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What if we have negative integral values for a random variable?Then is it possible to write a probability generating function for it? All definitions I have seen so far is for non negative integer values. I hope someone could assist me. Thanks

kjetil b halvorsen
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Heisenberg
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    I see no problem applying the definition $f(t)=\sum_{x\in\mathbb{R}} p(x)t^x$ to a discrete distribution with probabilities $p(x)$. Whether this converges (and therefore defines an actual function) or has any useful analytical properties is another matter, but clearly if the support of the distribution is integral it will be quite tractable (it's a [Laurent series](http://en.wikipedia.org/wiki/Laurent_series)). I wonder, therefore, what motivates this question: is there a particular difficulty raised by the possibility of negative support? – whuber Jul 11 '14 at 12:46

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As whuber stated above, there is really no problem here as long as the resultant sum is well-defined in some neighborhood of a finite point in $\mathbb{C}$ (we can always shift things around to find the moments, if they exist, or probabilities, no matter what the number is). There are a few things you can think about. First, if the probabilities $p_n$ have support on $\text{Supp}(p) = \{-N, 1-N,...\} \cup \mathbb{N}$, then there is clearly no problem at all since we will define the generating function as $$ G(z) = \sum_{n \geq -N}p_n z^n = \sum_{n \geq 0}q_n z^n, $$ where $q_n = p_{n - N}$, so that you are back to the familiar positive power case. Likewise, if your probability distribution is supported on $-\mathbb{N} \cap \{0,1,...,N\}$, define $r_n = p_{N-n}$ and a similar result is apparent. So the only real issue becomes when your series is a doubly-infinite Laurent series: $$ G(z) = \sum_{-\infty}^{\infty}p_n z^{n}. $$ with nonzero $p_n\ \forall n < 0$. This definitely can be an issue---see the wikipedia article on Laurent series for a discussion. The gist of the problem is that, in this case, $G$ might have an essential singularity or some nontrivial behavior on the inner disc of its annulus of convergence.

This gives (yet another) reason to prefer characteristic functions when possible...

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I believe it's basically because usually the treatment relies on results that apply to sums of non-negative powers.

An example of the sort of thing that's relied on would be Abel's theorem. With r.v.s that take negative values, you'd have to try to establish the radius of convergence without it.

So there are some issues to deal with when the X can be negative (though of course we still have MGFs, characteristic functions and so on in any case). You might find this [1] and some of its references useful. The discussion is of extending from random variables on the non-negative integers to more general cases (as an example, the treatment establishes that a connection between the characteristic function and pgf still holds for variables also taking negative values as long as the tails decay at least exponentially).

So it seems it can be extended in the sense you'd like, at least under certain conditions.

[1] Esquível, M.L. (2004),
Probability Generating Functions For Discrete Real Valued Random Variables,
(author's link)

Glen_b
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The (probability) generating function (a/k/a the factorial moment generatring function) is defined as $$h_X(t) = E\{t^X\}.$$ The garden-variety moment generating function is defined as $$M_X(t) = E\{e^{tX}\}.$$ For either to be useful it must exist in a neighborhood of 0 (to pull moments off) or in the case of $h_X(t)$ a neighborhood of 1 (to pull discrete probabilities off). In either case, if one exists, the other must also exist: What is $E\{e^{(\ln{u}\cdot X}\}$? This shows a backdoor to $h_X$: evaluate $M_X(\ln{t})$. Do be sure to check the neighborhoods of 0 and 1 once you've done the formalism. What is unclear to me at this point is how you would strip the pmf from this. For non-negative discrete random variables, $f_X(k) = h_X^{(k)}(1)$. Do you integrate $h$ to get to negative values of $X$? Or perhaps you don't or can't strip off the pmf values for $x<0$? It's late here, I'll have to think on this more tomorrow.

Dennis
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