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
$$
\begin{aligned}
{\rm E}[N] &= \sum_{N=0}^{\infty} \ N p(N)\\
&= \sum_{N=0}^{\infty} N \int_0^\infty dT_B \  p(N, T_B)\\
&= \sum_{N=0}^{\infty} N \int_0^\infty dT_B \ p(N \vert T_B) p(T_B)\\
&= \int_0^\infty dT_B \ p(T_B) \sum_{N=0}^{\infty} N p(N \vert T_B)\\
&= \int_0^\infty dT_B \ p(T_B) {\rm E}[N \vert T_B]\\
&= \int_0^\infty dT_B \ p(T_B) \lambda T_B\\
&= \lambda \int_0^\infty dT_B \ T_B p(T_B)\\
&= \lambda {\rm E}[T_B].
\end{aligned}
$$
