Question

Each of \(n\) skiers continually, and independently, climbs up and then skis down a particular slope. The time it takes skier \(i\) to climb up has distribution \(F_{i}\), and it is independent of her time to ski down, which has distribution \(H_{i}, i=1, \ldots, n\). Let \(N(t)\) denote the total number of times members of this group have skied down the slope by time \(t .\) Also, let \(U(t)\) denote the number of skiers climbing up the hill at time \(t\). (a) What is \(\lim _{t \rightarrow \infty} N(t) / t\) ? (b) Find \(\lim _{t \rightarrow \infty} E[U(t)]\). (c) If all \(F_{i}\) are exponential with rate \(\lambda\) and all \(G_{i}\) are exponential with rate \(\mu\), what is \(P\\{U(t)=k\\} ?\)

Short answer

The short answers to each part are: (a) \[\lim_{t \rightarrow \infty} \frac{N(t)}{t} = \sum_{i=1}^n \frac{1}{E[X_i]}.\] (b) \[\lim_{t \rightarrow \infty} E[U(t)] = \sum_{i=1}^n \frac{1}{E[T_i]}.\] (c) \[P\{U(t) = k\} = \frac{e^{-(\lambda+\mu)t}((\lambda + \mu)t)^{n-k}}{(n-k)!}.\]

Step-by-step solution

Calculate the average rate of skiing down the slope

Let's denote the random variable \(X_i\) as the time skier \(i\) takes to complete one round (climbing up and skiing down), which is the sum of two independent random variables: \(X_i = T_i + D_i\), where \(T_i\) and \(D_i\) are the time to climb up and ski down, respectively. The average rate of completing one round by skier \(i\) is \(\frac{1}{E[X_i]}\). Since all skiers are independent, the overall average rate of skiing down the slope will be the sum of individual average rates, i.e., \(\sum_{i=1}^n \frac{1}{E[X_i]}\).

Calculate the limit of the ratio

The total number of times members of the group have skied down the slope by time \(t\) can be denoted as \(N(t)\). Given the average rate calculated in step 1, we know that the overall average rate is constant, so as time \(t\) approaches infinity, the total number of times they ski down the slope will be proportional to \(t\), i.e., \(N(t) \approx \left(\sum_{i=1}^n \frac{1}{E[X_i]}\right) \cdot t\). Therefore, taking the limit, we have \[\lim_{t \rightarrow \infty} \frac{N(t)}{t} = \sum_{i=1}^n \frac{1}{E[X_i]}.\] (b) Calculate the limit of the expected number of skiers climbing up at time \(t\)

Express the expected number of skiers climbing up as a function of time

Let \(U_i(t)\) be an indicator function for skier \(i\) climbing up at time \(t\). We have \(U(t) \equiv \sum_{i=1}^n U_i(t)\). The expected number of skiers climbing up at time \(t\) is \[E[U(t)] = E\left[\sum_{i=1}^n U_i(t)\right] = \sum_{i=1}^n E[U_i(t)].\]

Calculate the individual expected value of skier \(i\) climbing up at time \(t\)

For each skier \(i\), we can calculate the expected value of \(U_i(t)\) as \(E[U_i(t)] = 1 \cdot P(U_i(t) = 1) + 0 \cdot P(U_i(t) = 0) = P(U_i(t)=1) = \frac{1}{E[T_i]}\).

Calculate the limit of the overall expected value as \(t\) approaches infinity

As the time \(t\) approaches infinity, the expected number of skiers climbing up at time \(t\) tends to a constant value, which is the sum of the individual probabilities of skiers climbing up: \[\lim_{t \rightarrow \infty} E[U(t)] = \sum_{i=1}^n \frac{1}{E[T_i]}.\] (c) Calculate the probability of having \(k\) skiers climbing up at time \(t\) when all \(F_i\) are exponential with the rate of \(\lambda\) and all \(G_i\) are exponential with the rate of \(\mu\)

Convert the problem into a Poisson process problem

When all \(F_i\) and \(G_i\) are exponential with rates \(\lambda\) and \(\mu\) respectively, we can model this problem as each skier following a Poisson process with a combined rate of \(\lambda+\mu\) when they are skiing and climbing up. Moreover, since all skiers are independent, we can model the entire system of \(n\) skiers as a Poisson process with a combined rate of \(n(\lambda + \mu)\).

Calculate the probability of having k skiers climbing up at time t

At time \(t\), we want to find the probability of having \(k\) skiers climbing up. In order to have \(k\) skiers climbing up, there must be \(n - k\) skiers skiing down. Therefore, we can write the probability as \[P\{U(t) = k\} = P\{N(t) = n - k\}.\] Since we are dealing with a Poisson process, we know that \[P\{N(t) = n - k\} = \frac{e^{-(\lambda+\mu)t}((\lambda + \mu)t)^{n-k}}{(n-k)!}.\] So, we have \[P\{U(t) = k\} = \frac{e^{-(\lambda+\mu)t}((\lambda + \mu)t)^{n-k}}{(n-k)!}.\]

Key concepts

Poisson Process

The Poisson process is a mathematical model used to describe random events that occur independently and at a constant average rate over time. Imagine events like emails arriving in your inbox or skiers descending a slope in our textbook exercise. One of the key properties of this process is its 'memorylessness,' meaning the probability of an event happening in a certain time interval is the same, no matter when you start counting.

In the context of the exercise, the climbing up and skiing down of skiers can be thought of as independent events occurring over time. If each skier follows an exponential distribution with rates \( \lambda \) and \( \mu \) for climbing and skiing, respectively, their combined activity mimics a Poisson process. This is a classic example of how the Poisson process helps us to model real-world phenomena, where events are happening continuously and randomly over time.

A common use of the Poisson process in real-world applications is queueing theory, where it helps forecast wait times and system efficiency for everything from checkout lines at a supermarket to network data packets transmitting over a server. Understanding this concept can give us profound insights into operational systems, optimizing business processes, and more.

Exponential Distribution

The exponential distribution, which we come across in the exercise in relation to the rate of skiers climbing and descending, is intrinsically linked to the Poisson process. It models the time between events in a Poisson process. Often represented by the symbol \( \lambda \), the rate parameter of the exponential distribution tells us the average number of occurrences in a given interval.

Now, why is the exponential distribution so special? It has the unique property of being memoryless, just like the Poisson process. This means if you want to predict the future without knowing the past, the exponential distribution is your go-to model. For example, if a skier is climbing up the hill, her time to reach the top does not depend on how long she's already been climbing - each moment is independent and equally likely for her to finish.

Moreover, when solving part (c) of the exercise, the use of the exponential distribution simplifies the process. The memoryless property allows us to see each skier as 'resetting' their clock after each descent, so their climbing times are independent of previous climbs. This concept is widely applicable in fields like telecommunications, survival analysis, and reliability engineering.

Expected Value

The expected value is a fundamental concept in probability theory, representing the average outcome if an experiment is repeated many times. Consider the expected value as the long-term average – it's what you would predict to happen over an extended period or over numerous trials. In our ski slope scenario, the expected value helps us to quantify the average behavior of the skiers over time.

For instance, to solve part (b) of the exercise, we need to calculate \( E[U(t)] \), the expected number of skiers climbing up at a certain time. Here, we are taking into account the average time it takes a skier to climb up, which over the long term, leads to a constant proportion of skiers ascending the slope at any given moment.

In real-life applications, the expected value is very useful for making predictions and informed decisions in situations involving uncertainty. Insurance companies use it to set premiums, investors use it to assess expected returns on their portfolios, and public policy experts use it to estimate the costs of social programs. By grasping the use of expected value, students can unlock a key tool used in a multitude of decision-making processes across various industries.