Every day, a friend chooses at random a coffee shop that is directly connected t

Question

Every day, a friend chooses at random a coffee shop that is directly connected to the shop visited on the previous day. That is, if he was at shop 3 yesterday, he can only be at 2 or 4 today with equal probabilities; see the figure below with shops marked by 1 - 4. Give the long run mean frequency of visits to the coffee shop 2. Suppose that the colleague buys an espresso in each coffee shop he visits. For concreteness, suppose that an espresso in the store t costs c, dollars, where c_1 = 2.00, c_2 = 2.50, c_3 = 3.00, c_4 = 3.00. Approximately, on average, how much our friend will spend on espresso during a whole year?

Explanation / Answer

a) Let the probability to be at Shop i be Pi

Therefore P1 , P2 , P3 and P4 are the respective probabilities to be at Shop 1 , 2 , 3 and 4.

Now ,

P1 + P2 + P3 + P4 = 1.

The person can go to shop 1 only if he has visited shop 2 and there are 3 ways to go to shop 2.

Therefore,

P1    = (1/3)P3

Similarly , P2 = P1 + (1/2) P3 + (1/2)P4

P3 = (1/3) P2 + (1/2)P4

P4 = (1/3)P2 + (1/2) P3

Equation above equations in

P1 + P2 + P3 + P4 = 1.

we get P2 = (3/8)

Therefore out of 8 visits 3 are done to coffee shop 2.

Hence Long run mean frequence of visits to coffee shop 2 is 3.

b)

We have

c1 = 2

c2 = 2.5

c3 = 3

c4 = 3

Average amount of money spend on Espresso = (c1P1 + c2P2 + c3P3 +c4P4) * 365

= (2 * (1/8) + 2.5 * (3/8) + 3 * (1/4) + 3 * (1/4) ) * 365

= 991$

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