Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The arrival of messages to a communications channel is modeled as a Poisson process, with rate $\lambda$ messages / unit time. Let $\{N(t), t \geq 0\}$ denote that process. Each message contains a random number of bytes; the probability mass function of the number of bytes $X_i$ in the $i$th message is given by $G_X(z) = \sum_{k=1}^\infty \alpha_k z^k$.

Let $Y(t)$ denote the number of bytes which arrived by time $t$.

How should we find the $z$-transform for $Y(t)$ in terms of the known quantities?

share|improve this question

1 Answer 1

up vote 0 down vote accepted

Thus, at every time $t\geqslant0$, $Y(t)=\sum\limits_{n=1}^{N(t)}X_n$. The invocation of the Z-transform of the continuous-time process $(Y(t))_{t\geqslant0}$ is rather mysterious here since Z-transforms apply to discrete-time processes.

One can however compute the generating function $g_t$ of each $Y(t)$, that is, the function defined, for (at least) every $|s|\leqslant1$, by $g_t(s)=\mathrm E\left(s^{Y(t)}\right)$. First note that, for every $k\geqslant0$, $$ \mathrm E\left(s^{Y(t)}\mid N(t)=k\right)=G_X(s)^k, $$ and that $N(t)$ is Poisson with parameter $\lambda t$, hence $$ g_t(s)=\sum_{k\geqslant0}\mathrm e^{-\lambda t}\frac{(\lambda t)^k}{k!}G_X(s)^k=\mathrm e^{-\lambda t(1-G_X(s))}. $$

share|improve this answer
    
If Z-transforms are really what you are after, you might wish to recall the definition of the notion you have in mind. –  Did Mar 17 '12 at 13:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.