12
$\begingroup$

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?

Improve this question
$\endgroup$
7
  • 4
    $\begingroup$ 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? $\endgroup$
    – cardinal
    Commented May 23, 2012 at 17:11
  • $\begingroup$ 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. $\endgroup$
    – Frank
    Commented May 23, 2012 at 17:20
  • $\begingroup$ 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.) $\endgroup$
    – jbowman
    Commented May 23, 2012 at 17:28
  • $\begingroup$ 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 $t_{i}$. $\endgroup$
    – Frank
    Commented May 23, 2012 at 17:36
  • 5
    $\begingroup$ The mistake is that the MGF of $a_iX_i$ is $exp(\lambda_i (exp(a_i t)-1))$ and not $exp(a_i\lambda_i (exp(t)-1))$ $\endgroup$
    – gui11aume
    Commented May 23, 2012 at 18:09

2 Answers 2

Reset to default
15
+50
$\begingroup$

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.

Improve this answer
$\endgroup$
9
  • 1
    $\begingroup$ (+1) Is the inversion formula purely theoretical or is it actually used in sometimes? $\endgroup$
    – gui11aume
    Commented May 23, 2012 at 18:28
  • 2
    $\begingroup$ @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. :) $\endgroup$
    – cardinal
    Commented May 23, 2012 at 18:31
  • $\begingroup$ So, presumably it's easier to work with the CFs than the MGFs? The MGFs don't always exist, right? Why bother with them? $\endgroup$
    – Frank
    Commented May 23, 2012 at 19:01
  • $\begingroup$ @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. $\endgroup$
    – cardinal
    Commented May 23, 2012 at 19:09
  • 1
    $\begingroup$ @Frank: These are the Levy $\alpha$-stable distributions and the only one with an MGF is the normal distribution. Indeed, CFs are the tool for this problem; the possible form of the CF is known for all such distributions, but closed-form corresponding pdfs are only know in a select handful of instances. $\endgroup$
    – cardinal
    Commented May 23, 2012 at 20:08
6
$\begingroup$

Poisson distributions are stable by sum. They are trivially not stable by linear combination because you can end up with noninteger values. For example, if $X$ is Poisson, $X/2$ is trivially not Poisson.

I am not aware of inversion formulas for MGF (but @cardinal seems to be).

Improve this answer
$\endgroup$
4
  • 2
    $\begingroup$ (+1) Because I like simple illustrative proofs and counterexamples that immediately bring to the fore the heart of the matter. $\endgroup$
    – cardinal
    Commented May 23, 2012 at 18:30
  • $\begingroup$ I have a question about terminology. In the statistics I studied stable dsitributions were the ones that were limits of distributions that satisfied a convergence condition called a stable law. These are continuous nonnormal distributions. The are a distribution for the limits of a normalized average Z but the central limit theorem does not applied to Z because of the tail behavior of the population distribution. Actually the central limit theorem can belong to the stable laws if a certain parameter alpha = 2. $\endgroup$
    – Michael R. Chernick
    Commented May 23, 2012 at 18:45
  • 1
    $\begingroup$ What you are calling stable here is closer under sums which seems to me more like the term infinitely divisible. In what fields is the term stable used for this? Is it becoming used in probability and statistics? $\endgroup$
    – Michael R. Chernick
    Commented May 23, 2012 at 18:46
  • 1
    $\begingroup$ (+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. $\endgroup$
    – gui11aume
    Commented May 23, 2012 at 21:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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