A previous investigation has shown that the number of imperfections in a fine copper wire averages $28$ imperfections per centimetre of length.

  1. What is the probability that there is a distance of $0.025$ centimetres between two imperfections in the wire?

  2. What is the probability that there is a maximum distance of $0.043$ centimetres between two wire imperfections?

  3. An experiment is defined as taking a measurement of the distance between two wire imperfections. If $100000$ experiments are simulated, the average of the distance between two imperfections is a value close to:

In this Poisson exercise, is $k$ the $0.025$? I don't know how to do it. The $\lambda$ would be $28$, wouldn't it?

  • $\begingroup$ What is the context of this exercise? Is it in a chapter on Poisson processes or just on the Poisson distribution? $\endgroup$ Nov 18, 2022 at 21:46
  • $\begingroup$ Also, is this the entire exercise as written? The first two questions have quite boring answers if there are no other assumptions $\endgroup$ Nov 18, 2022 at 21:52
  • $\begingroup$ @AsbjørnHolk Just on the Poisson distribution. This is the entire exercise. $\endgroup$
    – gijowo
    Nov 18, 2022 at 22:07
  • $\begingroup$ Is it from a book? $\endgroup$ Nov 18, 2022 at 22:16
  • $\begingroup$ @AsbjørnHolk No, it's from a list of exercises about an introductory course on statistics and probability. $\endgroup$
    – gijowo
    Nov 18, 2022 at 22:21

1 Answer 1


I will say that this solution seems a bit advanced for an introductory course, but I cannot see any other reasonable way to interpret these questions. Perhaps someone else can provide an answer for something, I am not seeing. In either case it can serve as a very short crash course introduction to Poisson processes.

We know that the number of imperfections on a $1$cm wire, call it $\mathsf{N}_1$, is Poisson with parameter $28$, i.e. $\mathsf{N}_1\sim \mathrm{Poisson}(\lambda)$ with $\lambda=28$. Let's say that you want to know the number of imperfection on a wire that is $t$ cm long for some $t\in[0, \infty)$, call it $\mathsf{N}_t$. Intuitively, you would expect this to be Poisson as well, but with a scaled parameter, i.e. $\mathsf{N}_t\sim \mathrm{Poisson}(t\cdot\lambda)$. Now you're not really interested in the number of faults, but rather the distances between them. Let's call these $\mathsf{X}_1, \mathsf{X}_2, \ldots$ and assume that they are independent. Now note the important relationship that $\mathsf{N}_t=k$ if and only if $\mathsf{X}_1+\mathsf{X}_2+\ldots+\mathsf{X}_k\le t$, but $\mathsf{X}_1+\mathsf{X}_2+\ldots+\mathsf{X}_k+\mathsf{X}_{k+1}>t$ (making a drawing might be helpful). Let's try to find the distribution of these $\mathsf{X}_i$'s by induction in $i$. First, note that $\mathsf{X}_1\le t$ if and only if $\mathsf{N}_t\ge1$, i.e. if and only if there is at least one fault in the first $t$ centimeters. As such, we have that $$ \mathbb{P}(\mathsf{X}_1\le t) =\mathbb{P}(\mathsf{N}_t\ge 1) =1- \mathbb{P}(\mathsf{N}_t=0) =1- \mathrm{e}^{-t\cdot\lambda}. $$ This is the CDF of the $\mathrm{Exponential}(\lambda)$-distribution! This means that $\mathsf{X}_1\sim \mathrm{Exponential}$, and I claim this to be true for all the $\mathsf{X}_i$'s. As such, assume that it is true for $\mathsf{X}_1,\ldots,\mathsf{X}_k$, and let's prove it for $\mathsf{X}_{k+1}$. For notation set $\mathsf{S}_k=\sum_{i=1}^{k}\mathsf{X}_i$. Using the important relationship from before, we find that $$ \mathbb{P}(\mathsf{N}_t=k) =\mathbb{P}(\mathsf{S}_k\le t, \mathsf{S}_k+\mathsf{X}_{k+1}>t) =\mathbb{P}(\mathsf{S}_k\le t, \mathsf{X}_{k+1}>t-\mathsf{S}_k). $$ Using the law of total probability and the independence of $\mathsf{X}_{k+1}$ and $\mathsf{S}_k$, we may write the right hand side as $$ \mathbb{P}(\mathsf{S}_k\le t, \mathsf{X}_{k+1}>t-\mathsf{S}_k) =\int_{0}^{t}\mathbb{P}(\mathsf{X}_{k+1}>t-s)f_{\mathsf{S}_k}(s)\mathrm{d}s, $$ and by our induction hypothesis, $\mathsf{S}_k\sim\Gamma(k, \lambda)$, i.e. $f_{\mathsf{S}_k}(s)=\frac{\lambda^k}{(k-1)!}s^{k-1}\mathrm{e}^{-\lambda s}$. Plugging in also our ansatz that $\mathsf{X}_{k+1}\sim \mathrm{Exponential}(\lambda)$, and hence $\mathbb{P}(\mathsf{X}_{k+1}>t-s)=\mathrm{e}^{-\lambda(t-s)}$, this becomes \begin{align*} \int_{0}^{t}\mathbb{P}(\mathsf{X}_{k+1}>t-s)f_{\mathsf{S}_k}(s)\mathrm{d}s &=\frac{\lambda^k}{(k-1)!}\int_{0}^{t}\mathrm{e}^{-\lambda(t-s)}s^{k-1}\mathrm{e}^{-\lambda s}\mathrm{d}s \\ &=\frac{\lambda^k \mathrm{e}^{-\lambda t}}{(k-1)!}\int_{0}^{t}s^{k-1}\mathrm{d}s \\ &=\frac{(t\lambda)^k \mathrm{e}^{-\lambda t}}{k!}, \end{align*} but this is exactly $\mathbb{P}(\mathsf{N}_t=k)$ as desired, and hence $\mathsf{X}_{k+1}\sim \mathrm{Exponential}(\lambda)$ as well.

But this is where your questions confuse me. Since the gaps between faults are continuous (exponential) the probability that they are exactly $0.025$ or $0.043$ are $0$, making the first two questions rather boring. To get a more fulfilling answer, they really should be inequalities (e.g. the probability of a distance of less than $0.025$cm). However for the final question, the average will converge to the mean of the distance, i.e. the mean of $\mathrm{Exponential}(\lambda)$, so $\frac{1}{\lambda}=\frac{1}{28}$.

I hope that this was more helpful than confusing.


Not the answer you're looking for? Browse other questions tagged .