(1/2,1]. The second one, in the edit, is partitioning (0,1] into the intervals (2-n,21-n].
You can construct XA as follows. Let T1,T2,… be independent with the Exp(k) distribution, and Sn=exp(-T1-…-Tn). The number of Sn in a subset A of (0,1] will be Poisson with parameter ∫Adx/x. If Y1,Y2,… are independent exponentially distributed then XA=∑n1{Sn∈A}SnYn has the correct moment generating function. (I'll leave you work through the details...). Alternatively, the set {(Sn,Yn):n≥1} is a Poisson point process with intensity ke-y dy ds/s.
A: Stella gives the example of X = 0 with probability 1 and Y a gamma distribution.  I wouldn't count this as a "true" counterexample, because one could think of X = 0 as a Gamma(k,0) random variable.
As for a counterexample: the characteristic function of the exponential(1) distribution is 1/(1-it).  The characteristic function of Gamma(k, θ) is (1-itθ)-k.  So the question is whether you can write 1/(1-it) as the product of two things which are characteristic functions of positive measures (i. e. random variables) other than in the obvious way.
A: The exponential (with mean 1) can be written as the sum of 2 independent chi-squared 1s. Of course this is just allowing fractional values of the "n" parameter.
A: No, take X=0 with probability 1. Y gamma 
X+Y 
