Question

Consider a pure death process where \(\mu_{n}=n \mu\) for \(n=1,2, \ldots\), i.e., \(P\\{X(t+h)\) \(=j \mid X(t)=k\\}=0\) for \(j>k\) and \(t\) and \(k\) positive. Assume an initial population of size \(i\). Find \(P_{n}(t)=P\\{X(t)=n\\}, E X(t)\), and \(V\) ar \(X(t)\).

Short answer

The short answer for this problem is as follows. The probability \(P_n(t)\) is given by: \(P_n(t) = e^{-(i-n)\mu t}\binom{i}{n}(\frac{\mu}{\mu+i-n\mu})^{n}(\frac{i \mu}{\mu+i-n\mu})^{i-n}\). The expected number of individuals in the process at time \(t\) is: \(E X(t) = \sum_{n=0}^{i} n e^{-(i-n)\mu t}\binom{i}{n}(\frac{\mu}{\mu+i-n\mu})^{n}(\frac{i \mu}{\mu+i-n\mu})^{i-n}\). The variance of the number of individuals in the process at time \(t\) is: \(Var X(t) = E(X^2(t)) - (E X(t))^2\), where \(E(X^2(t)) = \sum_{n=0}^{i} n^2 e^{-(i-n)\mu t}\binom{i}{n}(\frac{\mu}{\mu+i-n\mu})^{n}(\frac{i \mu}{\mu+i-n\mu})^{i-n}\).

Step-by-step solution

We are given that \(P\{X(t+h) = j | X(t) = k\} = 0\) for \(j>k\) and \(t\) and \(k\) positive. Since it is a pure death process, the transition probabilities \(p_{jk}(t)\) for \(j

Using the transition probabilities from Step 1, we get the probability \(P_n(t)\) as: \(P_n(t) = P\{X(t) = n | X(0) = i\} = p_{in}(t)\), which is given by: \(P_n(t) = e^{-(i-n)\mu t}\binom{i}{n}(\frac{\mu}{\mu+i-n\mu})^{n}(\frac{i \mu}{\mu+i-n\mu})^{i-n}\). #Step 3: Find the expected number of individuals \(E X(t)\)

The expected number of individuals in the process at time \(t\) is given by: \(E X(t) = \sum_{n=0}^{i} n P_n(t)\). Now substitute the expression for \(P_n(t)\) from Step 2: \(E X(t) = \sum_{n=0}^{i} n e^{-(i-n)\mu t}\binom{i}{n}(\frac{\mu}{\mu+i-n\mu})^{n}(\frac{i \mu}{\mu+i-n\mu})^{i-n}\). #Step 4: Find the variance \(Var X(t)\)

The variance is given by: \(Var X(t) = E(X^2(t)) - (E X(t))^2\). First, let's find \(E(X^2(t))\). Compute this as \(E(X^2(t)) = \sum_{n=0}^{i} n^2 P_{n}(t)\). Now substitute the expression for \(P_n(t)\) from Step 2: \(E(X^2(t)) = \sum_{n=0}^{i} n^2 e^{-(i-n)\mu t}\binom{i}{n}(\frac{\mu}{\mu+i-n\mu})^{n}(\frac{i \mu}{\mu+i-n\mu})^{i-n}\). Now, we can compute the variance using the expressions for \(E X(t)\) and \(E(X^2(t))\): \(Var X(t) = E(X^2(t)) - (E X(t))^2\).

Key concepts

Stochastic Processes

A stochastic process is a mathematical framework used to describe systems that evolve over time in a way that is not entirely predictable. When we use the term 'stochastic', we refer to processes that are random in nature. This randomness doesn't imply a lack of structure or patterns; rather, it means that the system's future state is determined both by predictable and unpredictable elements.

In the context of a pure death process—a specific type of stochastic process—the system's evolution depends on the occurrence of 'death' events. These events reduce the population size, and the process is characterized by having no births. The given exercise focuses on a situation where such events happen at rates proportional to the current population size, which is quite a common feature in natural and engineered systems.

To improve clarity for students, one might illustrate this with examples from real life, like radioactive decay (where particles 'die' at a rate proportional to the remaining quantity) or the decline in population of a species due to factors like disease or predation.

Transition Probabilities

Transition probabilities are a key concept in stochastic processes. They represent the likelihood of transitioning from one state to another within a particular time frame. For a pure death process, these reflect the chance of the population decreasing from size 'k' to size 'j' in a small time increment 'h'.

To explain this, you might think of each potential 'death' as a branch on a tree; transition probabilities are the chance you'll go down a particular branch. In our context, the transition probabilities are restricted by the rule that the population can't increase - it can only stay the same or decrease. Thus, the formula provided only applies when 'j' is less than or equal to 'k', aligning with the nature of a 'death' process.

To enhance comprehension for students, comparing these probabilities to stepping stones across a river could be useful. You can only step forwards (towards smaller population sizes), or stay on the same stone, but never back - similar to how a pure death process only allows for population decrease or stasis.

Expected Value

The expected value, often denoted as 'E', represents the long-term average outcome of a random process if it can be repeated many times. It's a measure of the central tendency of a probability distribution—a 'weighted average,' where each possible outcome is weighted by its probability.

For students, we could say it's like predicting the average result of flipping a biased coin over and over; some results will be more common than others, and the expected value helps you anticipate what's most likely over time. In a pure death process, the expected value tells us the average population size we'd expect at any given time 't', taking into account all the possible sizes the population could be and the likelihood of each size.

As an improvement, illustrating this concept with a simple scenario such as an uneven dice game where some numbers are much more likely to come up can help students grasp why some outcomes contribute more to the expected value than others.

Variance

Variance captures the spread of a probability distribution; it quantifies how much the outcomes of a process differ from the expected value. A high variance means that the actual outcomes deviate widely from the average, while a low variance indicates they are clustered closely around it.

To help students visualize this, imagine a bag filled with balls marked with different numbers. If most balls have a number close to the average, picking one at random will likely give you a number similar to the expected value—this is low variance. Meanwhile, if the numbers are all over the place, you're much less certain about what you'll get next—that's high variance.

In the context of a pure death process, the variance can provide insights into the stability of the population. Low variance might suggest a population with consistent, predictable decline, while high variance could suggest a more erratic population change, with periods of rapid decline contrasted against periods of relative stability.