Timeline for A convergence problem
Current License: CC BY-SA 4.0
14 events
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| Jan 28, 2023 at 10:16 | vote | accept | Star | ||
| Jan 21, 2023 at 20:05 | history | edited | Star | CC BY-SA 4.0 |
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| Jan 21, 2023 at 18:22 | history | edited | Star | CC BY-SA 4.0 |
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| Jan 9, 2023 at 14:36 | history | edited | Star | CC BY-SA 4.0 |
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| Nov 30, 2022 at 22:31 | comment | added | Anthony Quas | @TEX: Re "Is that possible?" Yes. That's exactly what happens in the example described by Christophe. You make a single choice at the outset, $Z$, (e.g. $Z=0$ with probability $\frac 12$ and 1 with probability $\frac 12$) and set all $a_n$'s equal to $Z$. So that $P(a_n=b)=P(Z=b)$ for each $b$. Now for every $N$, $x_N(b)$ is 1 if $Z=b$ and 0 otherwise. | |
| Nov 30, 2022 at 22:21 | comment | added | Christophe Leuridan | I relaxed the assumption of independence, see my post. | |
| Nov 30, 2022 at 21:39 | comment | added | Star | Yes, but is that possible as $n\rightarrow \infty$? | |
| Nov 30, 2022 at 21:37 | comment | added | Christophe Leuridan | You need an extra assumption. For example, imagine that all the draws give the same result as the first one. Then $x_n(b)$ will be $0$ or $1$. | |
| Nov 30, 2022 at 21:34 | history | edited | Star | CC BY-SA 4.0 |
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| Nov 30, 2022 at 21:33 | answer | added | Christophe Leuridan | timeline score: 4 | |
| Nov 30, 2022 at 21:33 | comment | added | Star | The draws are not independent but there are laws of large numbers also for non independent draws. Hence, I wouldn't say this is hopeless. | |
| Nov 30, 2022 at 21:27 | review | Close votes | |||
| Dec 15, 2022 at 3:06 | |||||
| Nov 30, 2022 at 21:27 | comment | added | Anthony Quas | You didn't say that the draws are independent between rounds. I assume this is the case. If not, it's hopeless. If yes, you should be able to prove what you want using second moment methods: it is clear that $\mathbb E x_n(b)=\frac 1N\sum_{n=1}^N P_n(b)$ which lies in $[v_l(b),v_r(b)]$ for each $b$. It suffices to show that the variance of $x_n(b)$ converges to 0. This should be doable by standard methods (using the independence). | |
| Nov 30, 2022 at 21:07 | history | asked | Star | CC BY-SA 4.0 |