Question

A viral linear DNA molecule of length, say, 1 is often known to contain a certain "marked position," with the exact location of this mark being unknown. One approach to locating the marked position is to cut the molecule by agents that break it at points chosen according to a Poisson process with rate \(\lambda .\) It is then possible to determine the fragment that contains the marked position. For instance, letting \(m\) denote the location on the line of the marked position, then if \(L_{1}\) denotes the last Poisson event time before \(m\) (or 0 if there are no Poisson events in \([0, m])\), and \(R_{1}\) denotes the first Poisson event time after \(m\) (or 1 if there are no Poisson events in \([m, 1])\), then it would be learned that the marked position lies between \(L_{1}\) and \(R_{1} .\) Find (a) \(P\left[L_{1}=0\right\\}\), (b) \(P\left(L_{1}x\right\\}, m

Short answer

\(E[R-L]\) can be found using the expected values of \(L\) and \(R\). Since \(L\) and \(R\) follow exponential distributions with rate \(\lambda\), we can use the expectations for the minimum and maximum of exponential distributions to find that: $$E[L] = \frac{-1}{(n - 1)\lambda}, \quad E[R] = \frac{1}{(n + 1)\lambda}$$ Then, we can find the expected value of \(R-L\): $$E[R-L] = E[R] - E[L] = \frac{1}{(n + 1)\lambda} - \frac{-1}{(n - 1)\lambda}$$ Simplifying this expression, we get: $$E[R-L] = \frac{2}{n\lambda } + O\left(\frac{1}{n^2\lambda}\right)$$ As \(n\) goes to infinity, the term \(O\left(\frac{1}{n^2\lambda}\right)\) has a negligible effect, which implies that: $$E[R-L] \sim \frac{2}{n\lambda}$$

Step-by-step solution

Part (a) - Finding the probability of \(L_1 = 0\)#

Since \(L_1\) is the last Poisson event time before \(m\), and \(L_1 = 0\) implies that there were no Poisson event times between \(0\) and \(m\). Thus, we need to find the probability that no event occurred in the interval \([0, m]\). Since the occurrences follow a Poisson process with rate \(\lambda\), the probability of no events is given by the following formula: $$ P\left\{L_1 = 0\right\} = e^{-\lambda m} $$

Part (b) - Finding the probability of \(L_1 < x, 0 < x < m\)#

We want to find the probability that the last Poisson event time before \(m\) is less than \(x\), where \(0<x<m\). This is the probability that there is at least one event in the interval \((x, m)\) and no event in the interval \((0, x)\). Since intervals of a Poisson process are independent, the probability of this event is given by: $$ P\left(L_1 < x\right) = e^{-\lambda (m-x)} \times (1-e^{-\lambda x}) $$

Part (c) - Finding the probability of \(R_1 = 1\)#

Since \(R_1\) is the first Poisson event time after \(m\), and \(R_1 = 1\) implies that there were no Poisson event times between \(m\) and \(1\). We need to find the probability that no event occurred in the interval \((m, 1]\). This probability is given by: $$ P\left\{R_1 = 1\right\} = e^{-\lambda (1-m)} $$

Part (d) - Finding the probability of \(R_1 > x, m < x < 1\)#

We want to find the probability that the first Poisson event time after \(m\) is greater than \(x\), where \(m x\right) = e^{-\lambda (x-m)} \times (1-e^{-\lambda (1-x)}) $$

Part (e) - Finding \(E[R-L]\) and showing that \(E[R-L] \sim \frac{2}{n\lambda}\)#

Using the previous results, we can restate the definitions of \(R\) and \(L\): $$ L = \max L_i, \quad R = \min R_i $$ The distribution of \(L\) and \(R\) is given by \(L \sim Exp(\lambda)\) and \(R \sim Exp(\lambda)\). The expectation for an exponential distribution with rate \(\lambda\) is given by \(\frac{1}{\lambda}\). Since the expectation of the minimum and maximum of an exponential distribution with rate \(\lambda\) is equal to \(\frac{1}{(n+1)\lambda}\) and \(\frac{-1}{(n-1)\lambda}\) respectively, we get: $$ E[L] = \frac{-1}{(n - 1)\lambda}, \quad E[R] = \frac{1}{(n + 1)\lambda} $$ Now, we find the expected value of \(R-L\): $$ E[R-L] = E[R] - E[L] = \frac{1}{(n + 1)\lambda} - \frac{-1}{(n - 1)\lambda} $$ Simplifying the expression, we get: $$ E[R-L] = \frac{2}{n\lambda } + O\left(\frac{1}{n^2\lambda}\right) $$ The term \(O\left(\frac{1}{n^2\lambda}\right)\) represents a more insignificant term as \(n\) increases. As it goes to infinity, the term's effect becomes negligible, implying that: $$ E[R-L] \sim \frac{2}{n\lambda} $$