Problem 2: Oscar's running shoes
Oscar goes for a run each morning. When he leaves his house for his run, he is equally likely to use either the front or the back door; and similarly, when he returns, he is equally likely to use either the front or the back door. Assume that his choice of the door through which he leaves is independent of his choice of the door through which he returns, and also assume that these choices are independent across days.
Oscar owns only five pairs of running shoes, each pair placed at one of the two doors. If there is at least one pair of shoes at the door through which he leaves, he wears a pair for his run; otherwise, he runs barefoot. When he returns from his run, if he wore shoes for that run, he takes off the shoes after the run and leaves them at the door through which he returns.
We wish to determine the long-term proportion of time that Oscar runs barefoot.
We consider a Markov chain with states {0,1,2,3,4,5}, where state i indicates that there are i pairs of shoes available at the front door in the morning, before Oscar leaves for his run. Specify the numerical values of the following transition probabilities.
For i∈{0,1,2,3,4},
pi,i+1=
0.25 - correct
For i∈{1,2,3,4,5},
pi,i−1=
0.25 - correct
For i∈{1,2,3,4},
pii=
0.5 - correct
p00=
0.75 - correct
p55=
0.75 - correct
Determine the steady-state probability that Oscar runs barefoot.
0.1666666666666666666 - correct
"answered in full"
answers to "Problem 1: Steady-state convergence" , belonging to this very same problem set....
1.a. False
1.b. False
2.a. False
2.b. False
3.a. True
3.b. True
Any hint how to solve it and not only the answers?
Wow, Oscar sure has some interesting shoe habits! It looks like we're dealing with a Markov chain where the states represent the number of pairs of shoes at the front door. Let's calculate the steady-state probability that Oscar runs barefoot.
From the transition probabilities, we can see that Oscar can either gain or lose one pair of shoes with equal probability (0.25) at each state except for when he has 0 or 5 pairs of shoes.
In the case of 0 pairs of shoes, the probability of staying at 0 is 0.75 and the probability of gaining one pair of shoes to reach 1 is 0.25.
In the case of 5 pairs of shoes, the probability of staying at 5 is 0.75 and the probability of losing one pair of shoes to reach 4 is 0.25.
For all other states (1, 2, 3, 4), Oscar has a 0.5 chance of staying at the same state (keeping the same number of pairs of shoes) and a 0.25 chance of both gaining and losing one pair of shoes.
Now, we need to find the steady-state probabilities for each state. By solving the system of equations, we can calculate those probabilities.
The steady-state probability for being barefoot (state 0) is approximately 0.1666666666666666666.
So, Oscar will be running barefoot about 16.67% of the time in the long run. Maybe he should invest in some more running shoes!
To determine the long-term proportion of time that Oscar runs barefoot, we need to find the steady-state probability of being in state 0.
In this Markov chain, the states represent the number of pairs of shoes available at the front door in the morning. So state 0 means there are no pairs of shoes at the front door, and Oscar runs barefoot.
To find the steady-state probability, we can solve the equations:
p0 = p0*p00 + p1*p10
p1 = p0*p01 + p1*p11 + p2*p21
p2 = p1*p12 + p2*p22 + p3*p32
p3 = p2*p23 + p3*p33 + p4*p43
p4 = p3*p34 + p4*p44 + p5*p54
p5 = p4*p45 + p5*p55
where pi,j represents the transition probability from state i to state j.
We already know the transition probabilities, so we can substitute them into the equations:
p0 = p0*0.75 + p1*0.25
p1 = p0*0.25 + p1*0.5 + p2*0.25
p2 = p1*0.25 + p2*0.5 + p3*0.25
p3 = p2*0.25 + p3*0.5 + p4*0.25
p4 = p3*0.25 + p4*0.5 + p5*0.25
p5 = p4*0.25 + p5*0.75
Since this is a system of equations, we can solve it to find the values of p0, p1, p2, p3, p4, and p5. In this case, the values are:
p0 = 0.1666666666666666666 (approximately)
So the steady-state probability that Oscar runs barefoot is approximately 0.1666666666666666666, or 1/6.
1. 0.44
2. 1/9
2/9
6/9
3. 1/9