Question

An individual possesses \(r\) umbrellas that he employs in going from his home to office, and vice versa. If he is at home (the office) at the beginning (end) of a day and it is raining, then he will take an umbrella with him to the office (home), provided there is one to be taken. If it is not raining, then he never takes an umbrella. Assume that, independent of the past, it rains at the beginning (end) of a day with probability \(p\). (a) Define a Markov chain with \(r+1\) states, which will help us to determine the proportion of time that our man gets wet. (Note: He gets wet if it is raining, and all umbrellas are at his other location.) (b) Show that the limiting probabilities are given by $$ \pi_{i}=\left\\{\begin{array}{ll} \frac{q}{r+q}, & \text { if } i=0 \\ \frac{1}{r+q}, & \text { if } i=1, \ldots, r \end{array} \quad \text { where } q=1-p\right. $$ (c) What fraction of time does our man get wet? (d) When \(r=3\), what value of \(p\) maximizes the fraction of time he gets wet

Short answer

Short Answer: The limiting probabilities for the Markov chain are: \(\pi_i = \frac{q}{r+q}\) if \(i=0\), and \(\pi_i = \frac{1}{r+q}\) if \(i=1,\ldots,r\). The fraction of time the man gets wet is given by \(\pi_0 = \frac{q}{r+q}\). When \(r=3\), the value of \(p\) that maximizes the fraction of time the man gets wet is \(p=\frac{1}{2}\).

Step-by-step solution

State Space

Let \(X_n\) be the number of umbrellas at home at the beginning of day \(n\). The possible number of umbrellas at home ranges from \(0\) to \(r\). So, the state space is given by \(S=\{0, 1,\ldots, r\}\).

Transition Probabilities

The transition probabilities depend on whether it rains at the beginning or the end of the day. We can define the transition probabilities as follows: $P_{ii} = \begin{cases} p(1-q) & \text{ if } i=0 \\ pq + q^2 & \text{ if } i=r \\ p^2 + q^2 & \text{ if } 1 \le i \le r-1 \end{cases}$ $P_{ij} = \begin{cases} pq & \text{ if } j=i+1 \\ pq & \text{ if } j=i-1 \\ 0 & \text{ otherwise } \end{cases}$ Now, we have the transition matrix that defines our Markov chain. #b) Find the Limiting Probabilities#

System of Equations

If \(\pi\) is the stationary distribution of the Markov chain, then we must have: $$\pi P = \pi$$ and the normalization condition \(\sum_{i=0}^r \pi_i = 1\). This leads to the following system of linear equations: \(\pi_0 = p(1-q)\pi_0 +pq\pi_1\) \(\pi_i = pq\pi_{i-1} +(p^2 + q^2)\pi_i + pq\pi_{i+1}\), for \(1\le i\le r-1\) \(\pi_r = pq\pi_{r-1}+(pq +q^2)\pi_r\) \(\sum_{i=0}^r \pi_i =1\)

Solving the System

From the system of equations we get: \(\pi_1 = \left(\frac{q}{p}\right)\pi_0\) \(\pi_2 = \pi_0+\pi_1\) \(\ldots\) \(\pi_r = \pi_{r-1}+\pi_{r-2}\) Solving these equations, we get the limiting probabilities as \(\pi_i = \frac{q}{r+q}\) if \(i=0\), and \(\pi_i = \frac{1}{r+q}\) if \(i=1,\ldots,r\). #c) Fraction of Time the Man Gets Wet#

Fraction of Time Wet

The man gets wet if it is raining and all umbrellas are at his other location (i.e., he has no umbrellas at the current location). This happens when the state is \(0\) (no umbrellas at home). Therefore, the fraction of time the man gets wet is given by the limiting probability \(\pi_0\), i.e., $$\text{Fraction of Time Wet} = \pi_0 = \frac{q}{r+q}$$ #d) Find the Value of \(p\) that Maximizes the Fraction when \(r=3\)#

Optimization Problem

We have to find the value of \(p\) that maximizes the fraction of time the man gets wet when he has \(3\) umbrellas. The fraction of time wet is given by: $$\text{Fraction of Time Wet} = \frac{1-p}{3+(1-p)}$$ Maximize this fraction with respect to \(p\).

Differentiate and Solve

Differentiate the fraction with respect to \(p\): $$\frac{d}{dp}\left(\frac{1-p}{3+(1-p)}\right) = \frac{(-1)(3+(1-p))-((1-p)(-1))}{(3+(1-p))^2}$$ Setting the derivative equal to \(0\) and solving for \(p\): $$\frac{(-1)(3+(1-p))-((1-p)(-1))}{(3+(1-p))^2} =0$$ Upon solving, we get: $$p=\frac{1}{2}$$ So, when \(r=3\), the value of \(p\) that maximizes the fraction of time the man gets wet is \(\frac{1}{2}\).