Timeline for Figuring out the variance of the birthday paradox
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| May 21, 2019 at 9:52 | comment | added | drhab | @InterstellarProbe Almost but not exactly. We are dealing with $X_1+\cdots+X_n$ where the $X_i$ have identical Bernoulli distribution. However the $X_i$ are not independent, and the distribution of $X_1+\cdots+X_n$ is not binomial. | |
| May 21, 2019 at 8:58 | vote | accept | EliT | ||
| May 21, 2019 at 7:57 | history | edited | EliT | CC BY-SA 4.0 |
added what I think is the calculation for $E[X_1\cdot X_2]
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| May 20, 2019 at 18:39 | answer | added | drhab | timeline score: 3 | |
| May 20, 2019 at 18:33 | comment | added | EliT | Yes there is - However, how does that relate back to the variance? | |
| May 20, 2019 at 18:28 | comment | added | drhab | Isn't there a connection $n$ and $p$? If you only count the birthdays of the $n$ people then $p=1-(364/365)^{n-1}$. | |
| May 20, 2019 at 18:04 | history | edited | EliT |
edited tags
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| May 20, 2019 at 18:01 | comment | added | EliT | How did you reach that conclusion? | |
| May 20, 2019 at 17:58 | comment | added | SlipEternal | It is probably $$V[X]=np(1-p)$$ | |
| May 20, 2019 at 17:53 | history | asked | EliT | CC BY-SA 4.0 |