Question

In a good weather year the number of storms is Poisson distributed with mean \(1 ;\) in a bad year it is Poisson distributed with mean 3. Suppose that any year's weather conditions depends on past years only through the previous year's condition. Suppose that a good year is equally likely to be followed by either a good or a bad year, and that a bad year is twice as likely to be followed by a bad year as by a good year. Suppose that last year-call it year 0 -was a good year. (a) Find the expected total number of storms in the next two years (that is, in years 1 and 2 ). (b) Find the probability there are no storms in year 3 . (c) Find the long-run average number of storms per year.

Short answer

The expected total number of storms in the next two years (years 1 and 2) is 4. The probability there are no storms in year 3 is approximately \(\frac{99e^{-1}+71e^{-3}}{216}\). The long-run average number of storms per year is \(\frac{9}{5}\).

Step-by-step solution

Define the random variables and probabilities

Let's denote the number of storms in a good year as \(X_1\), and the number of storms in a bad year as \(X_2\). Then, \(X_1 \sim Poisson(1)\) and \(X_2 \sim Poisson(3)\). We are also given the transition probabilities: \(P(Good_{n+1}|Good_n) = P(Bad_{n+1}|Bad_n) = \frac{1}{2}\) \(P(Good_{n+1}|Bad_n) = \frac{1}{3}\) and \(P(Bad_{n+1}|Good_n) = \frac{2}{3}\).

Calculate the expected total number of storms in the next two years

To find the total number of storms in years 1 and 2, we first need to consider the possible weather conditions for those years. Since year 0 was a good year, we have the following possibilities, with their respective probabilities: 1. Good, Good: with probability \(\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\) 2. Good, Bad: with probability \(\frac{1}{2} \times \frac{2}{3} = \frac{1}{3}\) 3. Bad, Good: with probability \(\frac{1}{2} \times \frac{1}{3} = \frac{1}{6}\) 4. Bad, Bad: with probability \(\frac{1}{2} \times \frac{2}{3} = \frac{1}{3}\) Now, we calculate the expected number of storms for each case and sum them up with their respective probabilities: Total expected storms = \(\frac{1}{4}E[X_1 + X_1] + \frac{1}{3}E[X_1 + X_2] + \frac{1}{6}E[X_2 + X_1] + \frac{1}{3}E[X_2 + X_2]\) For Poisson distributions, the sum of two independent Poisson random variables with parameters \(\lambda_1\) and \(\lambda_2\) is also a Poisson random variable with parameter \(\lambda_1 + \lambda_2\). Therefore, we have: \(E[X_1 + X_1] = E[Poisson(1+1)] = 2\), \(E[X_1 + X_2] = E[Poisson(1+3)] = 4\), \(E[X_2 + X_1] = E[Poisson(3+1)] = 4\), and \(E[X_2 + X_2] = E[Poisson(3+3)] = 6\). Total expected storms = \(\frac{1}{4}(2)+\frac{1}{3}(4)+\frac{1}{6}(4)+\frac{1}{3}(6)=\frac{1}{2}+\frac{4}{3}+\frac{2}{3}+2=4\) Thus, the expected total number of storms in the next two years is 4.

Calculate the probability there are no storms in year 3

The probability of no storms in year 3 depends on the weather conditions in year 2, which in turn depends on the weather conditions in year 1. We can calculate the probability of no storms in year 3 by considering both possible weather conditions in year 3: \(P(X_1=0)\) (Good) and \(P(X_2=0)\) (Bad): \(P(No\_storms\_in\_y3)=P(X_1=0|X_1=0)P(Year\_1|=Good)+P(X_1=0|X_1=1)P(Year\_1|=Bad)+P(X_2=0|X_2=0)P(Year\_2|=Good)+P(X_2=0|X_2=1)P(Year\_2|=Bad)\) Using the transition probabilities and the probability mass function of the Poisson distribution, we can substitute the values: \(P(No\_storms\_in\_y3)=e^{-1}\frac{1}{2}(1-\frac{1}{2}e^{-1})+e^{-1}\frac{1}{3}(1-\frac{1}{2}e^{-1})+e^{-3}\frac{1}{2}(1-\frac{1}{2}e^{-3})+e^{-3}\frac{2}{3}(1-\frac{1}{2}e^{-3})\) After simplifying, we get: \(P(No\_storms\_in\_y3)=\frac{99e^{-1}+71e^{-3}}{216}\)

Calculate the long-run average number of storms per year

We will need to calculate the stationary distribution of the transition probabilities matrix. This can be done by solving the following linear equations: \(\pi_G = \frac{1}{2}\pi_G + \frac{1}{3}\pi_B\) \(\pi_B = \frac{1}{2}\pi_G + \frac{2}{3}\pi_B\) \(\pi_G + \pi_B=1\) Solving the equations, we get: \(\pi_G=\frac{3}{5}\), and \(\pi_B=\frac{2}{5}\) The long-run average number of storms per year is then the sum of the expected number of storms per year in each weather condition, weighted by their stationary probabilities. Long-run average = \(\pi_GE[X_1] + \pi_GE[X_2] = \frac{3}{5}(1) + \frac{2}{5}(3) = \frac{9}{5}\) Thus, the long-run average number of storms per year is \(\frac{9}{5}\).

Key concepts

Probability Theory

Understanding probability theory is essential when dealing with stochastic phenomena like weather patterns and their impact on number of storms. Probability theory provides a mathematical framework for quantifying the uncertainty inherent in these phenomena. In our example, the number of storms in a given year is modeled using a Poisson distribution, which is commonly used for counting occurrences of events that happen independently at a constant average rate, such as storms in a year.

Considering the transition from one year to the next, we used given transition probabilities, which measure the likelihood of moving from one state (a good year or a bad year) to another. These probabilities help predict future weather patterns and, consequently, storms. To complete the exercise, we calculate expectations and probabilities using these transition probabilities which are fundamental components of probability theory.

Transition Probabilities

The concept of transition probabilities is a core component of Markov processes, where future states depend only on the current state and not on the sequence of events that preceded it. In our case, whether a year is good or bad depends only on the previous year, making this a simple Markov process. Given the transition probabilities, we can predict the likelihood of each sequence of weather conditions for the upcoming years, which in turn helps us determine the expected number of storms. To improve the understanding, creating a visual representation like a state diagram can be helpful in illustrating these transition probabilities between good and bad years.

Stationary Distribution

The stationary distribution of a Markov chain represents the long-term behavior of the system. It is the probability distribution over states that remains constant; in other words, once it is reached, the system tends to stay in that distribution over time, regardless of the initial state. In our exercise, the stationary distribution was found by solving a set of linear equations based on the given transition probabilities. This stationary distribution then gives us the long-run average number of storms per year, integrating over the fluctuations from year to year. Understanding the concept of stationary distribution is crucial for long-term predictions and planning in industries affected by weather, among other applications.

Expected Value

The expected value or mean of a random variable is a fundamental concept in probability and statistics, representing the average outcome if an experiment is repeated a large number of times. For Poisson distribution, the expected value is equal to its parameter, \( \lambda \). Therefore, for a good year (with \( \lambda = 1 \) ) the expected number of storms is 1, and for a bad year (with \( \lambda = 3 \) ), it's 3. The exercise involves calculating the expected number of storms over two years by weighting the expected values in good and bad years by their respective probabilities. Understanding how to find the expected value is vital in a wide range of fields, including insurance, finance, and any area dealing with risk and prediction.