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supplies customers bought for their new fish put Walgreens in the black for each Goldfish Giveaway Charles Walgreen Sr caught the decade s economic boom, too, though only on Chicago s South Side at first He had long sworn off starting stores inside Chicago s glamorous but pricey Loop, but ultimately he couldn t resist Walgreen made the leap in 1921 and located his first Loop store at 17 E Washington, in the Venetian Building To ensure the store s success, he sent a handful of his best people to work at the new branch, including fountain manager Ivar Pop Coulson Walgreen s belief in the power of soda fountains to give a store personality and charm ran deep, indeed As he told Chain Store Age (now Drugstore News), Fountains are the magnets drawing customers into the stores 35 A key to Walgreens success was a two-pronged approach: (1) The pharmacy gave the chain credibility, which explains why Walgreen was such a stickler for integrity behind the counter when others were selling snake oil and why he insisted on having the best pharmacies in town even when they were losing money The pharmacy was always the foundation of the business, Walgreens soul (2) The soda fountain, however, was pure fun If most drugstores of the day couldn t match Walgreens reputation for professionalism, those that could surely could not match Walgreens soda fountain excitement the heart of the business, the side that reached out to customers and made them feel welcome While the pharmacy fostered trust, the soda fountain generated warmth It was a tough combination to beat This explains why Walgreen spared no expense on his soda fountains When you entered a Walgreens in the 1920s, you d walk along the immaculate wooden floors to the booths built from sturdy hardwood, or to the tables in the middle of the room covered in white linen, or to one of the stools lining the long, gleaming, white marble counter (including a 78footer in Milwaukee) You d see another white marble counter running along the wall for the clerks to do their work, with a seemingly endless supply of confections neatly ordered on mahogany shelves and in glass cabi-.

(n) Proof For a transient state j we have by Lemma 323 that limn pij = 0 for all i I Using the well-known result that the Cesaro limit is equal to the ordinary limit whenever the latter limit exists, the results (331) and (332) follow

Apply charging GPRS (ACH-GPRS) Apply charging report GPRS (ACR-GPRS) Cancel GPRS (CAN-GPRS)

for transient states j Fix now a recurrent state j By the de nition of recurrence, we have fjj = 1 The times between successive visits to state j are independent and identically distributed random variables with mean jj In other words, visits of the Markov chain to state j can be seen as renewals Denote by N (t) the number of visits of the Markov chain to state j during the rst t transition epochs Then, by Lemma 222,

(333)

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This limiting result holds for both jj < and jj = In other words, the long-run average number of transitions to state j per time unit equals 1/ jj with probability 1 when the process starts in state j De ne the indicator variable Ik = 1 0 if the process visits state j at time k, otherwise

Since N (n) = I1 + + In , we can rewrite (333) as 1 lim n n Obviously,

Connect GPRS (CON-GPRS) Continue GPRS (CUE-GPRS)

(334)

Noting that (1/n) n Ik is bounded by 1 and using the bounded convergence k=1 theorem from Appendix A, it follows from (334) that 1 =E jj 1 lim n n

.

It remains to prove that (332) holds for any state i = j To do so, we use the relation (3212) which was derived in the proof of Lemma 323 Averaging this relation over n = 1, , m, interchanging the order of summation and letting m , the relation (332) follows in the same way as (3213) Another natural question is under which condition the effect of the initial state (k) of the process fades away as time increases so that limn (1/n) n pij does k=1 not depend on the initial state X0 = i for each j I We need some condition as the following example shows Take a Markov chain with state space I = {1, 2} and the one-step transition probabilities pij with p11 = p22 = 1 and p12 = p21 = 0 In (n) (n) (k) this example p11 = 1 and p21 = 0 for all n 1 so that limn (1/n) n pi1 k=1

.

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