Each day, one of n possible elements is requested, the ith one with probability  These elements are at all times arranged in an ordered list which is revised as follows: The element selected is moved to the front of the list with the relative positions of all the other elements remaining unchanged. Define the state at any time to be the list ordering at that time and note that there are n! possible states.

(a) Argue that the preceding is a Markov chain.

(b) For any state i₁, . . . , i_n (which is a permutation of 1, 2, . . . , n), let Ï€(i₁, . . . , i_n) denote the limiting probability. In order for the state to be i₁, . . . , i_n, it is necessary for the last request to be for i₁, the last non-i₁ request for i₂, the last non-i₁ or i₂ request for i₃, and so on. Hence, it appears intuitive that

 

Verify when n = 3 that the preceding are indeed the limiting probabilities.

Alpha Geek

(8)

Status NEW Posted 19 Jun 2017 12:06 PM My Price 5.00

-----------

Attachments

file 1497874578-683767_1_636331921459814385_683767.docx preview (1 words )

----------- ----------- -----------

Not Rated(0)

• Buy Answer