Remodelling random points on a circle: Arc length between points distributed on circle is uniform?

Question

An update to this: Modelling random points on a circle , but I hope my posts are self-contained.

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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 these points $A,B,C$.

There is a claim here:

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 $[0,1).$

Question: What exactly are $X$ and $Y$ (like the specific formulas in terms of the random variables $A,B,C$, and why are they iid Unif(0,1)? Please consider modifying what I did below to answer this question.

What I tried for identical distributed Unif(0,1): I tried thinking of something related to my wrong model.

1. For each $\omega \in \Omega$, we'll have either a line up of Case 1: $A(\omega),B(\omega),C(\omega)$ or Case 2: $C(\omega),B(\omega),A(\omega)$.

2. This model involves some point on the arc between A and C not containing B is selected as like the point 0/1 in viewing the circle as the unit interval with end points glued together. (Unfortunately, I think this choice of point kind of depends on a choice of $\omega$. Perhaps a choice of bijection $\phi$ from circle to unit interval doesn't depend on $\omega$ directly, but I think it depends on the choice of point.)

3. So, if we somehow are able to well-definedly imagine $A,B,C$ as members of the unit interval, then we have $A<B<C$ in Case 1 and then $C<B<A$ in Case 2.

4. Here, I might model $X$ as is something like $X=B-A$ in Case 1 and $X=1+(B-A)$ in Case 2. Similarly, $Y=B-C$ in Case 2 and $Y=1+(B-C)$ in Case 1.

5. To compute cdf of $X$: (btw, I'm assuming inequalities as strict or loose here are not really relevant)

   • 5.1. $F_X(x) = P(X \le x)$ is $1$ for $x \ge 1$ and $0$ for $x \le 0$

   • 5.2. For $x \in (0,1), F_X(x) = P(X \le x)$

   • $= P(X \le x, 1 > B-A > 0) + P(X \le x, -1 < B-A < 0)$

   • $= P(0 < B-A \le x < 1) + P(0 < 1+(B-A) \le x < 1)$

   • 5.3. If $P(0 < B-A \le x < 1) = x - \frac12 x^2$ and $P(0 < 1+(B-A) \le x < 1) = \frac12 x^2$, then I have proven $X \sim Unif(0,1)$. Similarly, $Y \sim Unif(0,1)$.

   • 5.4. I was able to make those computations in (5.3), but I'm not exactly sure what those assumptions were. Assuming the integrand is just '$1 \ da \ db$', the regions of integration have areas of, resp, $x - \frac12 x^2$ and $\frac12 x^2$ and so the areas are the values of the integrals. I think assuming $A,B$ are iid Unif(0,1), then the computation is correct, but I'm not sure that we can view $A,B$ as iid Unif(0,1) based on that my model was wrong. Well perhaps when converting $A,B,C$ from uniform on circle to uniform on $(0,1)$, we get that at least $A$ and $B$ are independent and that $B$ and $C$ are independent even if all 3 together are not independent (and even if they are not pairwise independent, i.e. we have pairwise independence except for independence of $A$ and $C$). So, perhaps the converted $A,B,C$ are not iid Unif(0,1), but maybe there is some partial independence here, and I believe the converted $A,B,C$ have $(0,1)$ (or $[0,1]$, whatever) as images.

What I tried for independence:

(Btw, since there's no measure theory here, I'm assuming the independence definition here is that the joint pdf splits up.)

I think this depends on the formulas for $X$ and $Y$, so I'm thinking to just wait on resolution on what the formulas are. Currently, there's a $B$ in each of the formulas for $X$ and $Y$, but I'm hoping this information with $B$ cancels out somehow. Something like how in the Gauss-Bonnet Theorem/Formula all the geometry information is cancelled out to get a topological quantity. Or maybe the $B$ isn't relevant after you compute the pdf's because...the joint pdf's are just constant whenever they are nonzero. Idk.

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

• Random points on a circle
• More random points on a circle
• Most random points on a circle
• Modelling random points on a circle
• Remodelling random points on a circle: Arc length between points distributed on circle is uniform?

Answer 1

As $A,B,C$ are independent, so are $X$ and $Y$ and by symmetry they are identically distributed.

Now for any $b$ (the position of a point being measured as the counterclockwise distance from a common origin), $a$ is uniform in $[0,1)$, $a-b$ is uniform in $[-b,1-b)$ and $x=(a-b)\bmod1$ is uniform in $[0,1)$.

Answer 2

There is actually nothing wrong with choosing an arbitrary fixed point on the circle as your "origin" and giving each point a coordinate equal to the distance from the origin to that point counterclockwise along the circle. In this coordinate system, three points $A,B,C$, each placed randomly on the circle with a uniform probability distribution on the circumference, each probability independent of the other two points, will be placed at three coordinates on the circle. If we also call those coordinates $A,B,C$ respectively, then the coordinates $A,B,C$ are real-valued random variables that are i.i.d. and uniformly distributed on $[0,1).$

Where you run into trouble is when you try to say $A < B < C$ or $C < B < A.$ It is certainly true that the three points will (with probability $1$) be placed in one of two ways: either $(1)$ a counterclockwise arc starting at $A$ will reach $B$ first and $C$ second, or $(2)$ a counterclockwise arc starting at $A$ will reach $C$ first and $B$ second. It is also true that in the case $(2)$ just described, you can equally well say that a counterclockwise arc starting at $C$ will reach $B$ first and $A$ second; provided all three points are distinct, the two descriptions of this case are equivalent. But none of this implies that $A < B$ or $B < C$ in case $(1),$ because you have forgotten that all three numbers $A,B,C$ are distances from a previously chosen fixed point we called the "origin." The statement $A<B<C$ is a statement about the relative placement of four points (the origin, $A$, $B$, and $C$): namely that by starting at the origin and going counterclockwise, you will encounter $A$ first, $B$ second, and $C$ last. And while it is true that there are really only two ways to sequence three points around a circle, there are much more than two ways to sequence four points around a circle.

I suggested here that if you really want to only deal with cases $A<B<C$ and $C<B<A$ you would have to give up the i.i.d. assumption. But I think a better approach is to keep the i.i.d. assumption and give up the idea of restricting your possible arrangements to only the cases $A<B<C$ and $C<B<A$.

I have spent an awful lot of text discussing the chains of inequalities $A<B<C$ and $C<B<A$ and the two cases $(1)$ and $(2)$, but only reason I have done this is to show why these ideas are wrong ways to think about this problem. I'm going to draw a line now (literally) under this paragraph and once we cross that line we will never see those inequalities or cases again.

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To recapitulate, we have three random points on a circle, and the locations of those three points (measured counterclockwise from some fixed point) are the three i.i.d. real random variables $A,B,C$ each uniformly distributed on $[0,1).$ The joint distribution of these three variables is a uniform distribution on the cube $[0,1) \times [0,1) \times [0,1),$ represented by the figure below:

Let $B$ be the coordinate along the vertical axis in this figure, and let $A$ and $C$ be the other two coordinates.

Now let's define $X$ as the length of the shortest counterclockwise arc starting at $B$ and ending at $A$. That's the definition of $X$; we'll work out the formulas as needed to satisfy that definition.

Consider the example where $A$ is $90$ degrees counterclockwise from $B$, that is, the counterclockwise distance from $B$ to $A$ is $0.25.$ If the coordinate of $B$ is $B = 0.3,$ the coordinate of $A$ is easily found: $A = 0.3 + 0.25 = 0.55.$ But if $B = 0.9,$ things work a little differently. The first $0.1$ unit of length in the counterclockwise direction gets us from $B$ to the origin; we still have to go another $0.15$ units of length from the origin in order to finish the arc from $B$ to $A$, and therefore the coordinate of $A$ is $A = 0.15.$ That is, simple addition would have given $0.9 + 0.25 = 1.15$, but when we passed the origin we instantly subtracted $1$ from the coordinate (because we start counting distance from the last time we crossed the origin), and therefore the result is $A = 0.9 + 0.25 - 1 = 0.15.$

The rule therefore is: if the arc from $B$ to $A$ goes past the origin, that is, if $A < B,$ then $A = B + X - 1$ and therefore $X = A - B + 1$; but if the arc does not go past the origin, that is, if $B < A,$ then $A = B + X$ and $X = A - B.$

Note that the inequality $B < A$ doesn't say anything about the positions of $A$ and $B$ relative to each other; it just says that we can get from $B$ to $A$ in a counterclockwise direction without passing through the arbitrary point we chose as the origin of coordinates.

Next, let's define $Y$ as the length of the shortest counterclockwise arc starting at $B$ and ending at $C$. Using the same reasoning we used for $X$, we can show that this definition implies that $Y = C - B + 1$ if $C < B$ and $Y = C - B$ if $B < C.$

By definition, since $X$ is the shortest counterclockwise arc from $B$ to $A$, $X$ cannot be less than $0$ (you have to go counterclockwise, not clockwise!) and $X$ cannot be greater than $1$ (because you will certainly reach $A$ before you circle all the way around to $B$ again). It should be clear that in fact the possible values of $X$ are exactly $[0,1),$ and likewise with $Y$.

There are at least two ways to prove that $X$ and $Y$ are i.i.d. and uniformly distributed on $[0,1).$

In my opinion, the easiest and most intuitive proof is that if the distribution of $X$ was not uniform, then if you found out the location of $B$ you would know that $A$ was more likely to occur in some arc of the circle than in in some other equal-sized arc. That is, the probability distribution of $A$ conditioned on a value of $B$ would be different from the unconditional distribution of $A.$ Therefore $B$ and $A$ would not be independent. But since $B$ and $A$ are independent, $X$ must be uniformly distributed on $[0,1).$ Likewise for $Y$.

The second part of the proof is that if $X$ and $Y$ were not independent, then if you knew the locations of $A$ and $B$ you would know $X$; the distribution of $Y$ conditioned on $X$ might not be the same as the unconditioned distribution, that is, it might not be uniform; and therefore you would be able to say (based on the knowledge of $A$ and $B$) that $C$ is more likely to occur in one arc of the circle than in some other arc of equal length. But that would mean $C$ would not be independent of $A$ and $B$. Since we know that $C$ is independent of $A$ and $B$, $Y$ must be independent of $X$.

Another, much longer proof can be done by constructing the joint distribution of $B,$ $X,$ and $Y.$

We start by considering a different joint distribution: the joint distribution of $B,$ $x = A - B,$ and $y = C - B.$ (Lower-case $x$ and $y$ for simple difference of coordinates rather than the length of a counterclockwise arc.) This distribution looks something like the graph below, where the vertex in the lower right of the figure represents the outcome $x = A-B=1-\epsilon,$ $y = C-B=1-\epsilon,$ $B=0$ and the vertex in the upper left represents the outcome $x=A-B=-1,$ $y=C-B=-1,$ $B=1-\epsilon.$ (I write $1-\epsilon$ instead of $1$ because technically it is not possible for any of these quantities to be exactly equal to $1$.)

If we take a horizontal slice of this distribution at any particular value of $B,$ the distribution in that slice will be uniform over the square $[-B,1-B) \times [-B,1-B)$, that is, the square such that $-B \leq x = A - B < 1 - B$ and $-B \leq y = C - B < 1 - B$. And since $B$ is uniformly distributed, any equally-very-thin horizontal slice is equally likely, so the distribution over the entire volume is uniform; the density is $1$ throughout.

But the only part of this distribution that directly represents part of the distribution of $X$ and $Y$ is the part where $A > B$ and $C > B,$ that is, the square-based pyramid below:

Within this pyramid, whatever $A - B$ and $C - B$ came out to be, $X$ and $Y$ come out to those values, respectively.

The case where $C > B$ but $A < B$ is represented by the following tetrahedron bounded by the equations $x = A - B \leq 0,$ $x + B = A \geq 0,$ $y = C - B \geq 0,$ $y + B = C \leq 1.$

But since $X = A - B + 1 = x + 1$ in this case, this part of the joint distribution of $X,$ $Y,$ and $B$ is translated $1$ unit in the $x$ direction so that it falls inside the cube $[0,1) \times [0,1) \times [0,1).$

The case where $A > B$ but $C < B$ is represented by the following tetrahedron bounded by the equations $x = A - B \geq 0,$ $x + B = A \leq 1,$ $y = C - B \leq 0,$ $y + B = C \geq 0.$

But since $Y = C - B + 1 = y + 1$ in this case, this part of the joint distribution of $X,$ $Y,$ and $B$ is translated $1$ unit in the $y$ direction so that it falls inside the cube $[0,1) \times [0,1) \times [0,1).$

The final case, where $A < B$ and $C < B$, is represented by the inverted square-based pyramid in the figure below, bounded by the equations $x = A - B \leq 0,$ $x + B = A \geq 0,$ $y = C - B \leq 0,$ $y + B = C \geq 0,$ $B < 1$:

In this case $X = x + 1$ and $Y = y + 1,$ that is, this pyramid is translated by $1$ unit in both the $x$ and $y$ directions in order to become the part of the joint distribution of $X,$ $Y,$ and $B$ for this case. Again this places the pyramid inside the cube $[0,1) \times [0,1) \times [0,1).$

When the four parts of the distribution of $X$ and $Y$ are assembled inside the cube $[0,1) \times [0,1) \times [0,1)$ in this way, they exactly fill the cube without overlapping. Moreover, since the distribution of $A-B,$ $C- B,$ and $B$ from which these pieces were taken has uniform density $1,$ so does the distribution of $X,$ $Y,$ and $B$ within the cube $[0,1) \times [0,1) \times [0,1).$

Therefore the variables $X,$ $Y,$ and $B$ are i.i.d. uniformly distributed over $[0,1).$ It follows a fortiori that the variables $X$ and $Y$ are i.i.d. uniformly distributed over $[0,1).$

Personally I like the first proof much better. The second proof takes a lot more work for the construction and requires careful checking to make sure that the four pieces of the distribution really fill the cube without overlapping.