Timeline for If the sample space is an Euclidean Space, we can use a different type of PDF
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Feb 12, 2016 at 19:54 | comment | added | Integral | This is a link to the book: springer.com/br/book/9783642388958 I'm about to take a one semester course on this book and I'm really concerned that this definitions are not well accepted. | |
| Feb 12, 2016 at 19:40 | comment | added | Iosif Pinelis | The excerpt from book at your link says it introduces the term "data space" because of "lacking some established name". I have never seen such a definition anywhere else. Your definition of an alternative pdf by formula $(1)$ does seem similar to the definition of the probability density in the book, except that there the sigma-algebra does not seem to be specified (or linked to a r.v.), whereas you seem to specify the sigma-algebra to be $\sigma(X)$. The book is not reviewed on MathSciNet. Even looking at the excerpt you posted, I don't think it should be taken as a mathematical book. | |
| Feb 12, 2016 at 18:34 | comment | added | Integral | Ok, I won't bring more questions from ME, didn't know about that. About data space, it's defined here: dropbox.com/s/dnx1itp96ky1suz/a.png?dl=0 Did I understand things wrongly or the author really defined a PDF like I was introducing here? One more time, thanks a lot for helping. | |
| Feb 12, 2016 at 18:08 | comment | added | Iosif Pinelis | A better, quite unabiguous term is a measurable space, instead of a sample space. Also, I now see that the term "data space" is indeed used at the link you provided. However, it is not defined there either, and even caused some confusion there as well. I also think it is generally not a good idea to bring questions from math.stackexchage to mathoverflow. | |
| Feb 12, 2016 at 17:46 | comment | added | Integral | I'm using the term "sample space" as people use in probability usually. It's a set (plus a sigma-algebra) of the outcomes we are interested. I don't know what is the problem. About the term "data space", I borrowed it from the link in my comment above, and I did this precisely so you could be able to understand the meaning before making your comment. | |
| Feb 12, 2016 at 16:27 | comment | added | Iosif Pinelis | In that comment above, you seem mainly concerned about notation and terminology. The term pdf is used to refer to the density of the distribution of a r.v. w.r. to the Lebesgue measure; see e.g. en.wikipedia.org/wiki/Probability_density_function . Also, it's better to have as few notions as possible, and make sure they are defined and/or anyway clearly understood by your audience. I have already advised you against using the vague term "sample space". You are still using it, and now also "data space", the meaning of which latter I can hardly be sure about. | |
| Feb 12, 2016 at 13:51 | vote | accept | Integral | ||
| Feb 12, 2016 at 13:51 | comment | added | Integral | I'm going to accept your answer, but I would be very glad if you enlighten me about my comment above. I feel I almost understand everything, almost. Thank you very much for your help. | |
| Feb 11, 2016 at 13:33 | comment | added | Integral | The author made a little mistake and should've write $\mu(A) = \int_Af\ dx$ (he doesn't write the $dx$ but this is the measure used, there is a image in a comment of mine there). The data space doesn't come from any random variable, it works as a sample space. It means their definition of (probability) density is as mine (if I didn't make a mistake), and then starts my confusion because it looks like they really define the PDF over the sample space (instead related to random variables). | |
| Feb 10, 2016 at 14:11 | comment | added | Integral | Yes, that's what I meant. My confusion started reading the post I mentioned first. This one: math.stackexchange.com/questions/1476568/… | |
| Feb 10, 2016 at 4:05 | comment | added | Iosif Pinelis | If by $\int_A f\, dx$ you mean the integral with respect to the Lebesgue measure $\lambda$ (say), then the condition that $\int_A f\, dx=\mu(A)$ for all Borel sets $A\subseteq\mathbb R^{n^2}$ means, by definition, that $f$ is the density of the measure $\mu$ with respect to $\lambda$. Of course, then one can construct any number of r.v.'s $X$ with values in $\mathbb R^{n^2}$ such that $f$ is the pdf of $X$, in the conventional sense. One such construction is letting $X$ be the identity mapping of $\mathbb R^{n^2}$ endowed with the Borel measure $\mu$. However, I don't see anything new here. | |
| Feb 10, 2016 at 3:40 | comment | added | Iosif Pinelis | I have added details on why the "alternative pdf" provides no information on the distribution of $X$. | |
| Feb 10, 2016 at 3:39 | comment | added | Iosif Pinelis | No, as I said, Diaconis considered (as a starting point) the standard Gaussian (not Haar) measure on $\mathbb R^{n^2}$. Also, the Haar measure is defined on the sigma-algebra of Borel subsets of the compact group $O_n$, rather than of $\mathbb R^{n^2}$. Further, I suggest that such imprecise phrases as "the sample space with ... a measure" be avoided. Why do you need a "sample space" if you can have a set? | |
| Feb 10, 2016 at 3:29 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
Added details on why the "alternative pdf" provides no information on the distribution of $X$.
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| Feb 10, 2016 at 2:26 | comment | added | Integral | Diaconis considered $\mathbb{R}^{n^2}$ as the sample space with the Haar measure $\mu$, if I understand correctly. For instance, if exists a continuous function $f:\mathbb{R}^{n^2}\to[0,\infty)$ such that $\int_{\mathbb{R}^{n^2}}f\ dx = 1$ and $\int_A f\ dx = \mu(A)$ (for all borel sets $A$), could I call that function a PDF? The PDF I'm contemplating goes in this lines. | |
| Feb 10, 2016 at 1:55 | history | answered | Iosif Pinelis | CC BY-SA 3.0 |
