Timeline for Constructing a discrete r.v. having as support all the rationals in $[0,1]$
Current License: CC BY-SA 3.0
30 events
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| Oct 4, 2023 at 21:29 | answer | added | Henry | timeline score: 1 | |
| Sep 19, 2022 at 18:34 | comment | added | Alecos Papadopoulos | @JohnBerryman Thanks for the pointers! | |
| Sep 19, 2022 at 16:02 | comment | added | John Berryman |
I went through a very similar thought process in this blog, though, for simplicity, I chose to use a uniform distribution for the numerator rather than a geometric distribution. The result is still not close to a uniform, but you can approach a uniform distribution as p -> 0. There's also this relevant publication about taking rational numbers at random.
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| Nov 21, 2020 at 22:29 | answer | added | Ben | timeline score: 2 | |
| Apr 13, 2017 at 12:44 | history | edited | CommunityBot |
replaced http://stats.stackexchange.com/ with https://stats.stackexchange.com/
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| Oct 22, 2014 at 13:00 | vote | accept | Alecos Papadopoulos | ||
| S Jun 27, 2014 at 23:28 | history | bounty ended | Alecos Papadopoulos | ||
| S Jun 27, 2014 at 23:28 | history | notice removed | Alecos Papadopoulos | ||
| Jun 21, 2014 at 12:24 | comment | added | whuber♦ | Re the bounty: Please notice that my response answers your question with $p=1/2$. The only difference is that your situation is a mixture of mine plus an atom at zero (corresponding to the possibility $X=0$). The generalization to arbitrary $p$ is so straightforward it did not seem worth mentioning, so I'll just sketch it here. Define $F(x,y)=C(p)^{-1}(1-p)^{x+y}$ where $C(p)$ is the normalizing constant $$C(p)=\sum_{x=0}^\infty\sum_{y=\max\{1,x\}}^\infty F(x,y)=-\frac{(p-1)^2}{(p-2) p^2}.$$ From this it follows $$G(x/y)=\frac{(p-2) p^2 (1-p)^{x+y-2}}{(1-p)^{x+y}-1}$$ for $\gcd(x,y)=1$. | |
| Jun 21, 2014 at 10:46 | history | tweeted | twitter.com/#!/StackStats/status/480300746379132928 | ||
| S Jun 21, 2014 at 10:10 | history | bounty started | Alecos Papadopoulos | ||
| S Jun 21, 2014 at 10:10 | history | notice added | Alecos Papadopoulos | Draw attention | |
| Jun 19, 2014 at 15:10 | answer | added | whuber♦ | timeline score: 25 | |
| Jun 19, 2014 at 13:08 | history | edited | Alecos Papadopoulos | CC BY-SA 3.0 |
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| Jun 19, 2014 at 12:39 | answer | added | Adrian | timeline score: 9 | |
| Jun 19, 2014 at 11:50 | history | edited | Alecos Papadopoulos | CC BY-SA 3.0 |
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| Jun 19, 2014 at 11:49 | comment | added | Alecos Papadopoulos | @Adrian Thanks for the continuing contribution here. I guess you are using the same notation as in the post, so in the equality you state, $Q$ is the ratio $X/Y$ and $Y$ the geometric r.v.-variant I? | |
| Jun 19, 2014 at 11:36 | comment | added | Alecos Papadopoulos | @JuhoKokkala A useful clarification - I updated the question accordingly. | |
| Jun 19, 2014 at 11:21 | comment | added | Adrian | Let $f: N \rightarrow \mathbb{Q}\cap[0,1]$ and Y the RV in your post. $Pr[Q =q] = Pr[Y = f^{-1}(q)]$ | |
| Jun 19, 2014 at 11:14 | comment | added | Juho Kokkala | Does the requirement to provide the pmf mean that a closed-form is required? Or is, e.g., @StéphaneLaurent's infinite sum enough to fulfill the condition? | |
| Jun 19, 2014 at 11:05 | history | edited | Alecos Papadopoulos | CC BY-SA 3.0 |
Some formal corrections
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| Jun 19, 2014 at 10:59 | comment | added | Stéphane Laurent | Sorry for my previous comment this is rather $\sum_k\Pr(X=kp)\times\Pr(Y=kq)$ | |
| Jun 19, 2014 at 10:57 | comment | added | Alecos Papadopoulos | @Adrian ...Meaning that the pmf of the random variable associated with what you describe will be...? | |
| Jun 19, 2014 at 10:56 | comment | added | Alecos Papadopoulos | @StéphaneLaurent Admittedly I used the conditioning notation loosely (and essentially incorrectly) -that's why I added also the verbal description. I will attempt a more standard definition. I think though that the definition of $Q$ in your comment describes a different random variable -one that takes the value zero when $X>Y$. So apparently, a lot more of probability mass will concentrate on zero. But this definition describes also a random variable that has the desired support, so it is fair game. | |
| Jun 19, 2014 at 10:55 | comment | added | Stéphane Laurent | anyway you have to calculate $Pr(X/Y=p/q)$ for every irreducible fraction $p/q$ and this is $\Pr(X=p, X=2p, \ldots)\times\Pr(Y=q, Y=2q, \ldots)$. | |
| Jun 19, 2014 at 10:41 | comment | added | Adrian | Your Q is countable: you know there exists a 1-1 correspondence between N={1, 2, ...} and Q. If you could find such a correspondence, the solution would be to pick any distribution over N and use it to pick the corresponding element of Q. | |
| Jun 19, 2014 at 10:36 | comment | added | Stéphane Laurent | Do you mean $Q = \frac {X}{Y} {\boldsymbol 1}_{\{X\leq Y\}}$ ? (defining a random variable conditionnally on something makes no sense, you could only define its distribution in this way) | |
| Jun 19, 2014 at 10:30 | history | edited | Alecos Papadopoulos | CC BY-SA 3.0 |
math typo
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| Jun 19, 2014 at 9:45 | history | edited | Alecos Papadopoulos | CC BY-SA 3.0 |
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| Jun 19, 2014 at 9:39 | history | asked | Alecos Papadopoulos | CC BY-SA 3.0 |