Assume a set of nodes is scattered over a 2D surface $\mathcal{S}$ so that for any given $\mathcal{A} \subset \mathcal{S}$, the number of nodes inside $\mathcal{A}$ follows a Poisson distribution with parameter $|\mathcal{A}| \rho$, where $|\mathcal{A}|$ shows the area of the subset $\mathcal{A}$ and $\rho$ is the intensity of the points (average number of points per unit area).

We are only interested in the points inside a given circle with radius $r$. The number of nodes inside the circle is a Poisson variable with parameter $\rho \pi r^2$. We pick two nodes from inside the circle at random. Let $d_1$ and $d_2$ show the distance of the first and the second node from the center of the circle.

2D Distribution

How can I compute the probability of the event:

$$ {d_1}^2 < \frac{{d_2}^2}{A(1+B{d_2}^2)} $$ where $A$ and $B$ are constants.


  1. Assume $A > 0$ and $B > 0$.

  2. I am interested in the process itself, not the points generated by the process (as whuber described in his answer below).

  3. How about the case that ${d_1}^2$ and ${d_2}^2$ is replaced with ${d_1}^\alpha$ and ${d_2}^\alpha$ for $\alpha > 2$ (I guess, this modifies the problem since ${d_1}^\alpha$ and ${d_2}^\alpha$ are not uniformly distributed anymore).

  • $\begingroup$ Given the formulation of the problem, it looks like the number of points if fixed, say $N$. Let $M<N$ be the number of points inside the circle. You can then check how many pairs satisfy that inequality, say $k$. Therefore, if you pick two points randomly, the probability that they satisfy the inequality is $k/\binom{M}{2}$. Note that I am considering that if $(p_1,p_2)$ satisfies the inequality, then $(p_2,p_1)$ also satisfies the inequality. $\endgroup$
    – user10525
    May 17, 2012 at 8:52
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    $\begingroup$ I wonder how are you defining a "uniform" distribution on ${\mathbb R}^2$? $\endgroup$
    – user10525
    May 17, 2012 at 9:25
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    $\begingroup$ Then I think that is a different problem, but a spatial Poisson process is now well-defined. You could try to formulate the problem in line with your new interests. $\endgroup$
    – user10525
    May 17, 2012 at 9:44
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    $\begingroup$ What is the difference with a Poisson point process? $\endgroup$
    – Xi'an
    May 17, 2012 at 10:37
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    $\begingroup$ @Xi'an: (+1) As far as I can tell from the description, it's not! :) $\endgroup$
    – cardinal
    May 17, 2012 at 13:06

2 Answers 2


There are at least two interpretations: one concerns the actual points generated by this process and the other concerns the process itself.

If a realization of the Poisson process is given and pairs of points are to be chosen from that realization, then there is nothing to be done except systematically compare all distances to all other distances (a double loop over the points).

Otherwise, if the procedure is intended to consist of (i) creating a realization of the process and then (ii) selecting a pair of points at random, then the assumptions imply the two points are selected uniformly and independently from the circle. The calculation for this situation can be performed once and for all.

Notice that the squared distances $r_1 = d_1^2$ and $r_2 = d_2^2$ are uniformly distributed, whence the desired probability is

$$p(a,b) = \Pr\left(d_1^2 \lt \frac{d_2^2}{a(1 + b d_2^2)}\right) = \int_0^1 d r_2 \int_0^{\max(0, \min(1, r_2 / (a(1 + b r_2))))} d r_1.$$

The $\max$ and $\min$ can be handled by breaking into cases. Some special values of $a$ and $b$ have to be handled. Because the integration is a square window over a region generically bounded by lines and lobes of a hyperbola (with vertical axis at $1/(ab)$ and horizontal axis at $-1/b$), the result is straightforward but messy; it should involve rational expressions in $a$ and $b$ and some inverse hyperbolic functions (that is, natural logarithms). I had Mathematica write it out:

$$\begin{array}{ll} \frac{b+1}{b} & \left(-1\leq a<0\land \frac{1}{a}-b\leq 1\land b<-1\right)\\ &\lor \left(a<-1\land \frac{1}{a}-b<1\land b<-1\right) \\ -\frac{1}{b (a b-1)} & \frac{1}{a}-b=1\land a<-1 \\ \frac{a^2 b+2 a b+a-2}{2 (a b-1)} & b=0\land a>0\land \frac{1}{a}-b>1 \\ \frac{b-\log (b+1)}{a b^2} & a>0\land \frac{1}{a}-b\leq 1\land b>-1 \\ \frac{a b^2+a b-a b \log (b+1)-b+\log (b+1)}{a b^2 (a b-1)} & a>0\land \frac{1}{a}-b\leq 1\land b\leq -1 \\ \frac{\log (1-a b)}{a b^2} & a>0\land \frac{1}{a}-b>1\land b\leq -1 \\ \frac{a b^2+a b+\log (1-a b)}{a b^2} & \left(-1<b<0\land a>0\land \frac{1}{a}-b>1\right) \\ & \lor \left(b>0\land a>0\land \frac{1}{a}-b>1\right) \\ \frac{b-\log ((-b-1) (a b-1))}{a b^2} & a<0\land \frac{1}{a}-b>1 \end{array}$$

Numeric integration and simulation over the ranges $-2 \le a \le 2$ and $-5 \le b \le 5$ confirm these results.


The modified question asks to replace $d_i^2$ by $d_i^\alpha$ and assumes $a$ and $b$ both positive. Upon making a substitution $r_i = d_i^\alpha$, the region of integration remains the same and integrand becomes $(2/\alpha)^2(r_1 r_2)^{2/\alpha-1}$ instead of $1$. Writing $\theta = \alpha/2$, we obtain

$$\frac{1}{2} a^{-1/\theta } \, _2F_1\left(\frac{1}{\theta },\frac{2}{\theta };\frac{\theta +2}{\theta };-b\right)$$

when $(a>0\land a<1\land a b+a\geq 1)$ or $a\geq 1$ and otherwise the result is

$$-a^{\frac{1}{\theta }} \left(\frac{1}{1-a b}\right)^{\frac{1}{\theta }}+\frac{1}{2} a^{\frac{1}{\theta }} (1-a b)^{-2/\theta } \, _2F_1\left(\frac{1}{\theta },\frac{2}{\theta };\frac{\theta +2}{\theta };1+\frac{1}{a b-1}\right)+1.$$

Here, $_2F_1$ is the hypergeometric function. The original case of $\alpha=2$ corresponds to $\theta=1$ and then these formulae reduce to the fourth and seventh of the eight previous cases. I have checked this result with a simulation, letting $\theta$ range from $1$ through $3$ and covering substantial ranges of $a$ and $b$.

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    $\begingroup$ Maybe you can clarify the wording of the sentence "Provided $B$..." since I don't believe the first condition implies the second if $A>0$ and $B < 0$, for example. (I'm likely just misunderstanding what you meant.) $\endgroup$
    – cardinal
    May 17, 2012 at 16:34
  • $\begingroup$ You're right, @cardinal: of course the inequality gets reversed for negative values of $AB$. That will cause us to replace the answer by its complement, which is simple enough to do. I will refrain from making any modifications, though, until more errors are pointed out by kind reviewers like you :-). $\endgroup$
    – whuber
    May 17, 2012 at 16:36
  • $\begingroup$ The other thing that doesn't makes sense to me is that the final answer is invariant to $B$. Just take $B \to \infty$. (Typo?) :) $\endgroup$
    – cardinal
    May 17, 2012 at 16:37
  • $\begingroup$ Also, any answer you get should be completely invariant to the norm chosen on $\mathbb R^2$, i.e., we can replace the circle by the disc induced by any norm without affecting the probability. This should provide another sanity check. $\endgroup$
    – cardinal
    May 17, 2012 at 16:39
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    $\begingroup$ Thanks whuber. Now I see why the problem looks so unclear to others. Actually, I am looking for the second case you described: "the process itself". $A$ and $B$ are both positive. $\endgroup$
    – Helium
    May 17, 2012 at 16:49

This problem can be solved by decomposing into parts and using the properties of a Poisson process.

It helps to recall how to generate a Poisson point process of intensity $\rho$ on a bounded subset $\newcommand{\A}{\mathcal A}$ of $\mathbb R^2$. We first generate a Poisson random variable $N$ with rate $\rho |\mathcal A|$ where $|\cdot|$ denotes Lebesgue measure, and then we sprinkle these $N$ points uniformly at random inside of $\A$.

This immediately tells us that as long as $N \geq 2$, if we choose two points (without replacement) at random, then these two points will be independent and uniformly distributed on $\A$. When $N < 2$, we have to do something and one natural choice is to define the desired probability as zero. Note that this happens with probability $$\renewcommand{\Pr}{\mathbb P}\Pr(N < 2) = (1+\rho|\A|) e^{-\rho|\A|} \>. $$ This is the only part of the problem that depends on the Poisson process intensity.

Probability conditional on $\{N \geq 2\}$

We are interested in the probability $$ p(A,B,r) := \Pr\Big( d_1^2 \leq \frac{d_2^2}{A(1+B d_2^2)}\Big) \>, $$ where $A > 0$, $B > 0$ and $\A = \{x : \|x\|_2 \leq r\}$. Here $d_1$ and $d_2$ are the radii of two of our uniformly distributed points that fall in $\A$.

Note that for a point randomly distributed in the disc of radius $r$, the distribution of the distance from the origin is $\Pr(D \leq d) = (d/r)^2$, from which we can see that $D^2$ has the same distribution as $r^2 U$ where $U \sim \mathcal U(0,1)$. From this, we can restate the probability of interest as $$ p(A,B,r) = \Pr\Big( U_1 \leq \frac{U_2}{A(1+B r^2 U_2)}\Big) = \iint 1_{(0 < x < 1)} 1_{(0 < y < 1)} 1_{(0 < y < x/(A+ABr^2 x))} \,\mathrm dy \, \mathrm dx \>. $$

This integral splits into two cases. To calculate it, we need the general integral $$ \int_0^t \frac{x}{a+bx} \,\mathrm d x = \frac{1}{b} (t - \frac{a}{b} \log(1+bt/a)) \>. $$

Case 1: $A(1+B r^2) \geq 1$.

Here we see that $u \leq A(1+B r^2 u)$ for $u \in [0,1]$, so $$ p(A,B,r) = \frac{1}{ABr^2}\Big(1 - \frac{\log(1+B r^2)}{B r^2}\Big) \>. $$

Case 2: $A(1+B r^2) < 1$.

Here the integral for $p(A,B,r)$ splits into two pieces since $u \geq A(1+B r^2 u)$ on $[A/(1-ABr^2),1]$. Hence we integrate up to $t = A/(1-A B r^2)$ using the general integral and then tack on an addition area of $1-A/(1-ABr^2)$ for the second piece. So, we get $$ \begin{align} p(A,B,r) &= \frac{1}{B r^2} \Big(\frac{1}{1-A B r^2} + \frac{\log(1-AB r^2)}{A B r^2}\Big) + 1 - \frac{A}{1 - A B r^2} \\ &= 1 + \frac{1}{B r^2} \Big(1 + \frac{\log(1-AB r^2)}{A B r^2}\Big) \>. \end{align} $$

Oftentimes a picture helps; here is one that shows an example of the integration region for each case. Note that $U_1$ is on the $y$-axis and $U_2$ on the $x$-axis.

Examples of each case

The final probability of interest is then, of course, $(1 - (1+\rho\pi r^2) e^{-\rho\pi r^2} ) p(A,B,r)$.

An easy generalization

We can easily generalize the result to use a different shaped ball. In fact, for any arbitrary norm on $\mathbb R^2$, the conditional probability $p(A,B,r)$ is invariant as long as we use the ball induced by the norm instead of the circle!

This is because no matter what norm we choose, the squared radius is uniformly distributed. To see why, let $\delta(\cdot)$ be a norm on $\mathbb R^2$ and $B_\delta(r) = \{x: \delta(x) \leq r\}$ the ball of radius $r$ under the norm $\delta$. Note that $rx \in B_\delta(r)$ if and only if $x \in B_\delta(1)$. The scaling up or down of the unit ball is a linear transformation and by a standard fact about Lebesgue measure, the measure of a linear transformation $T$ of $B_\delta(1)$ is $$ |B_\delta(r)| = |T B_\delta(1)| = |\det(T)| |B_{\delta(1)}| = r^2 |B_\delta(1)| \>, $$ since $T(x) = r x = (r x_1, r x_2)$ in this case.

This shows that if $D = \delta(X)$ for $X$ uniformly distributed in $B_\delta(r)$, then $$ \Pr(D \leq d) = \frac{|B_\delta(d)|}{|B_\delta(r)|} = (d/r)^2 \>. $$ The eagle-eyed reader will note that we've only used the homogeneity of the norm here, and so a similar result will hold in general for uniform distributions on classes of sets closed under a homogeneous transformation.

Here is a picture with two points selected. The norms shown are the Euclidean norm, $\ell_1$ norm, $\sup$ norm, and the $\ell^p$ norm for $p = 5$. Each unit ball is outlined in black, and the largest ball within which the two randomly selected points lie is drawn in the corresponding color.

The conditional probability $p(A,B,r)$ is the same for each picture when the distance is measured using the corresponding norm.

Four norms

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    $\begingroup$ +1 I was using similar pictures to solve this but in mine, $u_1$ is the horizontal axis, not the vertical :-). It helps to standardize the expression for the domain of integration; for positive $A$ and $B$, it's $(x-1/(AB))(y+1/B) \lt -1/(A^2 B)$, immediately exhibiting the center at $(1/(AB), -1/B)$ and showing the scaling with $A$ and $B$. $\endgroup$
    – whuber
    May 18, 2012 at 14:06
  • $\begingroup$ @whuber: (+1) I was on the fence about whether to do that or not. The reason I went with the figures I did was to avoid having to introduce the inverse mapping, which I thought would look messier. Flipping the axes from what seemed most natural allowed me to avoid that. :) $\endgroup$
    – cardinal
    May 18, 2012 at 16:55

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