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.
The book Directional Statistics by Mardia and Jupp discusses concrete examples of distributions on:
- Surface of the unit hypersphere
- On Stiefel Manifolds
- Maybe some others
EDIT In particular, have a look at §13.4.2 that discusses distributions on more general manifolds (e.g., on compact Riemannian manifolds). That section also provides several useful references.
I also recalled another source that might be useful to you:
Matrix Variate Distributions by Gupta and Nagar. In particular, see chapter 8.
Best Answer
The following link on the stats.SE answers your question in detail.
You might also find this wikipedia link useful.