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Is there anything nice I can say about the sum of two independent generalized Laplace variables, with different scales and sizes? i.e. are they distributed same as another generalized Laplace variable with some function of the moments, etc.

Edit: the PDF of generalized {Laplace or Gaussian} is: $f(x) = C\exp\{-|(x-\mu)/a|^b\}$ where $a$ is the scale and $b$ is the shape ($C$ is a normalization constant).

Thanks.

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    $\begingroup$ How generalised? There are lots of competing definitions floating about, so please specify the functional form for the pdf you are working with. Also, what does the or Guassian mean? Do you mean some generalised Gaussian distribution (again many competing definitions) or just a common general $N(\mu, \sigma^2)$ Gaussian? $\endgroup$ – wolfies Jun 18 '14 at 14:35
  • $\begingroup$ Can you be more specific about what precisely you mean by "generalized Laplace"? $\endgroup$ – Glen_b -Reinstate Monica Jun 19 '14 at 8:37
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    $\begingroup$ Heuristically, the sums should not be generalized Gaussian. (1) The shape parameter controls the heaviness of the tail, and the sums should in some sense have the same tail behavior (so $a$ must be the same) but (2) the CLT suggests that the mean will be asymptotically normal (i.e. Gaussian shaped near the mean) but the shape parameter also determines the shape of the distribution near the location parameter - so the shape parameter would need to change if you summed them up. $\endgroup$ – guy Jun 23 '14 at 1:01
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I'm not sure there's much nice to say about it: I don't recognize the characteristic of $X + Y$, and can't find a way to manipulate it into the same generalized form.

Taking the characteristic function of generalized normal random variable $X$ as given in (2.2) of Pogany and Nadarajah, noting a change in parameter names to match those used in the question:

$\phi_X(t) = \frac{e^{it\mu}}{\Gamma(1 / b)}\sum\limits_{m=0}^\infty\frac{\Gamma(2m/b+1/b)}{\Gamma(2m+1)}(-1)^m(at)^{2m}$

(I've changed the formula slightly for ease of differentiation, and replaced a factor of 2 that the paper lost in the algebraic manipulation.)

Then finding the characteristic of the sum as the product of their respective characteristic functions:

$\phi_{X+Y}(t) = \frac{e^{it\mu_X}}{\Gamma(1 / b_X)}\sum\limits_{m=0}^\infty\frac{\Gamma(2m/b_X+1/b_X)}{\Gamma(2m+1)}(-1)^m(a_Xt)^{2m}\frac{e^{it\mu_Y}}{\Gamma(1 / b_Y)}\sum\limits_{m=0}^\infty\frac{\Gamma(2m/b_Y+1/b_Y)}{\Gamma(2m+1)}(-1)^m(a_Yt)^{2m}$

$=\frac{e^{it(\mu_X + \mu_Y)}}{\Gamma(1/b_X)\Gamma(1/b_Y)}\sum\limits_{m=0}^\infty\frac{\Gamma(2m/b_X+1/b_X)}{\Gamma(2m+1)}(-1)^m(a_Xt)^{2m}\sum\limits_{m=0}^\infty\frac{\Gamma(2m/b_Y+1/b_Y)}{\Gamma(2m+1)}(-1)^m(a_Yt)^{2m}$

We can use that to find the moments of $X+Y$, but as guy points out, that tells us nothing we don't already know from the independence of $X,Y$.

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    $\begingroup$ If $X$ and $Y$ are independent with finite variance, $E[X + Y] = E[X] + E[Y]$ and $\mbox{Var}[X + Y] = \mbox{Var}[X] + \mbox{Var}[Y]$ no matter how the RVs are distributed, so unless I'm missing something I don't see how this answer tells us anything we didn't already know. $\endgroup$ – guy Jun 23 '14 at 0:58
  • $\begingroup$ @guy Yes, come to think of it, I should have stopped at the product of characteristics and asked if anyone could see a way to manipulate it into something recognizable. $\endgroup$ – Sean Easter Jun 23 '14 at 3:50
  • $\begingroup$ The paper you refer to does not contain the answer -it does not have any results related to the sum of two such r.v.'s. By the way, this distribution is nothing more than the "Power Exponential" or "Exponential Power" or "Subbotin" distribution, that has been around from 1923. So there may be older relevant papers. $\endgroup$ – Alecos Papadopoulos Jun 23 '14 at 18:10
  • $\begingroup$ @AlecosPapadopoulos Tx! Edited to remove the (pointless) reference. I did another search using the other names, and still couldn't find an explicit result. $\endgroup$ – Sean Easter Jun 23 '14 at 20:35
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I have actually encountered a paper that answers my question, and the answer is, as users have noted, that the result is not GGD. However, one can say something about the similarity of the sum to a GGD.

Qian Zhao; Hong-wei Li; Yuan-tong Shen, "On the sum of generalized Gaussian random signals," Signal Processing, 2004. Proceedings. ICSP '04. 2004

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