# Probability generating function zero implies random variable is infinite

Let $V$ be a random variable supported on the nonnegative integers (including $\infty$) and $f(x) = \mathbf E x^V$ be the probability generating function. In our model $V$ is the number of visits to the root of a graph by an interacting particle system. We prove $V = \infty$ a.s. by showing that $f$ satisfies a recurrence relation

$$f(x) =\frac{x+2}{3}f\Bigl(\frac{x+1}{2}\Bigr)^2 +\frac{x+1}{3}f\Bigl(\frac{x}{2}\Bigr)\biggl(1 -f\Bigl(\frac{x+1}{2}\Bigr)\biggr)$$

which through analytic methods we prove can only be satisfied when $f \equiv 0$ on $[0,1)$. Though our technique works we are somewhat baffled and are hoping to, in our upcoming paper, give some context by providing examples of this type of argument occurring in probability literature.

So, the question is are there other examples of proving a r.v. is a.s. infinite by proving the generating function is identically zero?

• This doesn't seem to have much to do with generating functions. For any fixed $x\in [0,1)$, $x^V$ is a nonnegative random variable. If $f(x) = \mathbb{E}x^V=0$ then $x^V = 0$ a.s., and so $V = \infty$ a.s. Apr 14, 2014 at 23:10
• I included a little more detail about the calculation. Working with the generating function is essential to obtain the recurrence relation described in the question. Apr 15, 2014 at 22:57
• I assume you mean $V$ is supported on the union of the positive integers and infinity, otherwise you're contradicting yourself. Apr 16, 2014 at 2:24
• Does the recurrence only hold on the interval $[0,1)$? Otherwise how do you make sense of the infinite values that appear (your conclusion means that $f(x) = \infty$ for $x>1$)? Apr 16, 2014 at 2:26
• $f$ is only defined on [0,1] Apr 16, 2014 at 4:36