Timeline for Inverses of probability generating functions: positivity of derivatives
Current License: CC BY-SA 3.0
13 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Feb 7, 2016 at 6:57 | history | edited | James Martin | CC BY-SA 3.0 |
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| Feb 7, 2016 at 6:51 | comment | added | James Martin | Thanks Brendan! - some less precise calculations are also getting me to $H''(0)=\infty$ in that case. | |
| Feb 7, 2016 at 5:02 | comment | added | Brendan McKay | 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. | |
| Feb 7, 2016 at 4:52 | comment | added | Brendan McKay | 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. | |
| Feb 7, 2016 at 4:46 | comment | added | James Martin | OK, sounds interesting - can you explain why? | |
| Feb 7, 2016 at 4:43 | comment | added | Brendan McKay | Precisely (contrary to what I wrote before), I think $H(x)$ has no second derivative at $x=0$. | |
| Feb 7, 2016 at 4:28 | comment | added | Brendan McKay | I might not have identified the problem precisely, but I still think it is a problem. In the example I gave, I think that $H$ has no Taylor series at the origin. But every pgf has a radius of convergence at least 1 since the coefficients are bounded by 1. If $H$ is a pgf, you should be able to identify the probabilities it corresponds to. | |
| Feb 7, 2016 at 4:13 | history | edited | James Martin | CC BY-SA 3.0 |
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| Feb 7, 2016 at 4:11 | comment | added | James Martin | 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.... | |
| Feb 7, 2016 at 4:06 | comment | added | James Martin | 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 $H\equiv G$. | |
| Feb 7, 2016 at 3:47 | comment | added | Brendan McKay | 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 $[0,1]$. | |
| Feb 7, 2016 at 3:04 | history | asked | James Martin | CC BY-SA 3.0 |