Let $\mathbf{\Pi}$ be a homogeneous Poisson point process on the real number line with intensity $\lambda$. Let $r^{+}$ denote the distance from the origin to the closest point of $\mathbf{\Pi}$ on the +'ve real axis and let $r^{-}$ denote the distance from the origin to the closest point of $\mathbf{\Pi}$ on the -'ve real axis.

I want to find the distribution of these functions of the randoms variables associated with the process, namely:

$$\mathbf{P}(r^{+} \leq x), $$

$$\mathbf{P}(r^{-} \leq x),$$

$$\mathbf{P}(\textrm{min}(r^{-},r^{+}) \leq x).$$

I know that the probability of points generated by a Poisson point process defined on the real line with intensity $\lambda \gt 0$ in a region $[a,b]$ where $a \leq b$ is given by:

$$\mathbf{P} \{N(a,b] = n\} = \frac{[\lambda(b-a)]^n}{n!}e^{-\lambda(b-a)}.$$

My solution to find $\mathbf{P}(r^{+} \leq x)$ is to consider a region $[0,r^+]$ where no points exist, i.e $N=0,$ which gives:

$$\mathbf{P}(r^{+} \leq x) =1- e^{-\lambda r^{+}}$$


$$\mathbf{P}(r^{-} \leq x) = 1- e^{-\lambda r^{-}} $$

$$\mathbf{P}(\textrm{min}(r^{-},r^{+}) \leq x) =(1-\lambda e^{-2\lambda x})^2. $$

Could anyone comment on the validity of my method to tackle this?

  • $\begingroup$ If $r^{+} = x \geq 0$ and $r^{-} = x \leq 0$ don't I recover the same result? Where do the intensities $\lambda$ multiplying the exponentials go? Thank you for your input. $\endgroup$ – keeran_q789 Feb 7 at 12:44
  • $\begingroup$ Okay I see now. Would I be correct in saying $\mathbf{P}(r^{+} \leq x) = 1 - \mathbf{P}(\Pi [0,x]) = 1-e^{-\lambda x}$ ? $\endgroup$ – keeran_q789 Feb 7 at 13:07
  • $\begingroup$ Could you please tell me why my formula $\mathbf{P}(r^{+} \leq x) = 1 - \mathbf{P}(\Pi[0,x]) = 1-e^{-\lambda x}$ is "syntactically absurd" and how to improve it rather than berate me when I am a student trying to learn? $\endgroup$ – keeran_q789 Feb 7 at 14:19
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    $\begingroup$ @Did If you are not willing to be civil in the comments, please refrain from engaging. The OP may have typos on their notation, but this does not entitle you to engage with them in this way. Any other comments of yours that are not civil will be deleted. $\endgroup$ – Pedro Tamaroff Feb 7 at 15:04
  • $\begingroup$ @PedroTamaroff Uncivil comment (example): "The answer is already in my last comment". $\endgroup$ – Did Feb 7 at 18:40

There is some confusion in your post; $r^+$ is a random variable, and $x$ is a fixed number, so the formula for $P(r^+\le x)$ cannot involve $r^+$. It will be some deterministic function of $x$.

In particular, $r^+\le x$ if and only if there is some arrival in the interval $[0,x]$, so $$ P(r^+\le x)=1-e^{-\lambda x} $$ The corresponding probability for $r^-$ is the same. Finally, since $r^+$ and $r^-$ are independent, \begin{align} P(\min(r^+,r^-)\le x) &=P(r^+\le x\cup r^-\le x) \\&=P(r^+\le x)+P(r^-\le x)-P(r^+\le x\cap r^-\le x) \\&=P(r^+\le x)+P(r^-\le x)-P(r^+\le x)P(r^-\le x) \\&=(1-e^{-\lambda x})+(1-e^{-\lambda x})-(1-e^{-\lambda x})^2 \\&=1-e^{-2\lambda x} \end{align} Another way to arrive at this; we see that $\min(r^+,r^-)>x$ if and only if there are no arrivals in $[-x,x]$.


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