An individual has three umbrellas, some at her office, and some at home. If she is leaving home in the morning (or leaving work at nigh) and it is raining she will take an umbrella, if one is there. Otherwise, she gets wet. Assume that independent of the past, it rains on each trip with probability 0.2. To formulate a Markov chain, let Xn be the number of umbrellas at her current location. (a) Find the transition probability for this Markov chain. (b) Calculate the limiting fraction Stout onowy di of time she gets wet. Mo)

Step-by-step explanation:

The number of umbrellas at her current location = Xn

The state space is X = {0,1,2,3,4,…r}

Probability of rain = p = 0.2

Probability it does not rain q = 1-p = 1-0.2 = 0.8

r = number of umbrellas she possess

let umbrellas at current place = i

Therefore umbrellas at other place = r-i

When she gets to the other place, there will still be r-i umbrellas if it does not rain

Pi,r-I = q = 0.8 (it does not rain)

Pi, r-i+1 = p = 0.2 (it rains) for i = 1,2,3…r

P0r = 1

Pij = 0, if i+j > r+2 or i+j<r

The markov chain in this case cannot be reduced and is finite and is therefore a positive recurrent. The only stationary distribution in this case is the limiting distribution.

m

∑     πiPi0= πrPr0 =q / (r+q) =π₀

i=0

m

∑    πi Pij=πr-j Pr-j,j +πr -j +1 Pr - j=1,j= q/(r + q) +p/(r+q) =1/(r+q)=πr

i=0

When i =0, no umbrella at current place, then \pi = q/(r+q) = 0.8/(3+0.8) = 0.210

When i = 1,2r, carries umbrella at current place, then \pi = 1/(r+q) = 1/(3+0.8) = 0.263

Therefore the π's given is the stationary distribution and is a limiting distribution

The Professor can get wet only when she has no umbrella at her current location and it rains. Therefore the probability is

π0p=pq/(r+q) = (0.2*0.8)/(3+0.8) = 0.042