Timeline for CDF raised to a power?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| May 20, 2011 at 19:11 | comment | added | cardinal | @G. Jay, yes it does. As a check, $x = g^{-1}( g(x) )$ for all $x$ in the domain. ($g^{-1}$ is composed of a set of functions all of which have inverses. So, that should have been clear[er] from the start.) | |
| May 20, 2011 at 18:02 | comment | added | user1108 | @cardinal does it not? I added material to my answer which suggests to me that it does? | |
| May 17, 2011 at 13:21 | comment | added | cardinal | @JMS: It was a good observation, nonetheless. :) | |
| May 16, 2011 at 23:59 | comment | added | JMS | @cardinal It's fantastic that I wrote that comment whilst oblivious to that fact... | |
| May 16, 2011 at 14:41 | comment | added | cardinal | @G. Jay Kerns, perhaps I'm having a "slow" day, but I don't believe your candidate $g$ works as you intended. | |
| May 16, 2011 at 13:37 | comment | added | user1108 | Continuing @cardinal's thought - in the case of a strictly increasing CDF (for example) we could just solve for $g$ directly to get $g = F^{-1}(F^{1/\alpha})$. The probability integral transform gives that the inner quantity is $U^{1/\alpha}$, that is, a Beta. The inverse CDF is closed form sometimes (exponential, pareto....) but even when it isn't we can calculate it pretty easily in many cases. Exponential works out nicely. | |
| May 15, 2011 at 12:05 | comment | added | cardinal | @JMS, Yes, but $\mathrm{Beta}(a,1)$ essentially are the transformations of $\mathcal{U}[0,1]$ of the type described in the question. :) | |
| May 15, 2011 at 3:11 | comment | added | JMS | @cardinal I never would have thought of such a rare distribution... but now that you mention it a $Beta(a, 1)$ should work in general, giving you back a $Beta(a\alpha, 1)$. | |
| May 15, 2011 at 2:38 | comment | added | cardinal | @JMS: $Z \sim \mathcal{U}[0,1]$ would be one positive example. | |
| May 14, 2011 at 23:28 | comment | added | JMS | @brianjd - What @cardinal said :) I couldn't even think of a special case for $F_Z$ where you'd get a closed form (not to say there isn't one of course). | |
| May 14, 2011 at 16:17 | comment | added | cardinal | @brianjd: I don't believe so. Let $g$ be a continuous strictly monotonic function (hence, having a well-defined inverse $g^{-1}$) that satisfies your conditions. Then, it must be that $\renewcommand{\Pr}{\mathbb{P}}\Phi^{\alpha}(u) = \Pr(g(Z) \leq u) = \Pr( Z \leq g^{-1}(u)) = \Phi(g^{-1}(u))$ and so $g^{-1}(u) = \Phi^{-1}(\Phi^{\alpha}(u))$. So the inverse is identified fairly explicitly, but not $g$ itself. This is what I meant in my previous comment about $g$ being found implicitly. | |
| May 14, 2011 at 15:45 | comment | added | lowndrul | @JMS: What about $Z \sim N(0,1)$ ? | |
| May 13, 2011 at 22:55 | history | answered | JMS | CC BY-SA 3.0 |