6

(Again this is based on pp240 - 242 of the 1966 edition of Cox and Miller's "The Theory of Stochastic Processes").

So we have, for a queue in equilibrium/stationary a probability density function for the delay (in virtual waiting time):

$$p_0 + \int_0^\infty p(x)dx = 1$$

Where $x$ is the time taken for arriving customer to begin service and $p_0$ the probability that the wait will be zero.

Now, the authors then state:

$$p_0 + \int_0^\infty p(x)dx = p_0 + p^*(0) =1$$

Where the Laplace transform $\mathcal{L}\{p(x)\}$ = $p^*(s)$.

That all seems fine to me, but they go on to say:

$$w^*(s)=p_0+p^*(s) =\frac{...}{...}$$

(The rightmost term is not important here). They state "we denote the m.g.f. of the equilibrium process by $w^*(s)$".

But isn't the mgf the (double sided) Laplace transform evaluated at $-s$? I don't suppose the double sized aspect matters much here: but does the $-s$ stipulation matter? (Perhaps not?)

And how does $p_0$ fit in here? It doesn't seem to have been subject to any transform process, so why is it included in the mgf?

adrianmcmenamin
  • 418

1 Answers1

9

The moment-generating function of a random variable X is $$w_{X}(s)=\mathbb{E} \!\left[e^{sX}\right]= \int_0^{\infty}p(x) e^{sx}dx$$ for $s\in \mathbb{R}$, if there is a probability density function $p$ on the nonnegative real line. The Laplace transform of a distribution with probability density function $p$ on the nonnegative real line is $${\mathcal{L}}\{p\}(s)=\mathbb{E}\!\left[e^{-sX}\right]=\int_0^{\infty}p(x) e^{-sx}dx.$$ You're almost right, $w_{X}(s)={\mathcal{L}}\{p\}(-s)$. However, in your case there is a Dirac delta with weight $p_0$ at zero (for immediate service), and the moment-generating function $w^*(s)$ is $$w^*(s)=\mathbb{E} \!\left[e^{sX}\right]= p_0 e^{s\times 0} + \int_0^{\infty}p(x) e^{sx}dx=p_0+{\mathcal{L}}\{p\}(-s).$$ The probability density function $p$ is actually that of a sub-probability measure.

Bence Mélykúti
  • 495