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Chapter 6: Problem 6

Let \(N\) be a birth process with intensities \(\lambda_{0}, \lambda_{1}, \ldots\), and let \(N(0)=0\). Show that \(p_{n}(t)=\) \(P(N(t)=n)\) is given by $$ p_{n}(t)=\frac{1}{\lambda_{n}} \sum_{i=0}^{n} \lambda_{i} e^{-\lambda_{i} t} \prod_{j=0 \atop j \neq i}^{n} \frac{\lambda_{j}}{\lambda_{j}-\lambda_{i}} $$ provided that \(\lambda_{i} \neq \lambda_{j}\) whenever \(i \neq j\).

Short Answer

Expert verified
Use differential equations for birth processes and verify by induction.

Step by step solution

01

Understand the Birth Process

A birth process is a continuous-time Markov chain where transitions occur from state \(n\) to \(n+1\) with rate \(\lambda_n\). It starts in state 0, i.e., \(N(0) = 0\). We are asked to find the probability \(p_n(t)\) that the process is in state \(n\) at time \(t\).
02

Formulate the Differential Equations

The probability \(p_n(t)\) satisfies the differential equations given by \[\frac{d}{dt} p_n(t) = \lambda_{n-1} p_{n-1}(t) - \lambda_n p_n(t)\]\(p_0(t)\) satisfies\[\frac{d}{dt} p_0(t) = -\lambda_0 p_0(t)\]with initial condition \(p_0(0) = 1\).
03

Solve the Base Case

The solution for \(p_0(t)\) is derived from the differential equation:\[p_0(t) = e^{-\lambda_0 t}\]This is because it鈥檚 the solution to the equation \(\frac{d}{dt} p_0(t) = -\lambda_0 p_0(t)\) with initial condition \(p_0(0) = 1\).
04

Solve for Higher Order Probabilities Using Induction

Assume \(p_k(t)\) for \(k < n\) is known, and use the recurrence relation to solve for \(p_n(t)\):\[p_n(t) = \frac{1}{\lambda_n} \int_0^t \lambda_{n-1} p_{n-1}(s) e^{-\lambda_n (t-s)} ds\]This integral form from the differential equation helps us derive \(p_n(t)\).
05

Verify the Given Formula Using the General Solution

Apply the assumed form for \(p_n(t)\) by plugging the expression \[p_{n}(t)=\frac{1}{\lambda_{n}} \sum_{i=0}^{n} \lambda_{i} e^{-\lambda_{i} t}\prod_{j=0 \atop j eq i}^{n} \frac{\lambda_{j}}{\lambda_{j}-\lambda_{i}} \]back into the differential equations to verify the solution satisfies both the probabilities and initial conditions by computation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Continuous-time Markov chain
A continuous-time Markov chain is a type of mathematical model that describes systems which transition between different states continuously over time. In these models, the process moves randomly from one state to another as time progresses, based on given rates or intensities.
In the context of a birth process, the chain represents a system where births (like arrival of customers or new system entries) happen over time. Each state corresponds to the number of births that have occurred.
  • The system starts at state 0, meaning no births have occurred yet at time 0.
  • The transition rate from state \(n\) to \(n+1\) is denoted by \(\lambda_n\), which is a defining characteristic of the birth rate.
  • This marks the continuous transitions from one state to the next in a Markov chain setting, driven by these \(\lambda\) rates.
Markov chains like this assume 'memoryless' property. It means that the future state depends only on the current state and not on the sequence of events that preceded it.
Differential equations
Differential equations are mathematical tools used to model the evolution of systems over time. They express how the rate of change of a quantity depends on other quantities. In the birth process scenario, differential equations help us figure out the probability that the system is in a particular state at any given time.
For finding the probability \(p_n(t)\) that the process is in state \(n\) at time \(t\), we use the following differential equations:
  • \(\frac{d}{dt} p_n(t) = \lambda_{n-1} p_{n-1}(t) - \lambda_n p_n(t)\), which tells how the probability of being in state \(n\) at time \(t\) changes with respect to time.
  • \(p_0(t)\) follows \(\frac{d}{dt} p_0(t) = -\lambda_0 p_0(t)\) with the initial condition \(p_0(0) = 1\).
Differential equations allow us to define an initial condition and then can be integrated over to solve for the probabilities as time increases.
Probability functions
Probability functions provide the likelihood of different states in a system described by a Markov chain. In the case of a birth process, the probability function \(p_n(t)\) defines the probability that the system is in state \(n\) at a particular time \(t\).
The formula \(p_n(t)\) given by:\[p_{n}(t)=\frac{1}{\lambda_{n}} \sum_{i=0}^{n} \lambda_{i} e^{-\lambda_{i} t}\prod_{j=0 \atop j eq i}^{n} \frac{\lambda_{j}}{\lambda_{j}-\lambda_{i}}\]presents a way to compute these probabilities.
  • Exponentials \(e^{-\lambda_i t}\) depict the decay in probabilities over time, influenced by each rate \(\lambda_i\).
  • The product term adjusts for the difference between rates, enhancing accuracy when \(\lambda_i eq \lambda_j\).
This function sums contributions from each possible rate, adjusted with products specific to the rate鈥檚 differences, offering a comprehensive model for the birth process.
Inductive reasoning
Inductive reasoning is a logical process used to prove statements for all natural numbers. In mathematical contexts, it is especially useful for proving results like the probability function in our birth process problem.
In this exercise, we solve for the base case and prove the general case using induction:
  • Base Case: Establish \(p_0(t) = e^{-\lambda_0 t}\) directly from the differential equation.
  • Inductive Step: Assuming \(p_k(t)\) for all \(k < n\), we derive \(p_n(t)\) with the integral form \(\frac{1}{\lambda_n} \int_0^t \lambda_{n-1} p_{n-1}(s) e^{-\lambda_n (t-s)} ds\).
With induction, we systematically build from our established base, verifying each step meets the condition from the level before it. This approach provides a full proof by considering individual cases and proving their generality.

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Most popular questions from this chapter

Flies and wasps land on your dinner plate in the manner of independent Poisson processes with respective intensities \(\lambda\) and \(\mu\). Show that the arrivals of flying objects form a Poisson process with intensity \(\lambda+\mu\).

Cars arrive at the beginning of a long road in a Poisson stream of rate \(\lambda\) from time \(t=0\) onwards. A car has a fixed velocity \(V>0\) which is a random variable. The velocities of cars are independent and identically distributed, and independent of the arrival process. Cars can overtake each other freely. Show that the number of cars on the first \(x\) miles of the road at time \(t\) has the Poisson distribution with parameter \(\lambda \mathrm{E}\left[\boldsymbol{V}^{-1} \min (x, V t)\right]\).

Let \(A\) and \(\mathbf{B}=\left(B_{0}, B_{1}, \ldots, B_{n}\right)\) be a discrete random variable and vector, respectively. The conditional entropy of \(A\) with respect to \(\mathrm{B}\) is defined as \(H(A \mid \mathbf{B})=\) \(\mathbb{E}(\mathbb{E}|-\log f(A \mid \mathbf{B})| \mathbf{B} \mid)\) where \(f(a \mid \mathbf{b})=\mathrm{P}(A=a \mid \mathbf{B}=\mathbf{b})\). Let \(X\) be an aperiodic Markov chain on a finite state space. Show that $$ H\left(X_{n+1} \mid X_{0}, X_{1} \ldots, X_{n}\right)=H\left(X_{n+1} \mid X_{n}\right) $$ and that $$ H\left(X_{n+1} \mid X_{n}\right) \rightarrow-\sum_{i} \pi_{i} \sum_{j} p_{i j} \log p_{i j} \quad \text { as } n \rightarrow \infty $$ if \(X\) is aperiodic with a unique stationary distribution \(\pi\).

Let \(\left\\{S_{n}: n \geq 0\right\\}\) be a simple random walk with \(S_{0}=0\), and show that \(X_{n}=\left|S_{n}\right|\) defines a Markov chain; find the transition probabilities of this chain. Let \(M_{n}=\max \left\\{S_{k}: 0 \leq k \leq n \mid\right.\), and show that \(Y_{n}=M_{n}-S_{n}\) defines a Markov chain. What happens if \(S_{0} \neq 0\) ?

Let \(B\) be a process of simple birth with immigration \((6.8 .11 c)\) with parameters \(\lambda\) and \(v\), and with \(B(0)=0\); the birth rates are \(\lambda_{n}=n \lambda+v .\) Write down the sequence of differential-difference equations for \(p_{n}(t)=\mathbb{P}(B(t)=n)\). Without solving these equations, use them to show that \(m(t)=\) \(\mathbb{E}(B(t))\) satisfies \(m^{\prime}(t)=\lambda m(t)+v\), and solve for \(m(t)\)
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