# Proving the expectation of a variable in a stochastic process

## 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!

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}
• 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?
• 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. Apr 10, 2022 at 19:18