The typical user does not reside in the typical cell

The analysis of cellular networks usually focuses on the typical user in the downlink and the typical base station (or, equivalently, the typical cell) in the uplink. It is important that if base station and user point processes are independent, the two notions of “typical” are not compatible – the typical user’s cell is statistically different from the typical cell. The difference is caused by the effect of size-biased sampling. The typical user’s performance corresponds to that of the average of all users, and there are more users in larger cells. Since a user model is not needed in the downlink as explained in this post, we can equivalently say that an arbitrary location is more likely to fall in a larger cell than a smaller cell.
The typical user’s cell, the so-called 0-cell, is the cell containing the origin, i.e., it is obtained by cell area-biased sampling, which gives larger cells more weight. As a result, the 0-cell is larger on average than the typical cell, which is the cell of the base station conditioned to be at the origin. The statistical properties of the typical cell correspond to the averages of all cells.

Such size-biased sampling is not restricted to cellular networks or stochastic geometry. If we throw a dart blindly on a world map until we hit land, the country we hit is quite likely to be a big one. In fact, there is a 50% chance that the dart lands on one of the 10 largest countries. Similarly, the typical country has 40 M inhabitants on average, but the typical person is likely to live in a country with more than 100 M people. The typical dollar is quite likely owned by a wealthy person, while the typical person is probably not rich. The typical human hair is likely to grow on a person with full hair, while the typical person has a 5-10% chance of being bald. The typical animal leg has a decent chance of belonging to a millipede or centipede, while the typical animal is very unlikely to have more than six feet.

Coming back to cellular networks, let us focus on a concrete example that is fully tractable in terms of the cell area distributions. Consider the lattice with holes shows in Fig. 1 below, obtained from a square lattice of density 1 by removing the four nearest neighbors of each 16th point. It is periodic with period 4 in both directions, its density is λ=3/4, and it has four different types of cells, with three different areas, 1, 3/2, and 2.

Fig. 1: Lattice hole process with density 3/4. Base stations (cell nuclei) are marked by red triangles, and those whose nearest neighbors were removed by stars, which makes these cell large. The grey box indicates one period of the lattice.

The typical cell has area 1 with probability 5/12, area 3/2 with probability 1/2, and area 2 with probability 1/12. The mean area follows as E(A)=5/12+1/2 3/2+1/12 2=4/3, which corresponds to 1/λ.

Now assume a stationary square lattice of density 1 as the user point process. Then the cells of area 1 always contain 1 user and those of area 2 always contain 2 users. Those of area 3/2 have 1 user or 2 users, each with probability 1/2. Deconditioning on the cell areas, we obtain the distribution of the number of users U in the typical cell as P(U=1)=2/3 and P(U=2)=1/3, for a mean number of users E(U)=4/3, which equals the mean area times the user density (chosen to be 1 here).

How about the typical user’s cell? This is where the size bias plays a role. The distribution of the area A₀ of the 0-cell is P(A₀=1)=5/16, P(A₀=3/2)=9/16, and P(A₀=2)=1/8. These are the fractions of the plane covered by cells of areas 1, 3/2, and 2. The mean area is E(A₀)=45/32, which is about 5.5% bigger than the mean area of the typical cell. The number of users U₀ in the 0-cell is distributed as P(U₀=1)=5/16+1/2 9/16=19/32, P(U₀=2)=1/2 9/16+1/8=13/32, resulting in a mean of E(U₀)=45/32, which is the user density times the mean area. The mean also follows from the general formula

$\displaystyle \mathbb{E}(f(V_0))=\frac{\mathbb{E}(A f(V))}{\mathbb{E}(A)}=\lambda\mathbb{E}(A f(V)).$

where V is the typical cell, V₀ the 0-cell, and f is a non-negative function on compact sets. Applied to our setting, where f(V) is the number of users in V, we obtain

$\displaystyle \mathbb{E}(U_0)=\frac{\mathbb{E}(A^2)}{\mathbb{E}(A)}=\lambda\mathbb{E}(A^2).$

Since the user density is 1, this is also the mean area of A₀. For the number of sides S, we have E(S)=19/4, but E(S₀)=155/32, which is bigger by 3/32.
At the end of this post are three more examples of similarly constructed lattices. In each case, the points within a certain distance of a sub-lattice are removed.

So the typical user is not served by the typical base station, and the typical base station does not serve the typical user. One way to reconcile the two is to define a user point process where a fixed number of users, say one, is placed uniformly at random in each cell. Such a user process is of course no longer independent of the base station process.

For Poisson distributed base stations, the 0-cell is 28% larger than the typical cell. Its mean number of sides is 6.41, whereas the typical cell has 6 sides on average. Hence the 0-cell is not just an enlarged version of the typical cell but also has a different shape. Accordingly, the distance from the nucleus of the typical cell to a random point in the cell is not Rayleigh distributed as it is in the 0-cell. Also, if users form a PPP of density 1, the typical user’s cell has 1+1.28/λ users on average (there is one extra user due to the conditioning of a user to be at the origin), while the typical cell only has a mean of 1/λ users.

Size-biased sampling is important in other wireless networks as well. If a vehicular network is modeled by placing one-dimensional Poisson point processes (cars) on line segments (streets) of independent random length (which is a Cox process supported on line segments), then the typical vehicle’s street length distribution f_L₀ is different from the length distribution f_L of the streets. By length-biased sampling, the two are related as

$\displaystyle f_{L_0}(x)=\frac{xf_L(x)}{\mathbb{E}(L)}.$

For example, if L is exponential with mean 1, then L₀ is gamma distributed with mean 2. The same situation arises in the interarrival intervals of a one-dimensional PPP (of density 1). The typical such interval is exponential with mean 1, but the interval containing the origin (or any other deterministic time instant) has a mean length of 2. This is sometimes referred to as the waiting time paradox, although there is nothing paradoxical about it – it is just size-biased sampling.

Lastly, as promised, here are three more examples of lattices with increasingly large holes.