Number of imperfections in a fine copper wire [closed]

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?

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

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 $$\frac{1}{\lambda}=\frac{1}{28}$$.