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Let $\mathcal{G}$ be the set of probability generating functions of random variables taking positive integer values, considered as functions on $[0,1]$.

So $G\in\mathcal{G}$ can be written $G(x)=\sum_{k=1}^\infty p_k x^k$ where $p_1, p_2, p_3, \dots$ are non-negative and sum to 1. Such a $G$ is a bijection from $[0,1]$ to $[0,1]$ whose derivatives exist and are non-negative everywhere on $[0,1)$.

Suppose $G\in\mathcal{G}$, and define $H(x)=1-G^{-1}(1-x)$. Is it necessarily the case that $H\in\mathcal{G}$?

(The motivation is closely related to that described in the question Involutions on $[0,1]$ given by power series (related to probability generating functions), which concerned the case where $H\equiv G$.)

EDIT: Thanks to Brendan McKay, who gives in the comments below the example $G(x)=\sum_{k\geq 2} \frac{x^k}{k(k-1)}$, for which one can show that $H''(0)$ diverges at $0$. I wonder if there are any natural general conditions on $G$ under which it does hold that $H$ must also be a probability generating function.

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  • $\begingroup$ Consider the Cauchy-like distribution $G(x) = \sum_{i\ge 1} 6\pi^{-2}i^{-2}x^i$. All its derivatives diverge as $x\to 1$, which I think means that all the derivatives of $1-G^{-1}(1-x)$ at $x=0$ are 0, which means it isn't a pgf. In fact I question your characterisation of pgfs: I think they have to have a radius of convergence at least 1, which means they equal their Taylor series in $[0,1]$. $\endgroup$
    – Brendan McKay
    Commented Feb 7, 2016 at 3:47
  • $\begingroup$ I don't think the derivatives diverging at 1 is a problem. (This just means that the mean is infinite, right?) I'm not sure about your particular example, but there are some that seem to work fine. For example, $G(x)=1-(1-x)^{1/2}$, which is a pdf with all derivatives diverging at 1, and which gives $H(x)=x^2$ which is a very well-behaved pdf. Or even $G(x)=[1-(1-x)^{1/2}]^2$, which gives $H\equiv G$. $\endgroup$
    – James Martin
    Commented Feb 7, 2016 at 4:06
  • $\begingroup$ I'm not completely sure about the characterisation (I intended it to say that all derivatives exist and are non-negative everywhere on $[0,1)$ ) so I've replaced it by something weaker.... $\endgroup$
    – James Martin
    Commented Feb 7, 2016 at 4:11
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    $\begingroup$ When I calculated the second derivative of the inverse of $G(x)$ I got $$\frac{\pi^4((1-x)\log(1-x)+x)x}{36(1-x)\log(1-x)^3}.$$ That's in terms of the original variable $x$. It diverges as $x\to 1$, which corresponds to $H''(y)$ at $y=0$. You need to check this as I'm supposed to be doing something else and I'm not being very careful. $\endgroup$
    – Brendan McKay
    Commented Feb 7, 2016 at 4:52
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    $\begingroup$ It might be easier to work with $G(x)=\sum_{i\ge 2} \frac{1}{i(i-1)}x^i=(1-x)\ln(1-x)+x$. The second derivative of the inverse of $G$ is $$\frac{1}{(1-x)\log(1-x)^3},$$ which diverges at both ends of the interval. $\endgroup$
    – Brendan McKay
    Commented Feb 7, 2016 at 5:02

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