Timeline for When is beta distribution bell-shaped or concave?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Sep 2, 2014 at 10:36 | comment | added | Glen_b | @amoeba I agree, it can be confusing; if you don't know which was intended it could be very easy to read it either way (and it was probably my fault it got written that way). Fortunately Alecos has removed the ambiguity now. | |
| Sep 2, 2014 at 10:24 | history | edited | Alecos Papadopoulos | CC BY-SA 3.0 |
added 7 characters in body
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| Sep 2, 2014 at 10:22 | comment | added | Alecos Papadopoulos | @amoeba You are right -the notation can be legitimately interpreted both ways. | |
| Sep 2, 2014 at 10:21 | comment | added | amoeba | @Glen_b: Thank you. Now I feel embarrassed... (But I also think that this notation, even if standard, can be really confusing.) | |
| Sep 2, 2014 at 10:15 | comment | added | Glen_b | @amoeba But $\alpha=5$ doesn't satisfy $\{1<\alpha,\beta \leq 2\}$. That doesn't mean "$1<\alpha$ and $\beta \leq 2$", but "$\alpha$ and $\beta$ are both between 1 and 2". | |
| Sep 2, 2014 at 10:04 | comment | added | amoeba | Counter-example: when $\alpha=5, \beta=1.5$, the distribution is not concave down everywhere. | |
| Sep 2, 2014 at 9:45 | history | edited | Alecos Papadopoulos | CC BY-SA 3.0 |
added 67 characters in body
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| Sep 2, 2014 at 8:05 | comment | added | Alecos Papadopoulos | @Glen_b Thanks Glen, I had the sign wrong. Corrected. | |
| Sep 2, 2014 at 8:04 | history | edited | Alecos Papadopoulos | CC BY-SA 3.0 |
corrected important mistake and conclusions
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| Sep 2, 2014 at 6:27 | comment | added | Glen_b | (... though I should probably check that carefully. If I post an answer, I will do so.) | |
| Sep 2, 2014 at 4:46 | comment | added | Glen_b |
To be specific, I think the term $(\beta-1)(\beta-2)x^2$ is wrong. And I think that changes the region where the curve is concave. (Of course, I could have made a mistake, but taking derivatives in R (via D) and making it a function also suggests there's a problem with your answer, since yours changes sign for cases when doing it via D doesn't.) -- I think that $1<\alpha,\beta\leq 2$ does it
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| Sep 2, 2014 at 4:33 | comment | added | Glen_b | hmm. I think you have an error in your calculation. I kind of suspected this one will actually turn out to be self study, which is why I held off. | |
| Sep 2, 2014 at 3:20 | history | edited | Alecos Papadopoulos | CC BY-SA 3.0 |
deleted 39 characters in body
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| Sep 2, 2014 at 3:12 | history | answered | Alecos Papadopoulos | CC BY-SA 3.0 |