Alternating Renewal Process

1. Definition

Consider a process $X(t)$ with “on” periods and “off” periods

Let $Z_j equiv$ ON time of cycle j
Let $Y_j equiv$ OFF time of cycle j

Then $Z_j$ and $Y_j$ must satisfy the following property in order for $X(t)$ to be regenerative

The pair $(Z_j, Y_j)$ must be i.i.d; in particular, $(Z_j, Y_j)$ independent of $(Z_i, Y_i)$ for $i neq j$ . That is

$Z_j$ are i.i.d
$Y_j$ are i.i.d
However, $Z_j$ and $Y_j$ may be independent for the same j

Cycle time $X_j = Z_j + Y_j$ .

Then by the regenerative process, we have

the average “up” time = $frac{mbox{average up time in a cycle}}{mbox{average time of one cycle}} = frac{E[Z_j]}{E[Z_j] + E[Y_j]}$

2. Example

Cars pass a point on highway according to Poisson process with rate $lambda = 2/min$ . $1/5$ of cars are speeding (>10 mph over the pretend speed limit). Assume time to issue a ticket~UNIF[10,14] minutes(one officer)

Question: What fraction of speeding cars pass when the officer is busy.

Answer: The answer to the question is equivalent to the fraction of time the officer is busy.

Let $Y_j$ = time spent waiting to give a ticker
Let $Z_j$ = time spent giving a ticker

Speeders arrive according to Poisson process with rate $2 cdot 1/5 = 2/5$ per min.

Fraction of time busy = $frac{E[Z_j]}{E[Z_j] + E[Y_j]} = frac{12}{12+5/2} = frac{12}{14.5}$

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Alternating Renewal Process