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Please assist with this problem.

Suppose 3 (distinct) points are uniformly and independently distributed on a circle of unit length (smaller than a unit circle!). This is really circle and not disc. Call one of these points $B$. Let $W$ be the minimum distance of the point B to its nearest neighbour, either clockwise or anti-clockwise, whichever is smaller.

Find the pdf of $W$. (Well there's no measure theory for this problem, but I assume this pdf exists. Of course we can see for ourselves by computing the cdf $F_W(w)= P(W \le w)$ 1st and then hope the cdf is absolutely continuous.)

A. My model:

  • A.1. The circle is bijective with $[0,1)$, so let's call these 3 points $A,B,C$ s.t. they are iid $\sim \ Unif(0,1)$ (or $[0,1)$ or whatever).

  • A.2. (Not sure if any measure theory needed here, but they are all distinct $\mathbb P$-almost surely.)

  • A.3. Let $A$ be the anti-clockwise neighbour and $C$ clockwise.

Question:

Well, I hope to find the pdf of $W$ via its cdf, which I think I'm able to compute if I know what $W$ is. Is it $W = \min\{B-A,A-C\}$? Or is it $W=\min\{Z,U\}$, where $Z$ is the distance from $B$ to $A$ (anti-clockwise of course) and $U$ is the distance from $B$ to $C$ (clockwise of course; which I think is equivalent to saying 'distance from $C$ to $B$ anti-clockwise') ? Or something else? (Note: I asked about $Z$ here, but I hope I made this post self-contained.)

These questions are all related, but I hope I made each self-contained

BCLC
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3 Answers3

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I will assume that the unit circle means a circle of unit length. Otherwise you should rescale the quantities by the factor $2\pi$.

Let the distance to the closest point be $w$. It leaves the other point the configuration space of $1-2w$.

Therefore the pdf of the least distance is: $$ \begin{cases} 4(1-2w),&w<0.5\\ 0,&w>0.5 \end{cases} $$

user
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  • Thanks user. But 1 - Where did you get the 2 in 2w and the factor of 4? 2 - with the substitution $u=2w$, why isn't this just the same thing as in part 1 (Random points on a circle)? – BCLC Mar 12 '21 at 10:38
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    Setting the closest distance to $w$ you forbid the third point to be inside the arc $\pm w$ from the point $B$. This reduces the configuration space to $1-2w$ ($1$ is the length of the circle). The simplest practical way to compute the prefactor is to treat it as a normalization constant $A$, so that $\int A \operatorname{pdf}(x),dx=1$. – user Mar 12 '21 at 10:57
  • thanks. ok I'm just avoid the ansatz gonna do the cases thing you said: A<B<C, C<B<A. and then in each case 2 points. – BCLC Mar 14 '21 at 08:23
  • user, is it that $W=\min{B-\min{A,C},|C-A|-(B-\min{A,C})}$? I ask about this here: Modelling random points on a circle – BCLC Mar 14 '21 at 08:23
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    If you would like to avoid it just count the possibilities: there are two points and there are two directions to measure. – user Mar 14 '21 at 08:26
  • ok thanks as for $W$ explicitly...? – BCLC Mar 14 '21 at 08:28
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Let $X$ be the anticlockwise distance along the circle's circumference from $B$ to $A$. Let $Y$ be the anticlockwise distance from $B$ to $C$.

Since the three points are independently uniformly distributed along the circumference of the circle, $X$ and $Y$ are iid variables with uniform distributions on $[0,1).$

Since the clockwise arc from $B$ to $A$ is $1 - X,$ the distance from $B$ to $A$ is the smaller of $X$ and $1-X$. That is, if we let $X_m = \min\{X, 1 - X\}$, then $X_m$ is the distance from $B$ to $A$.

Similarly, let $Y_m = \min\{Y, 1 - Y\}$, then $Y_m$ is the distance from $B$ to $C$.

By symmetry, $X_m \in [0,\frac12]$ and $X_m$ is uniformly distributed (glossing over the fact that the zero-probability events $X_m=0$ and $X_m=\frac12$ can each happen in only one way). Likewise, $Y_m$ also is uniform on $[0,\frac12]$. Moreover, $X_m$ and $Y_m$ are iid.

The nearest neighbor of $B$ is $A$ if $X_m < Y_m$, but is $C$ if $X_m > Y_m$. So $W = \min\{ X_m, Y_m \}.$

So we want the pdf (or cdf) of the minimum of two iid variables uniformly distributed on $[0,\frac12]$. For $0 \leq w \leq \frac12$ we have

\begin{align} P(W < w) &= 1 - P(W \geq w)\\ & = 1 - P(X_m \geq w) P(Y_m \geq w) \\ & = 1 - \left(2\left(\frac 12 - w\right)\right)^2 \end{align}

with $P(W < 0) = 0$ and $P(W < 1) = 1.$ The cdf is therefore $$ F_W(w) = 1 - (1 - 2w)^2. $$

The pdf is easily found from this.

David K
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  • My God David K this is exactly what I was looking for. I still have to analyse this, but really thank you so much! Please consider helping here: Random points on a circle and Most random points on a circle – BCLC Mar 14 '21 at 06:53
  • Btw, is this right? I realised an error in my model: I can't just say $A$ is the anti-clockwise neighbour and then $C$ is the clockwise one. I have to take 2 cases: Case 1: $A$ is the anti-clockwise and then $C$ is the clockwise and then Case 2: $C$ is the anti-clockwise and then $A$ is the clockwise. – BCLC Mar 14 '21 at 06:54
  • Also, what do you think of user's answer please? user has similar answers in the other questions – BCLC Mar 14 '21 at 08:17
  • 3rd thing: Is it that $W=\min{B-\min{A,C},|C-A|-(B-\min{A,C})}$? I ask about this here: Modelling random points on a circle – BCLC Mar 14 '21 at 08:22
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    3rd thing first: In terms of your iid variables $A,B,C,$ we have $X_m = \min{|B-A|,1-|B-A|}$ and $Y_m = \min{|B-C|,1-|B-C|}$. So $W = \min{|B-A|,1-|B-A|,|B-C|,1-|B-C|}$. The formula you derived is not suitable for an iid joint distribution. – David K Mar 14 '21 at 16:05
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    user's answer works because if $W$ is the distance from $B$ to the closer of the two points $A$ or $C$, and $V$ is the counterclockwise distance from $B$ to the other point, then $(W,V)$ is uniformly distributed on a triangular region bounded by $w = 0,$ $v = w,$ and $v = 1-w$, with density $4$ on that region. I can see intuitively that this is correct, although proving it rigorously is another matter. – David K Mar 14 '21 at 16:17
  • Thanks a lot David K. 1 - btw i started to analyse your answer and actually is $Z=\min{X,Y}$? ($Z$ is the one from Random points on a circle ). 2 - Is the following correct? I think of $U= \min{1-X,1-Y}$ such that $W=\min{Z,U} = \min{X_m,Y_m}$, where we must make the change from $(Z,U)$ to $(X_m,Y_m)$ because $Z$ and $U$ are identically distributed but not independent. – BCLC Mar 15 '21 at 08:55
  • 3 - Do you have some $g$ or $h$ in mind here please? Actually, I think $M=\min{X,1-X,Y,1-Y,g(X),h(Y)}$ (maybe $h=g$; $M$ is from Most random points on a circle) and then we can group again just like is done for $W$: $M=\min{X_n, Y_n}$, where $X_n=\min{X,1-X,g(X)}$ and $Y_n=\min{Y,1-Y,h(Y)}$ – BCLC Mar 15 '21 at 08:57
  • David K, btw you said 'minimum of two iid variables uniformly distributed', so yeah i just noticed a pattern here $Z$ is minimum of 2 iid unif(0,1), $W$ is minimum of 2 iid unif(0,$\frac12$) (and the minimum of 2 minima of iid unif(0,1)) and so I guess $M$ is the minimum of 2 iid unif(0,$\frac13$). in this case, i'm hoping $X_n$ and $Y_n$ are those iid unif(0,$\frac13$). – BCLC Mar 15 '21 at 10:26
  • 4 - Oh also David K, I was able to figure out $W = \min{|B-A|,1-|B-A|,|B-C|,1-|B-C|}$...but what exactly are $A,B,C$ please in re your answer here: Modelling and another question here: Remodelling? I mean $A,B,C$ are not necessarily uniform on the unit interval even though they are uniform on the circle right? – BCLC Mar 15 '21 at 10:33
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    If you are using $A,B,C$ both as names of points and also as the numeric coordinates of the same three points measured counterclockwise from a fixed origin, $A,B,C$ in that sense are three iid variables that are uniform on $[0,1).$ Where things get tricky is how you do the arithmetic with those three random numbers. – David K Mar 15 '21 at 13:14
  • right makes a lot of sense. thanks so so much David K – BCLC Mar 17 '21 at 04:19
  • David K, why do we even need to do $X_m$ and $Y_m$? I think we can just do like $$P(W \ge w) = P(X \ge w, 1-X \ge w, Y \ge w, 1-Y \ge w) = P(w \le X \le 1-w, w \le Y \le 1-w) = P(w \le X \le 1-w) P(w \le Y \le 1-w)$$ Am I wrong? I posted answer here. – BCLC Mar 17 '21 at 06:31
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    Yes, that works too and it is fewer steps. – David K Mar 17 '21 at 12:23
  • thanks David K!! – BCLC Mar 17 '21 at 22:59
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I'm going to use David K's $X$ and $Y$ but compute $F_W(w)$ differently for $w \in (0,\frac12)$. I don't use $X_m$, $Y_m$.

$$P(W \ge w) = P(X \ge w, 1-X \ge w, Y \ge w, 1-Y \ge w) = P(w \le X \le 1-w, w \le Y \le 1-w)$$

$$= P(w \le X \le 1-w) P(w \le Y \le 1-w)$$

$$= \int_w^{1-w} 1_{(0,1)}(x) dx \int_w^{1-w} 1_{(0,1)}(y) dy$$

$$= (1-2w)^2$$

$$ \iff F_W(w) = 1-(1-2w)^2$$

BCLC
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