MHB Probability generating function

AI Thread Summary
A probability generating function (pgf) can indeed have a constant term, but adding a constant alters its fundamental properties. Specifically, including a constant term means the sum of the coefficients would no longer equal one, disqualifying it as a pgf. The discussion emphasizes the importance of adhering to the definition of pgfs when considering modifications. Additionally, the derivative of a pgf can yield the same result even if a constant is added, but this would not maintain the pgf's integrity. Thus, while technically possible, adding a constant to a pgf is not appropriate.
Poirot1
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My question is : can a pgf have a constant term?

The reason I ask is that I was asked to show the (time) derivative of a pgf was equal to some multiple of the pgf and hence show the pgf was as given. So naturally , I differentiated the given answer and showed it satisfied the equation. But surely I could stick a constant on the end and it would differentiate to the same thing.
 
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Poirot said:
My question is : can a pgf have a constant term?

The reason I ask is that I was asked to show the (time) derivative of a pgf was equal to some multiple of the pgf and hence show the pgf was as given. So naturally , I differentiated the given answer and showed it satisfied the equation. But surely I could stick a constant on the end and it would differentiate to the same thing.

The simple answer is yes.

The more complicated answer is: please post the actual question or more context.

Adding something onto a pgf will result in it no longer being a pgf, since the sum of the coefficients should be 1.

CB
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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