Timeline for Approximate expectation of a random variable that is the logarithm of a function of a binomial
Current License: CC BY-SA 4.0
15 events
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| Apr 24, 2021 at 23:04 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Mar 25, 2021 at 20:03 | comment | added | user64494 | $\left.6 p^2 (1-p)^2 \log \left(\alpha +\frac{2}{3}\right)+4 p^3 (1-p) \log \left(\alpha +\frac{3}{2}\right)+p^4 \log (\alpha +4)+(1-p)^4 \log (\alpha )+4 p (1-p)^3 \log \left(\alpha +\frac{1}{4}\right)\right\}$. | |
| Mar 25, 2021 at 20:00 | comment | added | user64494 |
The command of Mathematica Table[Mean[ TransformedDistribution[Log[x/(k - x) + \[Alpha]], x \[Distributed] BinomialDistribution[k - 1, p]]], {k, 2, 5}] produces $\left\{(1-p) \log (\alpha )+p \log (\alpha +1),p^2 \log (\alpha +2)+(1-p)^2 \log (\alpha )+2 p (1-p) \log \left(\alpha +\frac{1}{2}\right),3 p^2 (1-p) \log (\alpha +1)+p^3 \log (\alpha +3)+(1-p)^3 \log (\alpha )+3 p (1-p)^2 \log \left(\alpha +\frac{1}{3}\right),\right.$
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| Feb 26, 2021 at 17:07 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jan 27, 2021 at 14:09 | vote | accept | qwert | ||
| Jan 27, 2021 at 16:42 | |||||
| Dec 30, 2020 at 22:00 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Nov 30, 2020 at 20:18 | answer | added | qwert | timeline score: 1 | |
| Nov 30, 2020 at 18:01 | history | edited | qwert | CC BY-SA 4.0 |
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| Nov 30, 2020 at 12:47 | comment | added | Clement C. | After "adding that $\alpha>0$", you can use this answer and linearity of expectation: for $\beta>0$, $$\mathbb{E}\log \frac{X+\beta}{k-X} = \mathbb{E}\log(X+\beta) - \mathbb{E}\log(Y+1)$$ where $Y$ is Binomial with parameters $k-1$ and $1-p$. | |
| Nov 30, 2020 at 12:47 | history | edited | qwert | CC BY-SA 4.0 |
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| Nov 30, 2020 at 12:39 | comment | added | qwert | you're right. I add an $\alpha > 0$ | |
| Nov 30, 2020 at 12:23 | comment | added | Carlo Beenakker | the variable $X$ can take on the value 0 with nonzero probability, so wouldn't the expectation of $\log(X/(k-X))$ diverge? | |
| S Nov 30, 2020 at 12:17 | history | suggested | gmvh |
Added top-level tag
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| Nov 30, 2020 at 12:05 | review | Suggested edits | |||
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| Nov 30, 2020 at 11:56 | history | asked | qwert | CC BY-SA 4.0 |