2
$\begingroup$

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.

Improve this question
$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$
    – cardinal
    May 7, 2012 at 0:58
  • 2
    $\begingroup$ 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). $\endgroup$
    – whuber ♦
    May 7, 2012 at 17:12

2 Answers 2

Reset to default
1
$\begingroup$

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.

Improve this answer
$\endgroup$
14
  • 2
    $\begingroup$ 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. $\endgroup$
    – cardinal
    May 6, 2012 at 19:22
  • 3
    $\begingroup$ The orders sum (and give you a gamma distribution) whenever the scales are the same, but not otherwise. $\endgroup$
    – Dilip Sarwate
    May 6, 2012 at 19:52
  • 7
    $\begingroup$ 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. $\endgroup$
    – cardinal
    May 6, 2012 at 23:10
  • 3
    $\begingroup$ 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. $\endgroup$
    – jbowman
    May 8, 2012 at 0:12
  • 5
    $\begingroup$ 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. $\endgroup$
    – cardinal
    May 8, 2012 at 1:23
0
$\begingroup$

You could also sum the squares of independent standard normal variates, giving chi-square random variables, which are a special case of the gamma.

Improve this answer
$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – cardinal
    May 6, 2012 at 19:23
  • 1
    $\begingroup$ @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. $\endgroup$
    – Michael R. Chernick
    May 6, 2012 at 23:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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