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This is the constructivist sequel of this question.

If we can't have a discrete uniform random variable having as support all the rationals in the interval $[0,1]$, then the next best thing is:

Construct a random variable $Q$ that has this support, $Q\in \mathbb{Q}\cap[0,1]$, and that it follows some distribution. And the craftsman in me requires that this random variable is constructed from existing distributions, rather than created by abstractly defining what we desire to obtain.

So I came up with the following:

Let $X$ be a discrete random variable following the Geometric Distribution-Variant II with parameter $0<p<1$, namely

$$ X \in \{0,1,2,...\},\;\;\;\; P(X=k) = (1-p)^kp,\;\;\; F_X(X) = 1-(1-p)^{k+1}$$

Let also $Y$ be a discrete random variable following the Geometric Distribution-Variant I with identical parameter $p$, namely

$$ Y \in \{1,2,...\},\;\;\;\; P(Y=k) = (1-p)^{k-1}p,\;\;\; F_Y(Y) = 1-(1-p)^k$$

$X$ and $Y$ are independent. Define now the random variable

$$Q = \frac {X}{Y}$$

and consider the conditional distribution

$$P(Q\leq q \mid \{X\leq Y\})$$

In loose words "conditional $Q$ is the ratio of $X$ over $Y$ conditional on $X$ being smaller or equal than $Y$." The support of this conditional distribution is $\{0,1,1/2,1/3,...,1/k,1/(k+1),...,2/3,2/4,...\} = \mathbb{Q}\cap[0,1]$.

The "question" is: Can somebody please provide the associated conditional probability mass function?

A comment asked "should it be closed-form"? Since what constitutes a closed-form nowadays is not so clear cut, let me put it this way: we are searching for a functional form into which we can input a rational number from $[0,1]$, and obtain the probability (for some specified value of the parameter $p$ of course), leading to an indicative graph of the pmf. And then vary $p$ to see how the graph changes.

If it helps, then we can make the one or both bounds of the support open, although these variants will deprive us of the ability to definitely graph the upper and/or lower values of the pmf. Also, if we make open the upper bound, then we should consider the conditioning event $\{X<Y\}$.

Alternatively, I welcome also other r.v.'s that have this support(s), as long as they come together with their pmf.

I used the Geometric distribution because it has readily available two variants with the one not including zero in the support (so that division by zero is avoided). Obviously, one can use other discrete r.v.'s, using some truncation.

I most certainly will put a bounty on this question, but the system does not immediately permit this.

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    $\begingroup$ 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) $\endgroup$ – Stéphane Laurent Jun 19 '14 at 10:36
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    $\begingroup$ 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. $\endgroup$ – Adrian Jun 19 '14 at 10:41
  • $\begingroup$ anyway you have to calculate $Pr(X/Y=p/q)$ for every irreducible fraction $p/q$ and this is $\Pr(X=p, X=2p, \ldots)\times\Pr(Y=q, Y=2q, \ldots)$. $\endgroup$ – Stéphane Laurent Jun 19 '14 at 10:55
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    $\begingroup$ 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? $\endgroup$ – Juho Kokkala Jun 19 '14 at 11:14
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    $\begingroup$ Let $f: N \rightarrow \mathbb{Q}\cap[0,1]$ and Y the RV in your post. $Pr[Q =q] = Pr[Y = f^{-1}(q)]$ $\endgroup$ – Adrian Jun 19 '14 at 11:21
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Consider the discrete distribution $F$ with support on the set $\{(p,q)\,|\, q \ge p \ge 1\}\subset \mathbb{N}^2$ with probability masses

$$F(p,q) = \frac{3}{2^{1+p+q}}.$$

This is easily summed (all series involved are geometric) to demonstrate it really is a distribution (the total probability is unity).

For any nonzero rational number $x$ let $a/b=x$ be its representation in lowest terms: that is, $b\gt 0$ and $\gcd(a,b)=1$.

$F$ induces a discrete distribution $G$ on $[0,1]\cap \mathbb{Q}$ via the rules

$$G(x) = G\left(\frac{a}{b}\right) = \sum_{n=1}^\infty F\left(an, bn\right)=\frac{3}{2^{1+a+b}-2}.$$

(and $G(0)=0$). Every rational number in $(0,1]$ has nonzero probability. (If you must include $0$ among the values with positive probability, just take some of the probability away from another number--like $1$--and assign it to $0$.)

To understand this construction, look at this depiction of $F$:

[Figure of F]

$F$ gives probability masses at all points $p,q$ with positive integral coordinates. Values of $F$ are represented by the colored areas of circular symbols. The lines have slopes $p/q$ for all possible combinations of coordinates $p$ and $q$ appearing in the plot. They are colored in the same way the circular symbols are: according to their slopes. Thus, slope (which clearly ranges from $0$ through $1$) and color correspond to the argument of $G$ and the values of $G$ are obtained by summing the areas of all circles lying on each line. For instance, $G(1)$ is obtained by summing the areas of all the (red) circles along the main diagonal of slope $1$, given by $F(1,1)+F(2,2)+F(3,3)+\cdots$ = $3/8 + 3/32 + 3/128 + \cdots = 1/2$.

Figure

This figure shows an approximation to $G$ achieved by limiting $q\le 100$: it plots its values at $3044$ rational numbers ranging from $1/100$ through $1$. The largest probability masses are $\frac{1}{2},\frac{3}{14},\frac{1}{10},\frac{3}{62},\frac{3}{62},\frac{1}{42},\ldots$.

Here is the full CDF of $G$ (accurate to the resolution of the image). The six numbers just listed give the sizes of the visible jumps, but every part of the CDF consists of jumps, without exception:

Figure 2

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    $\begingroup$ Thanks! I am in the process of understanding the construction. Just two questions: a) $F$ is bivariate, but in the expression linking it to $G$ it appears as univariate. Am I missing something? and b) Since $G$ is univariate, I guess all the dots in the impressively looking first graph represent a different value on the horizontal axis (although of course this cannot be faithfully represented in such a scale), am I right? $\endgroup$ – Alecos Papadopoulos Jun 19 '14 at 15:33
  • $\begingroup$ I was just completing a figure that might address your comment, Alecos, and have added it to the answer. Note that I could have begun with any discrete distribution $F$ and constructed $G$ in the same way; this particular distribution was chosen to make the calculations easy. $\endgroup$ – whuber Jun 19 '14 at 15:40
  • $\begingroup$ Gets better and better, As for my first question in the previous comment, should it be $F\left(\frac ab,n\right)$ instead of $F\left(\frac abn\right)$? I.e. that $p=a/b$ and $q=n$? $\endgroup$ – Alecos Papadopoulos Jun 19 '14 at 15:57
  • $\begingroup$ This is a better answer than mine! I noticed two little things: I think your F(p, q) sums to 4 as written. Also in the equation below "F induces a discrete distribution G" you should have F(n a, n b) no? $\endgroup$ – Adrian Jun 19 '14 at 16:04
  • $\begingroup$ @Adrian, Alecos Thanks for catching those typos: the $1$ should be a $-1$ and the notation for $F$ obviously is incorrect. I'll fix them right away. $\endgroup$ – whuber Jun 19 '14 at 16:15
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I'll put my comments together and post them as an answer just for clarity. I expect you won't be very satisfied, however, as all I do is reduce your problem to another problem.

My notation:

$Q$ is a RV whose support is $\mathbb{Q}\cap\left[0,1\right]$ -- my $Q$ is not the same as the $Q$ the OP constructs from his $\frac{X}{Y}$. We'll define this $Q$ using $Y$ and $f$, which I introduce below.

$Y$ is any RV whose support is $\mathbb{N}\equiv\left\{1, 2, \ldots\right\}$ -- the $Y$ given by the OP would work, for example.

$f$ is any one-to-one correspondence $f:\mathbb{N}\rightarrow\mathbb{Q}\cap\left[0,1\right]$ and $f^{-1}$ is its inverse. We know these exist.

Now I claim I can reduce your problem to just finding an $f$ and its $f^{-1}$:

Just let $Q=f\left(Y\right)$ and you are done. The PMF of $Q$ is $\Pr[Q =q] = \Pr[Y = f^{-1}(q)]$.

Edit:

Here is a function g that plays the role of $f$, despite not being a one-to-one correspondence (because of duplicates):

g <- function(y) {
    y <- as.integer(y)
    stopifnot(y >= 1)
    b <- 0
    a <- 0
    for (unused_index in seq(1, y)) {
        if (a >= b) {
            b <- b+1
            a <- 0
        } else {
            a <- a+1
        }
    }
    return(sprintf("q = %s / %s", a, b))
    ## return(a / b)
}
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  • $\begingroup$ (+1) No, I consider your approach an excellent example of how one can think and use the abstract approach in order to arrive at very applicable results and algortihms. The main point as I now understand it, is that one can obtain the desired construction by using as a functional form the pmf of any discrete distribution having support $\mathbb{N}\equiv\left\{1, 2, \ldots\right\}$. Of course it remains to find $f$ and $f^{-1}$. Since you have a better understanding of this approach than I do, is the phrase "we know these exist" a polite way to say "but we have no idea how they look like"?:) $\endgroup$ – Alecos Papadopoulos Jun 19 '14 at 12:58
  • $\begingroup$ See jcu.edu/math/vignettes/infinity.htm: you could use a similar "diagonal pattern". The difficult part is getting an expression for $f^{-1}$. I'm not sure how to do that, but you could ask on math.stackexchange.com (or do some more googling first). $\endgroup$ – Adrian Jun 19 '14 at 13:10
  • $\begingroup$ In the link you provided it says at some point: "Note that it is not necessary to find a formula for the correspondence; all that is necessary is the certainty that such a correspondence exists. There are many other instances in mathematics that are like this--where the point is to show that something has to happen or that something exists, rather than to actually exhibit a formula." Well, the point in my question is to actually exhibit a formula : I called this question "constructivist" for a reason. $\endgroup$ – Alecos Papadopoulos Jun 19 '14 at 13:14
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    $\begingroup$ I think I can provide an algorithm that would work -- I'll think about it a bit more. $\endgroup$ – Adrian Jun 19 '14 at 13:15
  • $\begingroup$ I posted something -- lets you simulate Q, but doesn't solve the PMF issue. $\endgroup$ – Adrian Jun 19 '14 at 13:30

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