## Deconvolution of gamma distributions

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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.

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## 4 Answers

No, take X=0 with probability 1. Y gamma

X+Y

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Thanks Stella. I should have stated that in the problem I am dealing with the component random variables are non-trivial, well behaved, continuous, positive. Does this make a difference? Or put slightly differently, are there reasonable properties I could specify for the component rvs that would cause the proposition to be true? – Trevor Stewart Nov 17 2009 at 17:11

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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.

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Michael, thanks. This is essentially the way I have been trying to tackle the problem, so far without success.  – Trevor Stewart Nov 17 2009 at 17:18

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.

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Jonathan, Yes this is true and my application contemplates fractional shape parameters. But my question is the converse: given that the sum of the two rvs is gamma are the two components necessarily also gamma. – Trevor Stewart Nov 17 2009 at 20:23

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.

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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. – George Lowther Nov 17 2009 at 22:59
Actually, A is gamma-distributed here, with k = 1. But B isn't, so this is still a counterexample. – Michael Lugo Nov 17 2009 at 23:01
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. – George Lowther Nov 17 2009 at 23:05
Further to my comment above, the more general example in my second edit can also be understood in terms of local times of Brownian motion. X_A is the local time at 0 of a Brownian motion B while its maximum process B*(t)=max_{s<=t}B(s) is in A. – George Lowther Nov 18 2009 at 2:08
Thanks, George. I'll work through this. Looks like an ingenious solution. – Trevor Stewart Nov 18 2009 at 12:53
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