# Is the Poisson distribution stable and are there inversion formulas for the MGF?

First, I have a question about whether the Poisson distribution is "stable" or not. Very naively (and I'm not too sure about "stable" distributions), I worked out the distribution of a linear combination of Poisson distributed R.V.'s, using the product of the MGF. It looks like I get another Poisson, with parameter equal to the linear combination of the parameters of the individual R.V.'s. So I conclude that Poisson is "stable". What am I missing?

Second, are there inversion formulas for the MGF like there are for the characteristic function?

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It is closed under (independent) sums, but not arbitrary linear combinations. If you include your work, I suspect you'll end up seeing why in the process; and, if not, someone will be able to point it out. Yes, there are some inversion analogues to that of characteristic functions. What do you know about the Laplace transform and Bromwich contour integration? –  cardinal May 23 '12 at 17:11
OK, I'll go back to the drawing board. I have the MGF of the i-th Poisson as: exp(lambda_i (exp(t) - 1)). So the product of n Poisson MGF's gives me: exp(sum(i,0,n) alpha_i * lambda_i * (exp(t) - 1)) and I take the new lambda = sum(i,0,n) alpha_i * lambda_i. Now I'm afraid I'm going to look stupid for making an obvious mistake. - I know about the Laplace transform and contour integration in general, but not Bromwish contour integration. - Would you recommend working with the CFs rather than the MGFs in general? It seems more powerful. –  Frank May 23 '12 at 17:20
What is the $\alpha_i$ in your comment? Also, surround your math-LaTeX with dollar signs to get it to work (using \exp to make the "exp" come out looking right, and \lambda to make a $\lambda$, \sum for $\sum$, etc.) –  jbowman May 23 '12 at 17:28
Yes, I'm not very good at LaTex, but here goes. So, my linear combination of R.V.s is: $$\sum_{i=0}^{n}\alpha_{i} X_{i}$$, and the product of their MGFs is: $$\exp(\sum_{i=0}^{n}\alpha_{i} \lambda_{i} (\exp(t_{i}) - 1))$$, if I'm correct, if the R.V.s are distributed as $Poisson(\lambda_{i})$. I had used the same t for all the R.V.s, but I need to use $t_{i}$. –  Frank May 23 '12 at 17:36
The mistake is that the MGF of $a_iX_i$ is $exp(\lambda_i (exp(a_i t)-1))$ and not $exp(a_i\lambda_i (exp(t)-1))$ –  gui11aume May 23 '12 at 18:09

Linear combinations of Poisson random variables

As you've calculated, the moment-generating function of the Poisson distribution with rate $\lambda$ is $$m_X(t) = \mathbb E e^{t X} = e^{\lambda (e^t - 1)} \>.$$

Now, let's focus on a linear combination of independent Poisson random variables $X$ and $Y$. Let $Z = a X + b Y$. Then, $$m_Z(t) = \mathbb Ee^{tZ} = \mathbb E e^{t (a X + b Y)} = \mathbb E e^{t(aX)} \mathbb E e^{t (bY)} = m_X(at) m_Y(bt) \>.$$

So, if $X$ has rate $\lambda_x$ and $Y$ has rate $\lambda_y$, we get $$m_Z(t) = \exp({\lambda_x (e^{at} - 1)}) \exp({\lambda_y (e^{bt} - 1)}) = \exp(\lambda_x e^{at} + \lambda_y e^{bt} - (\lambda_x + \lambda_y))\>,$$ and this cannot, in general, be written in the form $\exp(\lambda(e^t - 1))$ for some $\lambda$ unless $a = b = 1$.

Inversion of moment-generating functions

If the moment generating function exists in a neighborhood of zero, then it also exists as a complex-valued function in an infinite strip around zero. This allows inversion by contour integration to come into play in many cases. Indeed, the Laplace transform $\mathcal L(s) = \mathbb E e^{-s T}$ of a nonnegative random variable $T$ is a common tool in stochastic-process theory, particularly for analyzing stopping times. Note that $\mathcal L(s) = m_T(-s)$ for real valued $s$. You should prove as an exercise that the Laplace transform always exists for $s \geq 0$ for nonnegative random variables.

Inversion can then be accomplished either via the Bromwich integral or the Post inversion formula. A probabilistic interpretation of the latter can be found as an exercise in several classical probability texts.

Though not directly related, you may be interested in the following note as well.

J. H. Curtiss (1942), A note on the theory of moment generating functions, Ann. Math. Stat., vol. 13, no. 4, pp. 430–433.

The associated theory is more commonly developed for characteristic functions since these are fully general: They exist for all distributions without support or moment restrictions.

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(+1) Is the inversion formula purely theoretical or is it actually used in sometimes? –  gui11aume May 23 '12 at 18:28
@gui11aume: It is used in places; but, the examples you'd commonly find in a text are usually precisely the examples for which you don't need it. :) –  cardinal May 23 '12 at 18:31
So, presumably it's easier to work with the CFs than the MGFs? The MGFs don't always exist, right? Why bother with them? –  Frank May 23 '12 at 19:01
@Frank: Pedagogically they're easier to introduce to students that know calculus, but have little or no background in complex variables. When they exist, they have entirely analogous properties to those of CFs. They play important roles in some parts of probability theory and theoretical statistics, e.g., large deviations and exponential tilting. –  cardinal May 23 '12 at 19:09
@cardinal - Thanks a lot. This is really helping me. And thanks a lot for your patience with all my mistakes. –  Frank May 23 '12 at 19:16
Poisson distributions are stable by sum. They are trivially not stable by linear combination because you can end up with non integer values. For example, if $X$ is Poisson, $X/2$ is trivially not Poisson.
(+1) According to Wikipedia "stable" distributions are such athat $aX_1 + bX_2$ has the same distribution as $cX + d$, which is not the case of the Poisson. I guess the only proper term (correct me if I am wrong) would be "The Poisson family is stable by sum". In general, this does not mean that the distribution is infinitely divisible (think of the binomial), but the Poisson happens to have this property. –  gui11aume May 23 '12 at 21:41