A little spider lives in a rectangular box of which the sides are 3 and 4 cm lon

Question

A little spider lives in a rectangular box of which the sides are 3 and 4 cm long. It can only sit in one of the four corners marked with the numbers 1,2,3 and 4 as shown on the diagram 4 From time to time the spider runs from the occupied corner to another one, chosen at random with probabilities inversely proportional to the distances to the corner from the current position of the spider. Denote by Xn the number of the corner the spider is at after the nth run. Comment on why the sequence {X] is a Markov chain. ( ). a) Find the transition probabilities matrix for the Markov chain {X (b) Argue why this Markov chain is irreducible. (c) Let the initial distribution of Xo be uniform: p - (0.25,0.25, 0.25,0.25). Find the probability P(X1 = 1, x2 = 4, X,-2).

Explanation / Answer

(a) Transition matrix is as such:-

States = c("1","2","3","4")
spdrMatrix = matrix(data = c(0,3/12,5/12,4/12,3/12,0,4/12,5/12,5/12,4/12,0,3/12,4/12,5/12,3/12,0), byrow = byRow, nrow = length(States), dimnames = list(States, States))

1 2 3 4
1 0.0000000 0.2500000 0.4166667 0.3333333
2 0.2500000 0.0000000 0.3333333 0.4166667
3 0.4166667 0.3333333 0.0000000 0.2500000
4 0.3333333 0.4166667 0.2500000 0.0000000

(b) Irreducible is when all states can be reached from all other states => Can see that in the given transition matrix above that there are NO closed loops that the spider won't be able to come out from a subset of states

(c) Initial state probability = c(0.25,0.25,0.25,0.25)

What we want is Spider moves like this: Starting_State --> 1 --> 4 --> 2

=> Total probability =
P(SS = 1 and path is followed) = (1/4)*( (0)*(1/3)*(5/12) ) +
P(SS = 2 and path is followed) = (1/4)*( (1/4)*(1/3)*(5/12) ) +
P(SS = 3 and path is followed) = (1/4)*( (5/12)*(1/3)*(5/12) ) +
P(SS = 4 and path is followed) = (1/4)*( (1/3)*(1/3)*(5/12) )

= 0 + 0.0086805 + 0.01446759259 + 0.01157407407 = 0.03472216666 [ANSWER]

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