Timeline for Finding conditions on unspecified CDF that permit a solution to an equation
Current License: CC BY-SA 3.0
11 events
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| May 20, 2013 at 8:35 | vote | accept | Martin | ||
| May 19, 2013 at 21:12 | comment | added | whuber♦ | I have no problem finding solutions by perturbing the uniform and exponential cases. As you have begun to see, conditions on $F'$ are of a wholly different nature than conditions on $F$ and have little to do with conditions involving moments of the distribution. | |
| May 19, 2013 at 20:36 | comment | added | Martin | Also, while we cannot give any conditions on $F$, my above argument shows that we can very well do so for $F'$. In particular, the "value adjusted density" ($\alpha F'(\alpha)$) has to be sufficiently large relative to the parameters $N$ and $\delta$. I wonder whether much more could be said if I imposed some regularity conditions on $F'$, like single-peakedness. | |
| May 19, 2013 at 20:31 | comment | added | Martin | Thanks a lot for your continued effort! The graphs are amazing and add lots of intution. I believe that I now understand your point pretty well -- in essence, we cannot give any meaningful conditions on $F$ whatsoever because an epsilon perturbation of the pdf can preserve all of the required properties (while letting $F$ almost unchanged), yet we can make the pdf as large as we want in order to guarantee a solution. I think it's interesting however that a solution fails to exist for such common distributions as uniform on $[0,2]$, or exponential with mean $1$. | |
| May 19, 2013 at 20:14 | history | edited | whuber♦ | CC BY-SA 3.0 |
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| May 19, 2013 at 19:55 | comment | added | whuber♦ | You are right : if $F$ is twice - differentiable on $[ 0, \overline {\alpha}] $, then necessarily $F' $ is defined *and continuous* in a neighborhood of $0$, whence $F' $ is bounded at zero and the limit of $xF' (x) $ as $x\to 0 $ must be $0$ anyway. But don't hope for sharper conditions. I trust that the figures might help you gain some intuition concerning how little can be said about $F$. | |
| May 19, 2013 at 19:55 | history | edited | whuber♦ | CC BY-SA 3.0 |
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| May 19, 2013 at 17:58 | comment | added | Martin | That is, some $\alpha > 0$ must exist such that $\alpha F'(\alpha) > (N-1)\frac{1-\delta}{\delta}$. In a very similar fashion, a sufficient condition for the existence of a solution is that $\alpha F'(\alpha) > \frac{N-1}{\delta}$ for some $\alpha > 0$. However, I'm wondering whether sharper conditions can be found. | |
| May 19, 2013 at 17:53 | comment | added | Martin | Moreover, while I can easily be convinced that no conditions on any moments of F can be given, both necessary and sufficient conditions on "other" peculiarities of F can be stated. I think it should usually (always?) hold that $\lim_{\alpha \rightarrow 0}\alpha F'(\alpha) = 0$. Hence, in your terminology, $L[F](0) = (N-1)(1-\delta) > 0$. A necessary condition for a solution is thus that $(N-1)[1-\delta + \delta F(\alpha)] - \delta \alpha F'(\alpha) < 0$ for some $\alpha > 0$. Hence, at the very least, some $\alpha$ must exist such that $(N-1)(1-\delta) - \delta \alpha F'(\alpha) < 0$. | |
| May 19, 2013 at 17:41 | comment | added | Martin | Many thanks for your detailed reply! I was suspecting something along those lines, although it's a bit hard for me to evaluate all the steps of your argument. In particular, it is not completely intuitive to me why "[...] making the first of these mixed-in distributions have sufficiently small support, we can cause it to increase F′ at x0 by any desired amount while changing F(x0) arbitrarily little" preserves the variance of F, or any other higher moments. I must admit though that I have a relatively limited understanding of the technical details involved. | |
| May 19, 2013 at 17:17 | history | answered | whuber♦ | CC BY-SA 3.0 |