Deconvolution of gamma distributions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T15:35:32Z http://mathoverflow.net/feeds/question/5834 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/5834/deconvolution-of-gamma-distributions Deconvolution of gamma distributions Trevor Stewart 2009-11-17T16:31:10Z 2009-11-19T21:03:55Z <p>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.]</p> <p>If this can be answered assuming that the convolution is the exponential distribution f(x) = exp(-x) that would be sufficient for my purposes.</p> <p>Any help with this would be much appreciated. </p> http://mathoverflow.net/questions/5834/deconvolution-of-gamma-distributions/5835#5835 Answer by Stella for Deconvolution of gamma distributions Stella 2009-11-17T16:38:05Z 2009-11-17T16:38:05Z <p>No, take X=0 with probability 1. Y gamma </p> <p>X+Y </p> http://mathoverflow.net/questions/5834/deconvolution-of-gamma-distributions/5838#5838 Answer by Michael Lugo for Deconvolution of gamma distributions Michael Lugo 2009-11-17T16:55:27Z 2009-11-17T16:55:27Z <p>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.</p> <p>As for a counterexample: the <a href="http://en.wikipedia.org/wiki/Characteristic_function_(probability_theory)" rel="nofollow">characteristic function</a> of the exponential(1) distribution is 1/(1-it). The characteristic function of Gamma(k, &theta;) is (1-it&theta;)<sup>-k</sup>. 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) <i>other than</i> in the obvious way.</p> http://mathoverflow.net/questions/5834/deconvolution-of-gamma-distributions/5862#5862 Answer by Jonathan Kariv for Deconvolution of gamma distributions Jonathan Kariv 2009-11-17T20:03:12Z 2009-11-17T20:03:12Z <p>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.</p> http://mathoverflow.net/questions/5834/deconvolution-of-gamma-distributions/5882#5882 Answer by George Lowther for Deconvolution of gamma distributions George Lowther 2009-11-17T22:39:27Z 2009-11-19T21:03:55Z <p>You can decompose the exponential distribution into a sum of two terms, which are not both gamma distributed.</p> <p>Let A,B,&epsilon; be independent where A,B are exponentially distributed and &epsilon; takes the values 0,1 each with probability 1/2, and set X=A/2, Y=&epsilon;B. 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(-&lambda;(X+Y)]=E[exp(-&lambda;X)]E[exp(-&lambda;Y)]=1/(1+&lambda;) to see that X+Y has the exponential distribution.</p> <p>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 A<sub>1</sub>,A<sub>2</sub>,... have the exponential distribution and &epsilon;<sub>1</sub>,&epsilon;<sub>2</sub>,... take values 0,1 each with probability 1/2 (and all these rvs are independent), then X=&sum;<sub>n</sub>2<sup>1-n</sup>&epsilon;<sub>n</sub>A<sub>n</sub> 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.</p> <p>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.</p> <p>For any k&gt;0 and measurable subset A of the interval (0,1], you can define a random variable X<sub>A</sub> with moment generating function E[exp(-&lambda;X<sub>A</sub>)]=exp(-&lambda;k&int;<sub>A</sub>dx/(1+&lambda;x)). If you partition (0,1] into two measurable sets A,B and X<sub>A</sub>,X<sub>B</sub> are independent, then X<sub>A</sub>+X<sub>B</sub> has the Gamma(k) distribution. If A and B are unions of finitely many intervals then the moment generating functions will be k<sup>th</sup> powers of rational functions of &lambda; and its easy to make sure that X<sub>A</sub>,X<sub>B</sub> 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<sup>-n</sup>,2<sup>1-n</sup>].</p> <p>You can construct X<sub>A</sub> as follows. Let T<sub>1</sub>,T<sub>2</sub>,&hellip; be independent with the Exp(k) distribution, and S<sub>n</sub>=exp(-T<sub>1</sub>-&hellip;-T<sub>n</sub>). The number of S<sub>n</sub> in a subset A of (0,1] will be Poisson with parameter &int;<sub>A</sub>dx/x. If Y<sub>1</sub>,Y<sub>2</sub>,&hellip; are independent exponentially distributed then X<sub>A</sub>=&sum;<sub>n</sub>1<sub>{S<sub>n</sub>&isin;A}</sub>S<sub>n</sub>Y<sub>n</sub> has the correct moment generating function. (I'll leave you work through the details...). Alternatively, the set {(S<sub>n</sub>,Y<sub>n</sub>):n&ge;1} is a <em>Poisson point process</em> with intensity ke<sup>-y</sup>&thinsp;dy&thinsp;ds/s.</p>