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This is a problem from Ross' Introduction to Probability Models (11th) I got stuck on studying for my finals. I would really appreciate any help to solve question b, which I really don't know how to approach.

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) $P(L_1=0)$,

(b) $P(L_1<x),\quad 0<x<m,$

(c) $P(R_1=1)$,

(d) $P(R_1>x),\quad m<x<1,$

My Attempt at the problem:

Question (a):

$P(L_1=0)$ is just the probability of the process not jumping on $[0,m]$.

So we just have to calculate: $P(N_m -N_0=0)=P(S_1>m)=e^{-\lambda m}$

where $S_1$ denotes the arrival time of the first jump of $\{N_t\}$

Question (c): We focus on the segment $[m,1]$:

$P(R_1=1)$ is just the probability of the process not jumping on $[m,1]$. That is,

$P(N_1-N_m=0)$, which is independent of the history of the process $\{N_t\}$ up to m.

So, $N_1-N_m\sim Poi(\lambda(1-m))$ and $P(R_1=1)=P(N_1-N_m=0)=e^{-\lambda(1-m)}$

I think we can also use the distribution of the arrival time of the first jump of the resetted process at time m, $S_1'$.

We know that $S_1'\sim exp(\lambda)$, so we have that,

$P(R_1=1)=P(S_1'>1-m)=e^{-\lambda(1-m)}$

Question (d):

Using the distribution of $S_1'$.

We need to calculate $ P(R_1>x)=P(S_1'>x-m)=e^{-\lambda(x-m)}, m<x<1$.

Question (b):

I have absolutely no clue. As $L_1 :=max\, S_i : S_i<m$, it seems I should approach this by computing the probability of the maximun of the arrival times that take place before m. However, I don't know how many jumps there are before time m.

The answer is just $P(L_1<x)=e^{-\lambda(m-x)}$, which seems to indicate that there must be some not-that-complicated way to reason in order to get the answer, so I think I may be over complicating this.

Can anyone please, please help me with this???

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1 Answer 1

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$P(L_1<x)$ is the probability of the process not jumping on $[x,m]$, which is $e^{-\lambda(m-x)}$.

If the process jumps at any time $x<t<m$ then either $L_1= t$ or the process jumps again at some time $t<t'<m$ and $L_1>t$.

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  • $\begingroup$ Thank you so much for your answer, Angela. I am not sure I totally understand it though. Are you stating that: $P(max_{i} S_i < x) = P(N(m)-N(x)=0)?$ where the random variable $N(m)-N(x)$ is a Poisson with parameter $\lambda(m-x)$, independent of $N(x)$? I am sorry if I don't get it, but the fact that we are dealing with the last of the jumps really confuses me. I am just used to deal with the distribution of the first jump / first arrival time. $\endgroup$
    – Javi
    Commented Apr 13, 2020 at 10:49
  • $\begingroup$ $L_i$ is the maximum of all $S_i<m$. If there exist any $x\le S_i<m$ then $L_i\ge x$. If there are no $x\le S_i<m$ then the maximum of all $S_i<m$ is less than $x$. $\endgroup$
    – Angela Pretorius
    Commented Apr 13, 2020 at 10:55
  • $\begingroup$ Thank you for your reply. I think I understand. Just one thing, do you mean $L_{1}$ instead of $L_i$, right? $\endgroup$
    – Javi
    Commented Apr 13, 2020 at 11:13
  • $\begingroup$ Yes. I meant $L_1$. $\endgroup$
    – Angela Pretorius
    Commented Apr 13, 2020 at 12:41
  • $\begingroup$ Thank you so much for your help. $\endgroup$
    – Javi
    Commented Apr 13, 2020 at 13:19

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