Example 5.4

Consider passengers arrive at Jhakarkati bus depot in Kanpur, India, in accordance with renewal process with a mean arrival time of  minutes. Now the scheduling manager of the bus depot decides that when ever there are  number of passengers the bus (which are also in a queue) leaves. Also assume that when the number of passengers are  then the bus depot incurs a costs like running the air conditioning system etc., and that this cost is  per unit time. Apart from this cost, an additional cost of  is incurred by the bus company when ever a bus leaves with passengers. Our main concer is to calculate the average cost per unit time incurred by the bus depot. Assume the cycle is such that it is completed when a bus leaves and this in very simple terms is a renewal process.

Thus . Now if we denote the time between  and  arrival in a cycle, then the expected cost of a cyle can be expressed as:

 

Thus average cost is

Example 5.4

Consider  and  are independent normally distributed random variables with zero means and unit variances and along with that let us refer to the diagram given below which is a circle with radius .

Figure 5.1: Circle with circumference  and radius

Let us also assume  is the circumference such that . Also assume:  and . Let us also denote  as a bivariate process. It is very clear from the example given above that (i)  and (ii)   since .
Since  and  have a joint normal distribution, hence we need to find their covariance. Moreover we can easily see that . Thus the distribution of  is the same as that of  for any value of  which proves that fact that  is a stationary process.