Geometry vs. Probability

A conundrum of infinities

Consider a disc of radius R comprising of infinitely many points distributed uniformly over its surface. What’s the average distance of a point from the centre of the disc?

A friend and I were exchanging puzzles in probability when he asked me this one. It has a straightforward geometric solution, which goes like this. Given a function f, which is well defined in region X, the average value of f over X can be calculated in two steps.

1. Compute f(x) for every point x ∈ X and sum these values up.
2. Divide the resulting sum by the area of X.

Formally, this is expressed as follows.

Here dA is the differential of the area of the region. For the present problem, we choose the disc for the region X, and represent points on it using polar co-ordinates (r, 𝜃). Then the function f is nothing but the distance r of a point from the centre. The solution turns out to be 2R/3. If you insist, it takes two lines to write it down.

Since my friend and I were discussing probabilities, you can think about this problem in terms of a random variable X which models the distance of a point on the disc from its centre. What we wish to do is to find out the expected value E[X] of X.

Here, P is the probability density function (PDF) denoting the likelihood that the distance of a random point from the centre is r. Clearly, the solution depends upon the choice of P. So what's the right choice?

One view is that the value P(r) is proportional to the distance r of any point from the centre. This means that the farther out from the centre a point lies, the higher is the probability of our randomly choosing that point. Intuitively, a disc is made up of infinitely many concentric circles. Moreover, even though each circle contains infinitely many points, circles of larger radii can accommodate "more" points and are therefore more likely to contribute to a random choice of a point.

So we can write P(r) = c·r, where in fact the constant c = 2/R² since the area under the plot of the PDF must equal 1. Using P(r) = 2·r/R² in the formula for expectation, we get E[X] = 2R/3, which is consistent with the geometric solution above.

This was the position taken by my friend. An intuition which he provided to support this choice of the PDF is this. Imagine a dart board on which one can throw darts at random. It's more likely that darts would land at larger radii than smaller ones simply because a circle with larger radius can accommodate more darts.

Darts on a dart board. Unlike points on the disc, each dart has a finite area. [Source: Pexel]

But there's a problem with this approach. Points on a disc cannot be compared to darts on a dart board. Unlike darts, points are infinitesimal and occupy no area. If we must use a dart board analogy, then we must imagine a dart board the size of a 10 storey building on which we shoot at random darts the size of photons of light. Now can you use your imagination and say with conviction whether or not photons are more likely to land on circles with larger radii? I can't. My imagination breaks down, which is often the case when thinking about the infinitesimal and the infinite.

Let me give you an easy proof to the contrary. That is, no matter how large the differences in radii, all circles contain exactly the same number of points.

Consider two concentric circles of different radii on the disc. Identify a point on the outer circle and connect it to the centre of the disc with the help of a line segment. This segment also passes through a point on the inner circle. Hold on to these points. Now make an infinitesimal move on the outer circle in clockwise direction to arrive at a new outer point. Again, connect this new point to the centre of the disc. You will get a new line segment, and it will pass through a new point on the inner circle, one located infinitesimally close to the previous inner point but slightly towards the clockwise direction.

For every point on the larger circle, there exists a point on the inner circle, and the other way round.

We have found a one-to-one correspondence between the points on two concentric circles; and therefore there must be equal number of points on both, regardless of the difference in their sizes.

Now, if there are equal number of points on any given circle, each circle makes equal contribution to the choice of a random point. So the probability of choosing a point at random on the disc is independent of the distance of the point from the centre. Therefore, the PDF P(r) = c, and in fact c = 1/R since the area under its plot equals 1. Substituting this in the formula for expectation, we obtain E[X] = R/2.

Alternatively, note that each point p can be represented using its polar co-ordinates (r, 𝜃); and since the question states that the points are distributed uniformly over the disc, the points p can be modelled using a random variable P = (X, ϴ), where X and ϴ are independent, uniform random variables whose values range over intervals [0, R] and [0, 2𝜋] respectively. Therefore, E[X] = R/2.

Clearly, both of these solutions cannot be correct. So what gives? I'll be grateful if you could help me find a resolution. So far I see two possibilities but cannot see how to follow either.

1. I am making a silly mistake in thinking about probability density functions.
2. The choice of the correct probability density function depends upon the precise wording of question.

I would greatly appreciate a formal explanation.

Update 22.05.2018: I provide a resolution here.