I have the next question:
A communication system consists of two parts, component and communication line. The component stores all signals that are coming from a particular source, and the line of communication transmits the signals to the receiver. The system works in one-second cycles. The number of signals coming from the source to the component every second is a Poisson process with a parameter $\lambda $. The maximum number of signals that can be stored in a component is an integer constant b, and at the end of every second, the system sends the signals stored in the component to the receiver through the communication line. The maximum amount of signals that can be sent per second through the communication line is an integer constant c ($0<c<b$). If more the b signals come so the remainder is lost (and not stored in the component), and c is the maximum number of signals that can be sent through the line of communication. Assume that in the initial state the system is empty.
a. Find the probability function of:
1. The number of signals stored at the end of the first cycle before sending.
2. The number of signals stored at the end of the first cycle after sending.
b.What is the probability that in the first cycle you will lose half of the signals that reached the system? An answer can be given as a series.
My solution for a:
The distribution of the number of signals is Poissonic as given and therefore in one cycle k=1 and we will get the number of signals that will come is: $\lambda e^{-\lambda }$. I marked this number as i.
For the number of signals in storage before sending the probability function is:
$B=\left\{\begin{matrix}
\lambda e^{-\lambda }-b & b<i\\
\lambda e^{-\lambda }& b>i
\end{matrix}\right.$
For the number of signals in storage after sending:
$A=\left\{\begin{matrix}
B-c & c<B\\
0& c>B
\end{matrix}\right.$
But I do not know how to use the result of question a to solve the question b. Can anyone help me, please? Thanks for the help in advance.
a.1
let $X$ be the number of signals prior to the sending. Let $Y$ be number of signals came ($X$ can't go beyond limit $b$, while $Y$ can).
Then,
$$\begin{equation*} p_X(m) = \begin{cases} e^{-\lambda} \frac{\lambda^m}{m!} & m < b \\ p\left(Y \ge b\right) = 1 - p\left(Y < b\right) = 1 - \sum_{k=0}^{b-1} \textbf{<can you finish here>} & m = b \end{cases} \end{equation*}$$
a.2
Let $Z$ be the number of signals after sending. There are cases: if $X \le c$ than $Z=0$, otherwise $Z = X - c$, however need carefully look at one more corner case when $X = b$:
$$\begin{equation*} p_Z(m) = \begin{cases} e^{-\lambda} \sum_{k=0}^c \frac{\lambda^k}{k!} & m = 0\\ p_X(m+c) = \textbf{<can you finish here>} & 0 < m < b - c\\ p_X(b) = \textbf{<can you finish here>} & m = b - c \end{cases} \end{equation*}$$
b The system loses only signals coming out of the limit $b$. Note that signals from $c + 1$ to $b$ are not lost, and will be eventually sent the following seconds. Thus, the event that in the first cycle you will lose half of the signals that reached the system is the same event as receiving at least $2b$ signals and probability of that is $P_x(m > 2b)$, can you complete from here?