Timeline for Calculating the parameters of a Beta distribution using the mean and variance
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Dec 15, 2020 at 19:30 | comment | added | David M. | Given arbitrary $\mu\in(0,1)$ and $\sigma^2\in(0,0.5^2)$, there exists a beta distribution with mean $\mu$ and variance $\sigma^2$ if and only if $\sigma^2\leq\mu(1-\mu)$. @assumednormal showed the "only if" part of this claim, and danno hints at the "if" part. The takeaway is that not all values of $(\mu,\sigma^2)\in(0,1)\times(0,0.5^2)$ lead to valid beta distributions. Hopefully this clears up a little confusion about negative $\alpha$ and $\beta$! | |
| Apr 12, 2018 at 8:40 | comment | added | Faur | I think this is because $\sigma^2 = \mu(1-\mu)$ | |
| Apr 12, 2018 at 8:34 | comment | added | Faur | When I use this with the empirical estimates of $\mu=0.712$ and $\sigma=0.205$ I get zero for $\alpha$ and $\beta$ | |
| S Jun 5, 2016 at 2:16 | history | suggested | CommunityBot | CC BY-SA 3.0 |
Changed the variance upper bound to strong inequality. The variance can't be equal to mean * (1 - mean) because it then implies alpha and beta are both equal to zero.
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| Jun 5, 2016 at 2:13 | review | Suggested edits | |||
| S Jun 5, 2016 at 2:16 | |||||
| Nov 4, 2015 at 3:49 | comment | added | Glen_b | @AmelioVazquez-Reina If you give your original data I expect it will quickly be obvious why a beta distribution isn't suitable. | |
| Nov 4, 2015 at 3:26 | history | edited | assumednormal | CC BY-SA 3.0 |
clarified ranges of $\mu$ and $\sigma^2$
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| Nov 4, 2015 at 3:10 | comment | added | assumednormal | @AmelioVazquez-Reina - There's no Beta distribution with that mean and variance, which is why the estimates you get in return are nonsensical. | |
| Nov 4, 2015 at 3:06 | comment | added | assumednormal | @danno - It's always the case that $\sigma^2\leq\mu\left(1-\mu\right)$. To see this, rewrite the variance as $\sigma^2=\frac{\mu\left(1-\mu\right)}{\alpha+\beta+1}$. Since $\alpha+\beta+1\geq1$, $\sigma^2\leq\mu\left(1-\mu\right)$. | |
| Oct 29, 2015 at 18:20 | comment | added | danno | These calculations will only work if the variance is less than the mean*(1-mean). | |
| Oct 29, 2015 at 18:11 | comment | added | danno | I have to second Amelio's concern, here. I am getting results from this code that should not be possible, and do not result in a beta distribution... | |
| Aug 18, 2014 at 18:12 | history | edited | assumednormal | CC BY-SA 3.0 |
deleted 153 characters in body
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| May 7, 2014 at 23:46 | comment | added | Amelio Vazquez-Reina |
When I call this function with estBetaParams(0.06657, 0.1) I get alpha=-0.025, beta=-0.35. How is this possible?
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| Aug 19, 2012 at 20:19 | comment | added | assumednormal | @stan This will give you the Beta distribution which has the same mean and variance as your data. It will not tell you how well the distribution fits the data. Try the Kolmogorov-Smirnov Test. | |
| Aug 19, 2012 at 17:29 | comment | added | abc | Can one use it for generating beta distribution based on sample Mean and Variance = SD^2? I mean I want to check whether my sample (with Mean and SD^2) belongs to beta distribution with Mu=Mean and SD^2=Var). Is this a correct way? | |
| Jun 22, 2011 at 18:49 | vote | accept | Dave Kincaid | ||
| Jun 22, 2011 at 18:09 | history | edited | assumednormal | CC BY-SA 3.0 |
deleted 77 characters in body; deleted 20 characters in body
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| Jun 22, 2011 at 18:00 | history | answered | assumednormal | CC BY-SA 3.0 |