Timeline for Sum or mean of several related hypergeometric distributions
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Dec 9, 2011 at 22:26 | comment | added | jebyrnes | FYI, I've realized that one way to simplify this is, instead of thinking about multiple urns, is to think about one urn with balls that can have multiple colors on them. This then stumped me for a while until, with some discussion, I arrived at this solution - math.stackexchange.com/questions/86847/… . It's not totally satisfying, as I wish there was a way that did not involve having to go through every single combination of colors, though. But it is a solution. | |
| Oct 23, 2011 at 20:57 | vote | accept | jebyrnes | ||
| Oct 3, 2011 at 21:14 | comment | added | Karl | @jebyrnes - it's certainly possible. my first job as a research assistant concerned mixtures of the binomial. | |
| Oct 3, 2011 at 16:58 | comment | added | jebyrnes | On the other hand, what if you had a probability distribution for m? Say it was poisson distributed? Or, better, geometrically distributed? Would one be able to work from there? | |
| Sep 25, 2011 at 14:07 | comment | added | Karl | @jebyrnes - You can write it down or calculate it numerically, but it won't simplify much. | |
| Sep 25, 2011 at 13:38 | comment | added | jebyrnes | That's my fear - although, even if N and n are constant? There's no way to at least derive a distribution of Pr(k=0) if you know the distribution of m - or all values of m? | |
| Sep 25, 2011 at 12:24 | comment | added | Karl | I don't think you'll be able to simplify the average of $\Pr(k=0)$ to something that involves the average of $m$. You're likely stuck with numerical results. Consider the binomial case. The average of $p_i^n$ for fixed $n$, varying $p_i$, won't reduce to something that involves just the average of the $p_i$. | |
| Sep 25, 2011 at 9:42 | comment | added | jebyrnes | Yup, I had fallen on something similar myself. My question is more how to get the average probability across all urns when k=0. I'm trying to see if I can come up with an expression that uses the average value (or any other distributional properties) of m, really. It's really m I'm interested in here. See my edit above. | |
| Sep 25, 2011 at 0:44 | history | answered | Karl | CC BY-SA 3.0 |