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Problem

Information packets arrive at a server with a poisson process having rate $\lambda = 2$ per hour.

The server processing time for a packet follows the distribution : $f(x) = 1, 0\leq x\leq1$

The status of the server is busy if it is processing a packet, otherwise it is waiting. If a packet arrives while the server is busy, that packet is lost.

Let $t_B$ denote the length of one busy period, $t_I$ denote the length of one idle period. Find the distrubutions of $t_B$ & $t_I$.

Let $N$ denote the total number of lost packets, $T_B$ denote the total busy time of the server up until time $T=10$ hours.

Show that : $$ \begin{align} E[N] = \lambda E[T_B] \end{align} $$

My attempt :

$t_B$ is simply the time taken for the server to process a packet, so its distribution is also $f(x) = 1, 0 \leq x \leq 1$

$t_I$ is the inter-arrival time between two packets which follows an $Exp(\lambda = 2)$ distribution, using the relation between poisson & exponential rv's.

In trying to prove the equation, this is what I have so far :

N = total no. of lost packets = no. of packets arriving during $T_B$(while the server is busy)

$$ \begin{align} E[N] &= E[\textrm{No. of arrivals during }T_B] \\ &=E[\textrm{No. of arrivals per unit time}]\cdot E[T_B] \\ &=\lambda E[T_B]~\textrm{(Shown)} \end{align} $$

Am I going about this the right way? If so is there a proper way to express this in math notation? Thank you!

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1 Answer 1

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Because the packets arrive as a Poisson process, with a given $T_{B}$, the number of missing packets $N$ also follows a Poisson distribution, $$ p(N \vert T_B) = {\rm Poisson}(\lambda T_B)(N). $$ The expected value of $N$, ${\rm E}[N]$ is then given by $$ \begin{aligned} {\rm E}[N] &= \sum_{N=0}^{\infty} \ N p(N)\\ &= \sum_{N=0}^{\infty} N \int_0^\infty dT_B \ p(N, T_B)\\ &= \sum_{N=0}^{\infty} N \int_0^\infty dT_B \ p(N \vert T_B) p(T_B)\\ &= \int_0^\infty dT_B \ p(T_B) \sum_{N=0}^{\infty} N p(N \vert T_B)\\ &= \int_0^\infty dT_B \ p(T_B) {\rm E}[N \vert T_B]\\ &= \int_0^\infty dT_B \ p(T_B) \lambda T_B\\ &= \lambda \int_0^\infty dT_B \ T_B p(T_B)\\ &= \lambda {\rm E}[T_B]. \end{aligned} $$

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  • $\begingroup$ Thank you for your answer! I think I understand. Could I check why $E(N|T_B) = \lambda T_B$? Since $p(N|T_B) = Poisson(\lambda T_B)(N)$ wouldn't $E(N|T_B) = \lambda T_B N$? Have I misunderstood something? $\endgroup$
    – C C
    Commented Apr 9, 2022 at 15:01
  • $\begingroup$ The notion of $E(N\vert T_B)$ is the expected value of $N$ given $T_B$. Because the distribution $N\vert T_B \sim {\rm Poisson}(\lambda T_B)$, the expected value is simply $\lambda T_B$ as it is the property of a Poisson distribution. $\endgroup$
    – Peter Pang
    Commented Apr 10, 2022 at 19:18
  • $\begingroup$ I understand. Thank you so much! :) $\endgroup$
    – C C
    Commented Apr 15, 2022 at 7:07

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