Timeline for Generating random variables from the Cantor Distribution
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jan 23, 2015 at 20:59 | comment | added | user42192 | @RobertIsrael thanks..that's how I interpreted it. You gave the Cantor variate in terms of a binary sequence. Of course, it's well known that the continuous uniform can also be constructed as such, but I didn't see the relevance. | |
| Jan 23, 2015 at 15:44 | comment | added | Robert Israel | You could go either way: given $U$, take its base-2 digits and generate $X$ (which is what I intended), or given $X$, take its base-3 digits and generate $U$. | |
| Jan 23, 2015 at 11:09 | comment | added | Jochen Wengenroth | If you agree that you can create a Bernoulli sequence from the Cantor distribution then $U=\sum_{j=1}^\infty 2^{-j}B_j$ is uniformly distributed. | |
| Jan 23, 2015 at 10:35 | comment | added | user42192 | @JochenWengenroth this method does not generate U[0,1], it generates a value from the cantor distribution. I don't understand your comment. | |
| Jan 23, 2015 at 9:46 | comment | added | Jochen Wengenroth | Wasn't the question the other way round? However, from a Cantor distribution you can get a Bernoulli sequence which gives you a uniform distribution and hence every distribution on $\mathbb R$. | |
| Jan 23, 2015 at 9:13 | comment | added | Ori Gurel-Gurevich | Yes. Notice that it's not different then getting a U[0,1] (or any other continuous) RV. | |
| Jan 23, 2015 at 5:27 | vote | accept | CommunityBot | ||
| Jan 23, 2015 at 5:27 | comment | added | user42192 | Thanks! So practically speaking, I can get arbitrarily good accuracy with this method depending on where I truncate the infinite series? | |
| Jan 23, 2015 at 5:24 | history | answered | Robert Israel | CC BY-SA 3.0 |