Deconvolution of gamma distributions - MathOverflow

Deconvolution of gamma distributions

2009-11-17T16:31:10Z

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.

Answer by Stella for Deconvolution of gamma distributions

2009-11-17T16:38:05Z

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

X+Y

Answer by Michael Lugo for Deconvolution of gamma distributions

2009-11-17T16:55:27Z

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.

Answer by Jonathan Kariv for Deconvolution of gamma distributions

2009-11-17T20:03:12Z

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.

Answer by George Lowther for Deconvolution of gamma distributions

2009-11-17T22:39:27Z

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=∑_n2^1-nε_nA_n 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 X_A with moment generating function E[exp(-λX_A)]=exp(-λk∫_Adx/(1+λx)). If you partition (0,1] into two measurable sets A,B and X_A,X_B are independent, then X_A+X_B has the Gamma(k) distribution. If A and B are unions of finitely many intervals then the moment generating functions will be k^th powers of rational functions of λ and its easy to make sure that X_A,X_B 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,2^1-n].

You can construct X_A as follows. Let T₁,T₂,… be independent with the Exp(k) distribution, and S_n=exp(-T₁-…-T_n). The number of S_n in a subset A of (0,1] will be Poisson with parameter ∫_Adx/x. If Y₁,Y₂,… are independent exponentially distributed then X_A=∑_n1_{{S_n∈A}}S_nY_n has the correct moment generating function. (I'll leave you work through the details...). Alternatively, the set {(S_n,Y_n):n≥1} is a Poisson point process with intensity ke^-y dy ds/s.