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} dmu
                                               = int^{infty}_x Pr{X_j > mu} dmu
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) dmu = frac{1}{lambda} int^{infty}_x e^{-lambda mu} dmu
                     = 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) dmu
                            = 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.

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