Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
add comment

4 Answers

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 ε12,... 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.

share|improve this answer
1  
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 '09 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 '09 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 '09 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 '09 at 2:08
    
Thanks, George. I'll work through this. Looks like an ingenious solution. –  Trevor Stewart Nov 18 '09 at 12:53
add comment

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.

share|improve this answer
    
Michael, thanks. This is essentially the way I have been trying to tackle the problem, so far without success. –  Trevor Stewart Nov 17 '09 at 17:18
add comment

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.

share|improve this answer
    
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 '09 at 20:23
add comment

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

X+Y

share|improve this answer
    
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 '09 at 17:11
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.