Calculating the probability of finishing service first












1














I have a job where the service time follows an exponential distribution with mean $1/mu$. At the same time, there are new customers coming in with Poisson rate $lambda$. As soon as they arrive, they start a job whose service time follows an exponential distribution with mean $1/mu$. I would like to know the probability that I finish the job first (i.e., there are no incoming customers, if any, finishes their job before me).



Things I tried:



(1) Conditional on my own service time, then integrate over my service time. This looks very complicated.



(2) Assume $x_k$ is the probability that when $k$ people (other than me) are currently doing their job and I am still the first to finish. The original question asks for $x_0$.



I can write the system of equations according to the exponential race:



$$
x_n=frac{mu}{mu+lambda+nmu}+frac{nmu}{mu+lambda+nmu}x_{n-1}+frac{lambda}{mu+lambda+nmu}x_{n+1}
$$

where the first term stands for the case where I finish the first, the second term stands for the probability that a customer other than me finishes the first, and the third term stands for a new arrival happens first. Then I am not sure how to solve this.



Any hints will be extremely helpful. Thank you!



Edit:
$$
x_n=frac{mu}{mu+lambda+nmu}+frac{lambda}{mu+lambda+nmu}x_{n+1}
$$










share|cite|improve this question




















  • 1




    I think you should delete the middle term, since if one of those other $n$ people win, we are done. So you can write $x_n$ in terms of $x_{n+1}$ only. You also know $lim_{nrightarrowinfty} x_n=0$.
    – Michael
    Nov 25 at 0:40












  • @Michael That's right! Thanks!
    – hdiuwhciu
    Nov 25 at 0:40






  • 1




    The sequence $x_n$ is interesting but looks hard to solve. Another (direct) way to solve is to do the first: If $X$ is your own service time and $N$ is the random number of arrivals during your service, condition on $X=t$ and $N=n$, then use the fact that, given $N=n$ and $X=t$, the $n$ arrival times are independently and uniformly distributed over the interval $[0,t]$.
    – Michael
    Nov 25 at 0:56


















1














I have a job where the service time follows an exponential distribution with mean $1/mu$. At the same time, there are new customers coming in with Poisson rate $lambda$. As soon as they arrive, they start a job whose service time follows an exponential distribution with mean $1/mu$. I would like to know the probability that I finish the job first (i.e., there are no incoming customers, if any, finishes their job before me).



Things I tried:



(1) Conditional on my own service time, then integrate over my service time. This looks very complicated.



(2) Assume $x_k$ is the probability that when $k$ people (other than me) are currently doing their job and I am still the first to finish. The original question asks for $x_0$.



I can write the system of equations according to the exponential race:



$$
x_n=frac{mu}{mu+lambda+nmu}+frac{nmu}{mu+lambda+nmu}x_{n-1}+frac{lambda}{mu+lambda+nmu}x_{n+1}
$$

where the first term stands for the case where I finish the first, the second term stands for the probability that a customer other than me finishes the first, and the third term stands for a new arrival happens first. Then I am not sure how to solve this.



Any hints will be extremely helpful. Thank you!



Edit:
$$
x_n=frac{mu}{mu+lambda+nmu}+frac{lambda}{mu+lambda+nmu}x_{n+1}
$$










share|cite|improve this question




















  • 1




    I think you should delete the middle term, since if one of those other $n$ people win, we are done. So you can write $x_n$ in terms of $x_{n+1}$ only. You also know $lim_{nrightarrowinfty} x_n=0$.
    – Michael
    Nov 25 at 0:40












  • @Michael That's right! Thanks!
    – hdiuwhciu
    Nov 25 at 0:40






  • 1




    The sequence $x_n$ is interesting but looks hard to solve. Another (direct) way to solve is to do the first: If $X$ is your own service time and $N$ is the random number of arrivals during your service, condition on $X=t$ and $N=n$, then use the fact that, given $N=n$ and $X=t$, the $n$ arrival times are independently and uniformly distributed over the interval $[0,t]$.
    – Michael
    Nov 25 at 0:56
















1












1








1







I have a job where the service time follows an exponential distribution with mean $1/mu$. At the same time, there are new customers coming in with Poisson rate $lambda$. As soon as they arrive, they start a job whose service time follows an exponential distribution with mean $1/mu$. I would like to know the probability that I finish the job first (i.e., there are no incoming customers, if any, finishes their job before me).



Things I tried:



(1) Conditional on my own service time, then integrate over my service time. This looks very complicated.



(2) Assume $x_k$ is the probability that when $k$ people (other than me) are currently doing their job and I am still the first to finish. The original question asks for $x_0$.



I can write the system of equations according to the exponential race:



$$
x_n=frac{mu}{mu+lambda+nmu}+frac{nmu}{mu+lambda+nmu}x_{n-1}+frac{lambda}{mu+lambda+nmu}x_{n+1}
$$

where the first term stands for the case where I finish the first, the second term stands for the probability that a customer other than me finishes the first, and the third term stands for a new arrival happens first. Then I am not sure how to solve this.



Any hints will be extremely helpful. Thank you!



Edit:
$$
x_n=frac{mu}{mu+lambda+nmu}+frac{lambda}{mu+lambda+nmu}x_{n+1}
$$










share|cite|improve this question















I have a job where the service time follows an exponential distribution with mean $1/mu$. At the same time, there are new customers coming in with Poisson rate $lambda$. As soon as they arrive, they start a job whose service time follows an exponential distribution with mean $1/mu$. I would like to know the probability that I finish the job first (i.e., there are no incoming customers, if any, finishes their job before me).



Things I tried:



(1) Conditional on my own service time, then integrate over my service time. This looks very complicated.



(2) Assume $x_k$ is the probability that when $k$ people (other than me) are currently doing their job and I am still the first to finish. The original question asks for $x_0$.



I can write the system of equations according to the exponential race:



$$
x_n=frac{mu}{mu+lambda+nmu}+frac{nmu}{mu+lambda+nmu}x_{n-1}+frac{lambda}{mu+lambda+nmu}x_{n+1}
$$

where the first term stands for the case where I finish the first, the second term stands for the probability that a customer other than me finishes the first, and the third term stands for a new arrival happens first. Then I am not sure how to solve this.



Any hints will be extremely helpful. Thank you!



Edit:
$$
x_n=frac{mu}{mu+lambda+nmu}+frac{lambda}{mu+lambda+nmu}x_{n+1}
$$







probability stochastic-processes






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share|cite|improve this question













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edited Nov 25 at 0:44

























asked Nov 25 at 0:17









hdiuwhciu

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  • 1




    I think you should delete the middle term, since if one of those other $n$ people win, we are done. So you can write $x_n$ in terms of $x_{n+1}$ only. You also know $lim_{nrightarrowinfty} x_n=0$.
    – Michael
    Nov 25 at 0:40












  • @Michael That's right! Thanks!
    – hdiuwhciu
    Nov 25 at 0:40






  • 1




    The sequence $x_n$ is interesting but looks hard to solve. Another (direct) way to solve is to do the first: If $X$ is your own service time and $N$ is the random number of arrivals during your service, condition on $X=t$ and $N=n$, then use the fact that, given $N=n$ and $X=t$, the $n$ arrival times are independently and uniformly distributed over the interval $[0,t]$.
    – Michael
    Nov 25 at 0:56
















  • 1




    I think you should delete the middle term, since if one of those other $n$ people win, we are done. So you can write $x_n$ in terms of $x_{n+1}$ only. You also know $lim_{nrightarrowinfty} x_n=0$.
    – Michael
    Nov 25 at 0:40












  • @Michael That's right! Thanks!
    – hdiuwhciu
    Nov 25 at 0:40






  • 1




    The sequence $x_n$ is interesting but looks hard to solve. Another (direct) way to solve is to do the first: If $X$ is your own service time and $N$ is the random number of arrivals during your service, condition on $X=t$ and $N=n$, then use the fact that, given $N=n$ and $X=t$, the $n$ arrival times are independently and uniformly distributed over the interval $[0,t]$.
    – Michael
    Nov 25 at 0:56










1




1




I think you should delete the middle term, since if one of those other $n$ people win, we are done. So you can write $x_n$ in terms of $x_{n+1}$ only. You also know $lim_{nrightarrowinfty} x_n=0$.
– Michael
Nov 25 at 0:40






I think you should delete the middle term, since if one of those other $n$ people win, we are done. So you can write $x_n$ in terms of $x_{n+1}$ only. You also know $lim_{nrightarrowinfty} x_n=0$.
– Michael
Nov 25 at 0:40














@Michael That's right! Thanks!
– hdiuwhciu
Nov 25 at 0:40




@Michael That's right! Thanks!
– hdiuwhciu
Nov 25 at 0:40




1




1




The sequence $x_n$ is interesting but looks hard to solve. Another (direct) way to solve is to do the first: If $X$ is your own service time and $N$ is the random number of arrivals during your service, condition on $X=t$ and $N=n$, then use the fact that, given $N=n$ and $X=t$, the $n$ arrival times are independently and uniformly distributed over the interval $[0,t]$.
– Michael
Nov 25 at 0:56






The sequence $x_n$ is interesting but looks hard to solve. Another (direct) way to solve is to do the first: If $X$ is your own service time and $N$ is the random number of arrivals during your service, condition on $X=t$ and $N=n$, then use the fact that, given $N=n$ and $X=t$, the $n$ arrival times are independently and uniformly distributed over the interval $[0,t]$.
– Michael
Nov 25 at 0:56

















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