Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Thanks, Michael, I fixed my post. I suppose you could further decompose X in a similar way to get a sum of as many independent rvs as you like, and them rearrange them into two terms neither of which are gamma distributed.
Some background on where my example came from: I know that the local time at 0 of a Brownian motion B at the first time it hits 1 has the exponential distribution. If you understand these concepts, then it is easy to see that it is the sum of the local time at 0 at which B first hits 1/2 plus the local time at 0 of the BM started at 1/2 when it first hits 1. These are independent, and the second has a prob of 1/2 of being 0, so can't be gamma distributed. I just converted this example into a simple argument using moment generating functions.
Interesting that you get a weaker condition for distributions supported on the positive reals, but I can see why that should be so. If X is a non-negative random variable then you can set Y=ε√X, where ε=+1,-1 each with probability 1/2 and independent of X. Then, the odd moments of Y are zero, the 2n'th moment of Y equals the n'th moment of X, and determining the distribution of Y is enough to find the distribution of X.
This sounds like a very interesting question. For an arbitrary Banach algebra, I would have thought that the answer is no. Have to think of a counterexample...
This seems like a reasonable response to the question. Because it relates to a physical problem doesn't prove the result mathematically, or help to prove it, unless the physical situation can be handled with rigorous mathematics. The mathematical theory of Brownian motion does this (although it wasn't understood mathematically in the 19th century).
Indeed. Start a regular Brownian motion from some point x in the interior. Take the expectation of the boundary value at the first boundary point it hits. This gives a rigorous proof, and any open bounded domain will work (and bounded & measurable boundary values).