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How can one obtain the gamma distribution through convolution of two different distributions? Could the gamma distribution be created as a non-trivial sum of $N$ random variables $X$ which have the same distribution and parameters?

Trivial case is summation of fixed number $N$ of variables with gamma distribution as described on Wikipedia.

cardinal
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Qbik
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  • Your title and the beginning of your question references two different distributions, but later you talk about a sum of random variables from the *same* distribution. Which are you interested in? – cardinal May 07 '12 at 00:58
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    This problem is solved for any positive integral $n$ by taking the $n^\text{th}$ root of the characteristic function of the given Gamma distribution. This yields a c.f. which--by inspection--is easily seen to correspond to (another) Gamma distribution (with the same scale parameter but a different shape parameter). – whuber May 07 '12 at 17:12

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Wikipedia answered the question for you. The sum of any two independent identically distributed gamma random variables is gamma. So if you sum two independent identically distributed gammas you get a gamma. Is that what you meant or did you want to sum two distributions that are identical but not gamma? If that is the case I beleive the answer is no. Also since you mentioned N possibly greater than 2, it is also true as others have pointed out that any two gammas that are independent but not necessarily identically distributed will sum to a gamma if they have the same scale parameters. So you can get a gamma as the sum of n iid gammas. You see if you can get 2 by summing two iid gammas you can get 4 by adding another 2 iid gammas with the same distribution as the first two and so on. The chi-square is a special case where this works. Distributions like the normal and the chi square that can be represented as the sum of n iid random variables of the same form (normal or chi-square respectively) are called infinitely divisible.

Michael R. Chernick
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  • You could sum iid exponential distributions to create a gamma distribution, but of course the exponential distribution is a special case of the gamma distribution. – Neil G May 06 '12 at 19:06
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    *The sum of any two independent gamma random variables is gamma*. Hmm, Michael. It might be worth considering whether that statement is true or not. – cardinal May 06 '12 at 19:22
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    The orders sum (and give you a gamma distribution) whenever the scales are the same, but not otherwise. – Dilip Sarwate May 06 '12 at 19:52
  • @Dilip: Precisely. – cardinal May 06 '12 at 22:06
  • @MichaelChernick See also this [question](http://stats.stackexchange.com/q/27908/6633) and its answers to see what is being complained about in your answer. – Dilip Sarwate May 06 '12 at 22:24
  • Thanks. I was a little careless with my language but the sarcasm was unnecessary. The point is that if the question is whether you can add independent identically distributed gammas and get a gamma the answer is yes. If you sum random variables from a different distribution family the answer is no. – Michael R. Chernick May 06 '12 at 22:54
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    Michael, sarcasm? People *do* insist on a certain degree of precision, for a reason: many, if not most, of our questioners come to get advice to solve their problems and are not, themselves, experts in statistics. As you gain more experience here, you'll notice that overgeneralizing is one of the most likely mistakes a questioner will make. Often to seek improvement in both questions and answers, a form of Socratic questioning is used. Cheers. – cardinal May 06 '12 at 23:10
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    Michael, I apologize for the impression of sarcasm. @Cardinal is *extremely* good at helping people improve the mathematical accuracy of their statements and ordinarily is able to do so in a remarkably gentle, tactful way. I see how his first remark could be taken as sarcastic but I am *certain* it was not intended as such. I am pleased to see that these comments indeed led to a minor but important improvement in your reply here. Might I further suggest that your reply appears to be incomplete? The question asks how to represent *a given* Gamma as a sum of iid distributions. – whuber May 07 '12 at 17:09
  • @whuber I think it is cardinal who should apologize and not you for him. I was not complaining about being downvoted for being imprecise and hence in error. That I understand and appreciate. But his exact words were "The sum of any two independent gamma random variables is gamma. Hmm, Michael. It might be worth considering whether that statement is true or not. " If that isn't sarcasm I don't know what is! It was unnecessary and rude. Regarding your point I had not noticed that in the question it was for (N which could be greater than 2) and that the ditributions had to be the same. – Michael R. Chernick May 07 '12 at 20:27
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    I'd have taken it as a good-humored hint that I'd made a mistake - the humor intended to take any possible sting out of what could, in an alternate universe, have been a direct "You're wrong and here's why" response. But that's just me. – jbowman May 08 '12 at 00:12
  • I really don't mind being corrected. I said something technically wrong and deserved the downvote I got. I have no beef with that. But I saw no humor in cardinal's response. He apparently doesn't think he said anything wrong. – Michael R. Chernick May 08 '12 at 00:42
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    Michael, let me be perfectly direct for a moment: (1) Absolutely no sarcasm was intended or implied by my first comment. It was meant to be *helpful*. You are welcome to peruse the (many) comments I've made on this site and weigh the evidence regarding the intended tone. (2) I did *not* downvote your answer. It would be good form to temper such comments unless you can be sure of the person to whom they should be addressed. (3) The OpenID site with which I log in has been down for several hours now; I fail to see how my *lack* of response could be used as evidence for the inference you draw. – cardinal May 08 '12 at 01:23
  • @cardinal I did not accuse you of downvoting me. I just said that I felt that a downvote was jsutified and if you wanted to be critical of me I didn't take offense to that either it was justified. Maybe you didn't intend to sarcastic and offensive but even if you can't see the sarcasm in your statement others surely did. Otherwise why would Bill Huber be apologizing for you. Can I take your statement above as an apology? – Michael R. Chernick May 08 '12 at 02:15
  • @cardinal I guess I shouldn't have presumed that there were any implications to the slowness of your response. But I am still not sure whether or not you apologized for the unintended sarcasm. Also in my response to Bill I said I wasn't complaining about being downvoted. I did not intend to imply that you were the one who downvoted me but that is what you inferred. Very similar to you not intending to be sarcastic but I took it that way. – Michael R. Chernick May 08 '12 at 02:37
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You could also sum the squares of independent standard normal variates, giving chi-square random variables, which are a special case of the gamma.

Mike Anderson
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  • I'm curious why it seems that a new account is created nearly every time you post an answer. I'll flag this so that your account is merged with your previous ones. Cheers. – cardinal May 06 '12 at 19:23
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    @mike I prefer to comment on your comment rather than your account. It is a cute answer but once you square the nrmals you have a chi square which is gamma and what you are doing is summing (getting a convolution) i.i.d. gamma. – Michael R. Chernick May 06 '12 at 23:30