Excess Distribution of Renewal Process

Excess Distribution of Renewal Process

- Course notes of Stochastic Process, 2014 Fall

1. Definition

Excess of renewal process is defined as $Y(t) = S_{N(t)+1} -t$ (time until next event)

In the example of average time waiting bus, we drived

$\lim_{t \to \infty} \frac{\int^T_0 Y(u)du}{T} = \frac{E[X_n]}{2} + \frac{car[X_n]}{2E[X_n]}$

Now we are going to derive $Pr(X(t) >x)$ for a random $t$.
Interpretation: You show up "at random". What is the probability that you wait more than x for the next event?

2. Derivation of $Pr(X(t) > x)$

As we want to determine the fraction of time that $Y(t) > x$. 

Let $ I(t) = 1$ if $Y(t) > x$, and let $I(t) = 0$ otherwise.

Interpretation: Fraction of time that $Y(t) > x$ = Fraction of "on" time for $I(t)$

Let $Z_j$ be "on" time during cycle $j$, then $Z_j = max(X_j - x, 0)$

• $Z_j = X_j -x$ if $X_j > x$
• $Z_j = 0$ otherwise

Note: the ON time and OFF time for each cycle are dependent. A longer ON time implies a shorter OFF time.

Then we have

$E[Z_j] = E[max(X_j -x, 0)] = \int^{\infty}_0 Pr(max(X_j- x, 0) > u) du$

                                               = $\int^{\infty}_0 Pr\{X_j - x > u\} du$

                                               = $\int^{\infty}_0 Pr\{X_j > x+ \mu\} d\mu$

                                               = $\int^{\infty}_x Pr\{X_j > \mu\} d\mu$

Since $I(t)$ is an alternating renewal process, fraction of "on" time is

$\frac{E[Z_j]}{E[X_j]} = \frac{1}{E[X_j]} \int^{\infty}_x F^c(u) dy$

This is sometimes called equilibrium distribution,

                                                

3. Example of different distribution of $X(t)$

Example: $X_j$ ~ exp$(\lambda)$

$Pr(Y(t) >x) = \frac{1}{E[X_j} \int^{\infty}_{x} F^c(\mu) d\mu = \frac{1}{\lambda} \int^{\infty}_x e^{-\lambda \mu} d\mu$

                     = $e^{-\lambda x}$

As expected (by memoryless property), excess distribution is an exponential distribution.

Example: Pareto Distribution 

$Pr(X_j > x) = (1+x)^{-\alpha}$

$\alpha$ is also called the tail distribution

Some properties: $E[X_j ] = \frac{1}{\alpha}$ (mean only exist if $\alpha >1$

$Pr(Y(t) > x) = \frac{1}{E[X_j]} \int^{\infty}_x F^c(u) du = (1+x)^{-(\alpha-1)}$

Example: Deterministic $X_j = D$

- Assume $F^c(\mu) = 1$ if $\mu \leq D$, otherwise $F^c(\mu) = 0$.

$Pr(Y(t) \geq x) = \frac{1}{E[X_j]} \int^{\infty}_x F^c(\mu) d\mu $

                            = $\frac{1}{D} \int^{D}_{x} d du$

                            = $1 - \frac{x}{D}$

This is the CCDF of a uniform distribution on $[0,D]$.

Note: the above assume that $x \leq D$, If $x > D$, then the result is 0.