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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?

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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 $$ g_t(s)=\sum_{k\geqslant0}\mathrm e^{-\lambda t}\frac{(\lambda t)^k}{k!}G_X(s)^k=\mathrm e^{-\lambda t(1-G_X(s))}. $$

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  • $\begingroup$ If Z-transforms are really what you are after, you might wish to recall the definition of the notion you have in mind. $\endgroup$ – Did Mar 17 '12 at 13:19

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