Author: Martin Haenggi

A case for T junctions

It has been established (for example, here) that the standard two-dimensional homogeneous PPP is not an adequate model for vehicular networks, since vehicles are mostly confined to streets. The Poisson line Cox process (PLCP) has naturally emerged as the model of choice. In this process, one-dimensional PPPs are placed on a street system formed by a Poisson line process. This model is somewhat tractable and thus has gained some traction in the community. With probability 1 each line (or street) intersects with each other line, so intersections are formed, and the communication performance at the typical intersection vehicles can be studied. This is important since vehicles at intersections are more accident-prone than other vehicles.

How about T junctions? Clearly, the PLCP has no T junctions a.s. But while not quite as frequent as (four-way) intersections, they are an important building block of the street systems in every city, and it is reasonable to assume that they inherit some of the dangers of intersections. However, the performance of vehicles at T junctions have barely been modeled and analyzed. The reason is perhaps not that it is not worthy of study but the lack of a natural model. Let’s say we wanted to construct a Cox model of vehicles that is supported on a street system that has no intersections but only T junctions, with the T junctions themselves forming a stationary point process (in the same way the intersections in the PLCP form a stationary point process). What is the simplest (most natural, most tractable) model?

One model we came up with is inspired by the so-called lilypond model. From each point of a PPP, a line segment grows in a random orientation in both directions. All segments grow at the same speed until one of their endpoints hit another segment. Once all growth has stopped, the lilypond street model is obtained. Here is a realization:

Figure 1. Realization of Lilypond street model, starting with a PPP of density 0.1.

Then PPPs of vehicles can be placed on each line segment to form a Lilypond line segment Cox process. Some results for vehicular networks based on this model are available here. The model has the advantage that it has only a single parameter – the density of the underlying PPP of the center points of each line segment. On the other hand, the distribution of the length of the line segments can only be bounded, and the construction naturally creates a dependence between the lengths of nearby segments, which limits the tractability. For instance, in a region with many initial Poisson points, segments will be short on average, while in a region with sparse Poisson points, segments will be long. Also, the construction implies that simulating this process takes significantly more time than simulating a PLCP.

Given the shortcoming of the model, it seems quite probable that there are other, simpler and (even) more natural models for street systems with T junctions. Let’s try and find them!

The typical user and her malfunctioning base station

Let us consider a cellular network with Poisson distributed base stations (BSs). We assume that in each Voronoi cell, one user is located uniformly at random (and independently across cells), and, naturally, the user is connected to the nucleus of that cell. This is the user point process of type I defined in this article. In this case, the typical user, named Alice, does reside in the typical cell since there is no size-biased sampling involved in defining the typical user. The downlink SIR performance of this network has been analyzed here (SIR distribution) and here (SIR meta distribution).

Suddenly and unfortunately, Alice’s serving BS is malfunctioning. Her downlink is, well, down, and she gets reconnected to the next-nearest one. How does that affect her SIR performance?
In another network, also with Poisson BSs, lives another type of typical user, namely that of a stationary point process of users that is independent of the base station process. This typical user’s name is Bob. Bob’s SIR performance is the same as the one measured at an arbitrary deterministic location on the plane, as discussed in this post. He resides in the 0-cell, not the typical cell. So in his case, the typical user does not reside in the typical cell.

Noticing that Alice’s original BS ceased to operate, Bob says: “I think now your SIR performance is the same as mine. After all, your cell was formed by adding a BS at the origin, while my network has no such BS. With that added BS removed, we are in the same situation.

Alice responds: “You may have a point, but I am not sure that my location is uniform in the 0-cell, as yours is.”

Bob: “Good point, but wouldn’t that be the natural conjecture?

Alice: “I am not sure. How about we verify? Let’s look at the distances.”

Bob: “Ok. For me, if Bn is the distance to my n-th nearest BS, we have

\displaystyle \pi\mathbb{E}(B_n^2)=n.

Alice: “For me, the distance A1 to the malfunctioning BS satisfies π𝔼(A12)≈10/13, by the properties of the typical cell. If you are correct, then the distribution of A2 should be the same as that of B1. But that’s not the case, see this figure.

Mean squared distances times π. For Alice, the distances are An, for Bob, Bn.

Bob: “I see – your new serving BS at distance A2 is quite a bit further away than mine at B1. So my conjecture was wrong.

Alice: “Yes, but the question of how resilient a cellular network is to BS outages is an interesting one. How about we compare the SIR performance with and without BS outage in different networks, say Poisson networks and lattice networks? I bet Poisson networks are more robust, in the sense that the downlink SIR statistics change less when there is an outage and users need to be handed off to the next-nearest BS.

Bob: “Hmm… that would make sense. But would that mean we should build clustered networks, to achieve even higher robustness?

Alice: “Possibly – if all we worry about is a small loss when a user is offloaded. But we should take into account the absolute performance also, and clustered networks are worse in this regard. If the starting performance is much higher, it is acceptable to have a somewhat bigger loss due to outage and handover.

Bob: “Makes sense. Sorry, I have to go. My SIR is so high that I just got a phone call.

Percentage games

Let us consider a hypothetical situation where the authors of a paper promise the following:

In the next figure, we compare our Quantum Ultra-Enhanced Superior-Throughput Intelligent Objectively-Novel Adaptive Beamformed Low-Latency Emission (QUESTIONABLE) scheme with the Arbitrarily-Massive Antenna Zero-Innovation Neural Grandiloquence (AMAZING) scheme, which is the best previously known scheme. As the figure shows, the QUESTIONABLE scheme outperforms the AMAZING scheme by 6122% at a threshold of 2 dB.”

They refer to this figure:

Fig. 1: Fair comparison of our QUESTIONABLE scheme with the AMAZING scheme reported elsewhere. As is clearly apparent, the reliability gain at 2 dB is very high, unquestionably.

A gain of more than 6,000% sounds, well, amazing. But it also questionable: Is it good practice to state this figure of 6122%? Is it good marketing? Is it acceptable? Should it be discouraged? Is it even unethical?
If claiming 6122% at 2 dB is acceptable, how about a 25119% gain at 10 dB? Here the absolute reliability of AMAZING is 6.3e-9, and that of QUESTIONABLE is 1.60e-6. Not a practical regime, but a more than 25000% gain nonetheless.

Conversely, at -1.5 dB, the improvement is from 0.996 (which is AMAZING) to 0.9999999 (which is QUESTIONABLE). This is a modest 0.4% improvement both in absolute and relative terms. However, focusing on outage instead of reliability, the factor between the two outage probabilities is 62763, or 6,276,267%. So the 0.4% gap becomes a 6M% gap (that’s mega-percent, admittedly an usual combination of a metric prefix with a unit).

Independently of such amazing and questionable transmission schemes, is an improvement from 10% to 13% a 3% improvement or a 30% improvement? Frequently we see such a gain reported as a 30% improvement. Is it fair to use the ratio to measure improvement, or should it be the gap (difference)? Equivalently, should the percentage improvement be measure in percent of the underlying quantity (in which case it is 3%) or in percent of the percentages (in which case it is 30%)? The problem is that terms like “gain”, “increase”, “improvement”, “boost”, “raise” are imprecise as they may refer to differences or ratios.

How about an improvement from 10 dB to 13 dB? Here, most people would agree that the gain is 3 dB. The reason why there is no ambiguity here is that 3 dB is both the difference 13 dB-10 dB and also the ratio of the underlying quantities expressed in dB, 20/10. Thanks to the logarithm in the dB scale, both lead to the same result. With percentages, this is not the case.

The ambiguity also manifests itself when we compare percentages with physical units such as meters or seconds. The difference between 10 m and 13 m is 3 m, and 13 m is 30% longer than 10 m. Percentage gains always measure ratios in this case. But if “meter” is replaced by “percent”, the situation becomes murky. By analogy then the difference between 10% and 13% is 3%, so 13% is 3% more than 10%, in the same way that 13 m is 3 m more than 10 m. But at the same time, 13% is 30% more than 10%, in the same way that 13 m is 30% more than 10 m.

So how do we resolve this? Here are some suggestions:

  • Always state whether we refer to ratios or difference. For instance, instead of saying “The increase is 30%” say the “increase is a factor of 1.3 or 30%”.
  • Avoid ambiguity by using fractions instead of percentages. For instance, the difference between 0.1 and 0.13 is 0.03, and the ratio is 1.3.
  • Avoid aggressive marketing when quantities are in very impractical regimes. For instance, in the example above, advertising huge gains at reliabilities below 1e-5 is not good practice. Conversely, in regimes of interest, such as the high-reliablity regime, it is perfectly acceptable to report a gain of 100 in the outage performance if the outage probability is reduced from 0.01 to 0.0001. (However, talking about 10,000% is not recommended.)
  • Consider using horizontal distances (gaps) instead of vertical ones. Especially with steep (almost step-like) curves like the one in Fig. 1, the vertical gap strongly depends on where it is measured along the x axis. Conversely, the horizontal gap may be fairly constant. In the case of Fig. 1, the horizontal gap is, in fact, constant. It is exactly 3 dB at all reliabilities. A quasi-constant gap is often observed in SIR distributions (cdfs) of cellular networks when different transmission schemes are compared. The theory behind is explained in this letter (here on IEEE Xplore) and in this paper (here on IEEE Xplore). In this case, the gains are either factors (in linear scale) or gaps (in the dB) scale. This method of reporting gains is the one adopted by the coding theory community – coding gains are reported in dB (horizontally), not as gains in the bit error performance (vertically).

On cell slicing

Network slicing is a warm topic these days. Here we discuss cell slicing, where a polygon is cut in three pieces (sub-polygons) by two lines through its nucleus and a random point, respectively. First, as a sequel to this post, we focus on the 0-cell in the Poisson-Voronoi tessellation, which is the Voronoi cell of a PPP that contains the origin.
As in that earlier post, we rotate and shift the 0-cell so that its nucleus is at the origin and the point located uniformly randomly in the cell is to the right (on the positive x axis). In a cellular network application, the nucleus (now at the origin) is the location of the base station, while the random point is the typical user. Then we draw vertical lines through the origin and the random point and calculate the areas of the polygons to the left of the origin and to the left of the random point. This movie shows a number of realizations of this setup:

Movie 1. Realizations of 0-cells in Poisson-Voronoi tessellation sliced into three sub-polygons.

For intensity λ=1, the 0-cell in the Poisson-Voronoi tessellation has a mean area of 1.280176, obtained by the numerical evaluation of an integral. The polygon to the left of the nucleus o, shown in green in the movie, has a mean area of 0.517649, also obtained by numerical integration. The red polygon, to the right of o and to the left of x, has a mean area of 0.529111, obtained by simulation. Relative to the total mean area, we thus have:

  • Mean area to the left of o: 40.4%
  • Mean area to the left of x: 81.8%
  • Mean area to the right of x: 18.2%

Hence, on average, almost 60% of the (other) users in the cell are on the same side of the base station as the typical user, and the mean area to the left of the typical user is larger than the entire mean area of the typical cell (which is 1 for λ=1). Also, more than 18% of the users are “behind” the typical user.

Not surprisingly, the area fractions are quite similar in the typical Poisson-Voronoi cell. (Note that the uniformly random point in the typical cell does not correspond to the typical user of a user point process that is independent of the base station process – see this post). These percentages (area fractions) for the typical cell are all simulated:

  • Mean area to the left of o: 39.8%
  • Mean area to the left of x: 81.3%
  • Mean area to the right of x: 18.7%

How unusual are these area fractions? Let us compare them with those in the disk, which is, in some sense, the ideal cell shape. In the disk, with the nucleus o at the center, the mean area fraction to the left of o is trivially 1/2, while the area to the left of a uniformly random point is easily determined to be 7/8. This shows that the (roughly) 2/5 – 2/5 – 1/5 split in the typical Poisson-Voronoi cells is relatively far from the 1/2 – 3/8 – 1/8 split of the disk. How about regular polygons with a finite number of sides? The second movie shows some realizations for 3,4, … ,10,12,15,20,32 sides.

Movie 2. Realizations of regular polygons sliced into three sub-polygons.

As expected, for a larger number of sides the area fractions approach those of the disk. Here is a plot of the relative deviations to the disk of regular polygons, where o is the centroid:

Figure 1. Percentage deviation of mean areas left of o and left of x in regular polygons relative to the disk.

For instance, for the triangle, the area fraction to the left of o is 0.516, which is 3.2% larger than for the disk. Hence the first blue point in Fig. 1 (top left), corresponding to 3 sides, is at 3.2. To better see how the deviations behave for 5 or more sides, here is another version of the figure that shows only pentagons and higher.

Figure 2. Same as Fig. 1 but starting at pentagons.

From the blue curve it is apparent that the area to the left of the center is always exactly one half if the number of sides is even. The reason is that with an even number of sides, the polygons to the left of o and to the right of o are always congruent irrespective of the location of the random point. We also see that the hexagon is quite close to the disk already, with a mere 0.15% deviation from the 7/8 area fraction (to the left of x) of the disk.

Averages, distributions, and meta distributions

In this post I would like to show how meta distributions naturally emerge as an important extension of the concepts of averages and distributions. For a random variable Z, we call 𝔼(Z) its average (or mean). If we add a parameter z to compare Z against and form the family of random variables 1(Z>z), we call their mean the distribution of Z (to be precise, the complementary cumulative distribution function, ccdf for short).
Now, if Z does not depend on any other randomness, then 𝔼1(Z>z) gives the complete information about all statistics of Z, i.e., the probability of any event can be expressed by adding or subtracting these elementary probabilities.
However, if Z is a function of other sources of randomness, then 𝔼1(Z>z) does not reveal how the statistics of Z depend on those of the individual random elements. In general Z may depend on many, possibly infinitely many, random variables and random elements (e.g., point processes), such as the SIR in a wireless network. Let us focus on the case Z=f(X,Y), where X and Y are independent random variables. Then, to discern how X and Y individually affect Z, we need to add a second parameter, say x, to extend the distribution to the meta distribution:

\displaystyle \bar F_{[\![Z\mid Y]\!]}(z,x)=\mathbb{E}\mathbf{1}(\mathbb{E}[\mathbf{1}(Z>z) \mid Y]>x).

Alternatively,

\displaystyle \bar F_{[\![Z\mid Y]\!]}(z,x)=\mathbb{E}\mathbf{1}(\mathbb{E}_X\mathbf{1}(Z>z)>x).

Hence the meta distribution (MD) is defined by first conditioning on part of the randomness. It has two parameters, the distribution has one parameter, and the average has zero parameters. There is a natural progression from averages to distributions to meta distributions (and back), as illustrated in this figure:

Figure 1: Relationship between mean (average), ccdf (distribution), and MD (meta distribution).

From the top going down, we obtain more information about Z by adding indicators and parameters. Conversely, we can eliminate parameters by integration (taking averages). Letting U be the conditional ccdf given Y, i.e., U=𝔼X1(Z>z)=𝔼[1(Z>z) | Y], it is apparent that the distribution of Z is the average of U, while the MD is the distribution of U.

Let us consider the example Z=X/Y , where X is exponential with mean 1 and Y is exponential with mean 1/μ, independent of X. The ccdf of Z is

\displaystyle \bar F_{Z}(z)=\frac{\mu}{\mu+z}.

In this case, the mean 𝔼(Z) does not exist. The conditional ccdf given Y is the random variable

\displaystyle U=\bar F_{Z\mid Y}(z)=\mathbb{E}\mathbf{1}(Z>z\mid Y)=e^{-Yz},

and its distribution is the meta distribution

\displaystyle \bar F_{[\![Z\mid Y]\!]}(z,x)\!=\!\mathbb{P}(U\!>\!x)\!=\!\mathbb{P}(Y\!\leq\!-\log(x)/z)\!=\!1\!-\!x^{\mu/z}.

As expected, the ccdf of Z is retrieved by integration over x∈[0,1]. This MD has relevance in Poisson uplink cellular networks, where base stations (BSs) form a PPP Φ of intensity λ and the users are connected to the nearest BS. If the fading is Rayleigh fading and the path loss exponent is 2, the received power from a user at an arbitrary location is S=X/Y, where X is exponential with mean 1 and Y is exponential with mean 1/(λπ), exactly as in the example above. Hence the MD of the signal power S is

\displaystyle \qquad\qquad\qquad\bar F_{[\![S\mid \Phi]\!]}(z,x)=1-x^{\lambda\pi/z}.\qquad\qquad\qquad (1)

So what additional information do we get from the MD, compared to just the ccdf of S? Let us consider a realization of Φ and a set of users forming a lattice (any stationary point process of users would work) and determine each user’s individual probability that its received power exceeds 1:

Figure 2: Realization of a Poisson cellular network of density 1 where users (red crosses x) connect to the nearest base station (blue circles o). The number next to each user u is ℙ(Su>1 | Φ).

If we draw a histogram of all the user’s probabilities (the numbers in the figure), how does it look? This cannot be answered by merely looking at the ccdf of S. In fact ℙ(S>1)=π/(π+1)≈0.76 is merely the average of all the numbers. To know their distribution, we need to consult the MD. From (1) the MD (for λ=1 and z=1) is 1-xπ. Hence the histogram of the numbers has the form of the probability density function πxπ-1. In contrast, without the MD, we have no information about the disparity between the users. Their personal probabilities could all be well concentrated around 0.76, or some could have probabilities near 0 and others near 1. Put differently, only the MD can reveal the performance of user percentiles, such as the “5% user” performance, which is the performance that 95% of the users achieve but 5% do not.
This interpretation of the MD as a distribution over space for a fixed realization of the point process is valid whenever the point process is ergodic.

Another application of the MD is discussed in an earlier post on the fraction of reliable links in a network.

The point closest to the origin is not typical

When simulating a point process to characterize the performance of the typical point (typical user or receiver), a conditioned version of the point process given a point at the origin o may not be available. It is then tempting to choose the “next best” point as a substitute, which may be the point closest to the origin. (Whether the coordinates are then shifted so that this point is at o is irrelevant.) The goal of this post is to show that this point is not typical, i.e., producing many realizations of the point process and evaluating the performance at this point does not yield the performance of the typical point. I call the point closest to o after averaging over the point process the 0-point. Put differently, the 0-point is the typical point among all points closest to o across the realizations of the point process. In a cellular network, the 0-point is the nucleus of the 0-cell (see this post), hence the term.

For simplicity, let us consider the homogeneous PPP of intensity 1 and focus on the probability that a disk of radius r centered at a point contains no other points, which we refer to as the NOPID (no other point in disk) probability. Equivalently, it is the probability that the nearest neighbor is at distance at least r. For the typical point, the NOPID probability is exp(-πr2). For the 0-point, let D be its distance from o. Given D, the disk of radius D centered at o, denoted as b(o,D), is empty, so the points excluding the 0-point form a PPP on ℝ2\b(o,D), and the NOPID probability is the probability that b((D,0),r)\b(o,D) is empty. This region is shown in blue in the movie below for different r given that the 0-point is at (1,0), i.e., D=1. For r<2D, it is moon- or crescent-shaped, while for r>2D, it is a disk with a hole.

Movie: Illustration of relevant region for the NOPID probability of the 0-point.

Letting A(r,d)=|b((d,0),r)\b(o,d)|, the (unconditioned) NOPID probability is 𝔼(exp(-A(r,D)), where D is Rayleigh distributed with mean 1/2. It can be expressed as

\displaystyle \qquad\qquad F_0(r)=\frac{\pi}{4}r^2 e^{-\pi r^2}+\int_{r/2}^\infty e^{-A'(r,u)}2\pi u e^{-\pi u^2}{\rm d}u,\qquad\qquad (1)

where

\displaystyle A'(r,d)=\pi r^2-r^2\cos^{-1}\left(\frac{r}{2d}\right)-d^2\cos^{-1}\left(1-\frac{r^2}{2d^2}\right)+\frac{r}{2}\sqrt{4d^2-r^2}.

is the area A(r,d) for r<2d. For r>2d, A(r,d)=π(r2d2), which results in the first term in (1).

The NOPID probabilities of the 0-point and the typical point are compared below. It is apparent that the 0-point is more isolated than the typical point.

Figure: Comparison of NOPID probabilities.

By integrating the NOPID probability of the 0-point, we obtain the mean nearest-neighbor distance as 0.5953. This is almost 20% larger than that of the typical point, which is 1/2. The difference between the two NOPID probabilities is not just in the mean, though. They differ qualitatively in the tail. For large r, it follows from (1) that the ratio of the two NOPID probabilities approaches πr2/4. This implies that a Rayleigh distribution with adjusted mean will not provide a good fit to the NOPID probability at the 0-point.

The difference is even more pronounced if we consider directional nearest neighbors. If we consider a sector of angle π/2, then the nearest neighbor of the typical point is at distance 1 on average, irrespective of the orientation of the sector. For the 0-point, in the direction opposite from o, the mean distance is also 1, since on that side, the PPP is unaffected by the empty disk b(o,D). In the direction towards o, however, the distance is significantly larger, with a mean of 1.4205. The plot below shows the pdf of the directional nearest-neighbor distance of the 0-point oriented towards o (red) and the pdf of the directional nearest-neighbor distance of the typical point (blue), given by (π/2)r exp(-(π/4)r^2). The pdfs are the negative derivatives of the NOPIS (no other point in sector π/2) probabilities.

Figure: pdf of the distance to the directional nearest neighbor of the 0-point in π/2 sector oriented towards o, in comparison with that of the typical point.

When applied to cellular networks (with nearest-base station association), the 0-point is the base station serving the typical user (at o). The discussion here reveals that the 0-base station behaves differently from the typical base station. In particular, the point process of the other base stations viewed from the 0-base station is highly non-isotropic. In the direction of the typical user, the nearest other base station is much further away than in the opposite direction. This fact is consistent with the conclusions from this post on the shape of the 0-cell in the Poisson-Voronoi tessellation.

The curious shape of Poisson-Voronoi cells

In this blog we are exploring the shape of two kinds of cells in the Poisson-Voronoi tessellation on the plane, namely the 0-cell and the typical cell. The 0-cell is the cell containing the origin, while the typical cell is the cell obtained by conditioning on a Poisson point to be at the origin (which is the same as adding the origin to the PPP).

The cell shape has an important effect on the signal and interference powers at the typical user (in the 0-cell) and at the user in the typical cell. For instance, in the 0-cell, which contains the typical user at a uniformly random location, about 1/4 of the cell edge is at essentially the same distance to the base station as the typical user on average). Hence it is not the case that edge users necessarily suffer from larger signal attenuation than the typical user (who resides inside the cell).

The cell shape is determined by the directional radii of the cells when their nucleus is at the origin. To have a well-defined orientation, we select a location uniformly in the cell and rotate the cell so that this location falls on the positive x-axis. In the 0-cell, this involves first a translation of the cell’s nucleus to the origin, followed by a rotation until the original origin (which is uniformly distributed in the cell) lies on the positive x-axis. This is illustrated in Movie 1 below. In the typical cell, it involves adding a Poisson point, selecting a uniform location, and a rotation so that this uniform location lies on the positive x-axis. This is illustrated in Movie 2.

Movie 1. Rotated and translated 0-cell.

Movie 2. Rotated typical cell.

As indicated in the movies, the distances from the nucleus to the uniformly random location are denoted by D0 and D, respectively, and the directional radii by R0(ϕ) and R(ϕ), respectively. This way, the boundary of the cells is described in polar coordinates as (R0(ϕ),ϕ) and (R(ϕ),ϕ), ϕ ∈ [0,2π). In a cellular network model, the uniform random location could be that of a user, while the PPP models the base stations. In this case D0 is the link distance from the typical user to its serving base station, while D is the link distance from the typical base station to a randomly located user it serves. The distinction between the typical user’s and the typical base station’s point of view is explained in this blog.

Let λ denote the density of the PPP. Three results are well known:

  • The distribution of D0 follows from the void probability of the PPP. It is Rayleigh with mean 1/(2√λ).
  • Since the mean area of the typical cell is 1/λ, we have ∫ 0π 𝔼(R(ϕ)2) dϕ = 1/λ.
  • The minimum of R(ϕ) is distributed with pdf f(r)=8λπr exp(-4λπr2). This is half the distance to the nearest neighboring Poisson point (base station).

In contrast, there is no closed-form expression for the distribution of D. Due to size-biased sampling, the area of the 0-cell stochastically dominates that of the typical cell and, in turn, D0 dominates D.

Analyzing the directional radii, we obtain these new insights on the cell shapes:

  • If Ψ is uniform in [0,π], R(Ψ) is again Rayleigh with mean 1/(2√λ).
  • R0(π) is also Rayleigh with the same mean. In fact, R0(π) and D0 are iid.
  • R0(0) has mean 3/(4√λ) and is distributed as

\displaystyle f_{R_0(0)}(y)=2(\lambda\pi)^2 y^3 \exp(-\lambda\pi y^2).

  • Hence R0(0) is on average exactly 50% larger than R0(π). For the typical cell, simulation results indicate that R(0) is about 55% larger on average than R(π).
  • The difference R0(0)-D0 is distributed as f(r)=π√λ erfc(r √(πλ)). Its mean is 1/(4√λ). Hence the typical user is no further from the cell edge than the base station on average.
  • The joint distribution of D0 and R0(ϕ) can be given in exact analytical form.
  • 3/4 of the typical cell is further away from the nucleus than the nearest point on the cell edge (i.e., the minimum directional radius). Expressed differently, a uniformly random user in the typical cell has a 75% chance of being further away from the base station than the nearest edge user. By simulation, D on average is 2.7 times larger than the minimum of the directional radii.

In conclusion, the 0-cell and the typical cell are quite asymmetric around the nucleus (base station) and the uniformly random point (user). In the direction away from the base station, the user is about 4 times closer to the cell edge than in the direction towards the base station, and many locations on the cell edge are closer to the base station than the user inside the cell. These results have implications on the design of efficient cellular network transmission schemes, such as beamforming, NOMA, and base station cooperation, in both down- and uplink.

More details are available in Section II of this paper.

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 fL0 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.

Fig. 2: Lattice with period 6. λ=1/6; P(A=1)=1/6; P(A=89/15)=2/3; P(A=169/15)=1/6; E(A)=6.
P(A0=1)=1/36; P(A0=89/15)=89/135; P(A0=169/15)=169/540; E(A0)=6047/810 ≈ 7.46.
Fig. 3: Lattice with period 8. λ=7/32; P(A=1)=5/14; P(A=7/2)=2/7; P(A=29/4)=2/7; P(A=16)=1/14; E(A)=32/7. P(A0=1)=5/64; P(A0=7/2)=7/32; P(A0=29/4)=29/64; P(A0=16)=1/4; E(A0)=2081/256. Here the 0-cell is almost twice as big on average than the typical cell. It corresponds to the largest cell with probability 1/4, but only 1/14 of the cells are largest ones.
Fig. 4: Lattice with period 16. λ=7/128; P(A=1)=5/14; P(A=15/2)=2/7; P(A=33)=2/7; P(A=89)=1/14; E(A)=128/7. P(A0=1)=5/256; P(A0=15/2)=15/128; P(A_0=33)=33/64; P(A0=89)=89/256; E(A0)=12507/256 48.9. The 0-cell is more than 2.5 times larger than the typical one on average.