Timeline for Do I have a misconception about probability?
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16 events
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| Dec 7, 2023 at 7:50 | comment | added | plm | @Henry, yes, thank you. I thought about adding it a couple of times, and got lazy. ^^ It is thus also the expectation of the absolute value of standard brownian motion (starting at $0$) up to any real time $N$. | |
| Dec 6, 2023 at 18:24 | comment | added | Henry | The asymptotic $\sqrt{\frac{2N}{\pi}}$ is also the mean of the corresponding half-normal distribution with scale parameter $\sqrt{N}$, related to the central limit theorem applied to the signed distance | |
| Aug 13, 2023 at 6:29 | audit | First answers | |||
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| Aug 4, 2023 at 18:02 | audit | First answers | |||
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| Jul 27, 2023 at 23:33 | comment | added | galaxy-- | @plm: thank you very much for your detailed answer. | |
| Jul 27, 2023 at 12:43 | history | edited | plm | CC BY-SA 4.0 |
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| Jul 25, 2023 at 23:42 | comment | added | plm | Hi @galaxy-- , you can find the derivation of the exact formula in the link to Mathworld i gave above, it uses the (Legendre) multiplication formula for the Gamma function. Then you have a quotient of Gamma functions with a factor $\frac{2}{\sqrt{\pi}}$ whence you get the estimate either with an asymptotic expansion of Gamma functions or via Stirling's formula. I do not think there is a name for the asymptotic but it is related to "projection constants", as you may find in the references -eg ams.org/journals/tran/1960-095-03/S0002-9947-1960-0114110-9/…. | |
| Jul 25, 2023 at 22:06 | comment | added | galaxy-- | @plm: could you please tell the name of the asymptotic estimate which you used at the beginning? I would like to look up how it is derived. Thanks | |
| Jul 25, 2023 at 16:50 | audit | First answers | |||
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| Jul 25, 2023 at 11:32 | history | edited | plm | CC BY-SA 4.0 |
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| Jul 25, 2023 at 11:26 | comment | added | plm | @Džuris thank you, good remark, i should have highlighted it. This follows from the explicit formula, but we can see it directly: note that if after $N$ steps you stand at $0$ then you can only increase your distance from it at the next step, but if after $N$ steps you stand away from $0$ then you have as much probability to get closer to it as to get farther from it. Now we can only come back to $0$ after an even number of steps $N$, thus $E|S_{N+1}|>E|S_N|$ only for even $N$. | |
| Jul 25, 2023 at 10:17 | history | edited | Džuris | CC BY-SA 4.0 |
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| Jul 25, 2023 at 10:14 | comment | added | Džuris | Wow, so the even numbered steps don't change the expected distance. Makes sense when you think a bit, but it surprised me! | |
| Jul 25, 2023 at 6:24 | history | edited | plm | CC BY-SA 4.0 |
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| Jul 24, 2023 at 12:19 | history | edited | plm | CC BY-SA 4.0 |
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| Jul 24, 2023 at 12:06 | history | answered | plm | CC BY-SA 4.0 |