Some questions about Brownian Motion and Birth-Death in Markov chain

In summary: A(t)=e^{-\frac{t^{2}}{2}}\ (4)$... let $B(t)$ be the Brownian motion corresponding to the equation $A(t) = B(t)$.... then $B(t)$ is a random variable with pdf... $\displaystyle B(t) = e^{-\frac{t^{2}}{2}}\ (5)$... and the distribution of $B(t)$ is... $\displaystyle B(t) = \frac{1}{\sqrt{2\ \pi\ t}}\
  • #1
power3173
5
0
Hi,

I need urgent answers. Basically, I don't have background in Markov and I don't need to learn it now actually. But I have to solve the questions below somehow. If somebody can give detailed answers to the questions below (From beginning to the final solution with explanations), then I will show a very big appreciation to that person.

1) An urn contains totally N balls. Some balls are black and the other balls are white. A ball is chosen at random at time points whose spacings are independent and exponentially distributed with intensity 3 per minute. The chosen ball is immediately replaced by a ball of the other color. Let Xt denote the number of white balls in the urn at time t.
a) The process {X(t)} is a birth-death process. Find the birth and death rates.
b) Find the stationary distribution. In particular, determine the proportion of time all balls in the urn are white.

2) Give an example of birthrates λ(i) > 0 such that a a birth process {X(t)}t≥0 with birthrates λ(i), i ≥ 0 explodes on average at time t = 17.

3) a) Let Z ∈ N(0, 1) be a standard normal random variable. The process X(t) = sqrt(t)*Z is distributed as a normal random variable at every time, X(0) = 0, and it has continuous trajectories. Is {X(t)} a Brownian motion? (motivate your answer)

b) Suppose {B(t)} and {B˜(t)} are two independent standard Brownian motions and ρ is a constant, −1 < ρ < 1. The process Y(t) = ρB(t) + sqrt(1- ρ^2)*B˜(t) is distributed as a normal random variable at every time, Y0 = 0, and it has continuous trajectories. Is {Y(t)} a Brownian motion? (motivate your answer)
 
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  • #2
power3173 said:
Hi,

I need urgent answers. Basically, I don't have background in Markov and I don't need to learn it now actually. But I have to solve the questions below somehow. If somebody can give detailed answers to the questions below (From beginning to the final solution with explanations), then I will show a very big appreciation to that person.

1) An urn contains totally N balls. Some balls are black and the other balls are white. A ball is chosen at random at time points whose spacings are independent and exponentially distributed with intensity 3 per minute. The chosen ball is immediately replaced by a ball of the other color. Let Xt denote the number of white balls in the urn at time t.
a) The process {X(t)} is a birth-death process. Find the birth and death rates.
b) Find the stationary distribution. In particular, determine the proportion of time all balls in the urn are white.

2) Give an example of birthrates λ(i) > 0 such that a a birth process {X(t)}t≥0 with birthrates λ(i), i ≥ 0 explodes on average at time t = 17.

3) a) Let Z ∈ N(0, 1) be a standard normal random variable. The process X(t) = sqrt(t)*Z is distributed as a normal random variable at every time, X(0) = 0, and it has continuous trajectories. Is {X(t)} a Brownian motion? (motivate your answer)

b) Suppose {B(t)} and {B˜(t)} are two independent standard Brownian motions and ρ is a constant, −1 < ρ < 1. The process Y(t) = ρB(t) + sqrt(1- ρ^2)*B˜(t) is distributed as a normal random variable at every time, Y0 = 0, and it has continuous trajectories. Is {Y(t)} a Brownian motion? (motivate your answer)

Let's start with step 1) ... limiting ourselves to the population of white balls, calling $\lambda_{n}$ the birth probability and $\mu_{n}$ the death probability for each state with n=0,1,...,N You have...

$\displaystyle \lambda_{n} = \frac{N - n}{N}$

$\mu_{n} = \frac{n}{N}\ (1)$

Calling $p_{n} (t)$ the probability to be in the state n at the time t, the steady state value of the $p_{n}$ are given by...

$\displaystyle p_{0} = (1 + \sum_{n=1}^{N} \frac{\lambda_{0}\ \lambda_{1}\ ...\ \lambda_{n-1}}{\mu{1}\ \mu_{2}\ ...\ \mu_{n}})^{-1} = (\sum_{n = 0}^{N} \binom{N}{n} )^{-1} = 2^{- N}\ (2)$

$\displaystyle p_{n} = \frac{\lambda_{0}\ \lambda_{1}\ ...\ \lambda_{n-1}}{\mu{1}\ \mu_{2}\ ...\ \mu_{n}}\ p_{0} = \binom {N}{n}\ 2^{- N}\ (3)$

... so that the distribution is binomial...

Kind regards

$\chi$ $\sigma$
 
  • #3
Question 3 a) : Let Z ∈ N(0, 1) be a standard normal random variable. The process X(t) = sqrt(t)*Z is distributed as a normal random variable at every time, X(0) = 0, and it has continuous trajectories. Is {X(t)} a Brownian motion? ...(motivate your answer)

A Brownian motion is a stochastic process $W_{t}$ with p.d.f. ...

$\displaystyle f_{W} (x,t) = \frac{1}{\sqrt{2\ \pi\ t}}\ e^{- \frac{x^{2}}{2\ t}}\ (1)$

... i.e. with mean $\mu = 0$ and $\sigma^{2} = t$ ...

If Z is a standard r.v. with mean $\mu = 0$ and $\sigma^{2}=1$ and $X = a(t)\ Z$ is a stochastic, then X has mean $\mu=0$ and $\sigma^{2} = a^{2} (t)$, so that if $a(t)=\sqrt{t}$, then X is a Brownian motion...

Kind regards

$\chi$ $\sigma$
 
  • #4
Question 3 b): Suppose {A(t)} and {B(t)} are two independent standard Brownian motions and ρ is a constant, −1 < ρ < 1. The process Y(t) = ρ A(t) + sqrt(1- ρ^2)*B(t) is distributed as a normal random variable at every time, Y0 = 0, and it has continuous trajectories. Is {Y(t)} a Brownian motion? (motivate your answer)

If A
(t) and B(t) are both Brownian and independent, then $\rho\ A(t)$ has variance $\rho^{2}\ t$ and $\sqrt{1 - \rho^{2}}\ B(t)$ has variance $(1 - \rho^{2})\ t$, so that Y(t) has variance $(\rho^{2} + 1 - \rho^{2})\ t = t$ and it is also Brownian...

Kind regards

$\chi$ $\sigma$
 
  • #5
chisigma said:
Let's start with step 1) ... limiting ourselves to the population of white balls, calling $\lambda_{n}$ the birth probability and $\mu_{n}$ the death probability for each state with n=0,1,...,N You have...

$\displaystyle \lambda_{n} = \frac{N - n}{N}$

$\mu_{n} = \frac{n}{N}\ (1)$

Calling $p_{n} (t)$ the probability to be in the state n at the time t, the steady state value of the $p_{n}$ are given by...

$\displaystyle p_{0} = (1 + \sum_{n=1}^{N} \frac{\lambda_{0}\ \lambda_{1}\ ...\ \lambda_{n-1}}{\mu{1}\ \mu_{2}\ ...\ \mu_{n}})^{-1} = (\sum_{n = 0}^{N} \binom{N}{n} )^{-1} = 2^{- N}\ (2)$

$\displaystyle p_{n} = \frac{\lambda_{0}\ \lambda_{1}\ ...\ \lambda_{n-1}}{\mu{1}\ \mu_{2}\ ...\ \mu_{n}}\ p_{0} = \binom {N}{n}\ 2^{- N}\ (3)$

... so that the distribution is binomial...

Firstly, again thanks a lot for your answers.

Just some points I didn't understant completely.

I think P(0) and P(n) are the stationary distributions, right? Then what can we say about birth and death rates?

And what about the question 1.b, the proportion of time all balls in the urn are white?

Thanks...
 
  • #6
power3173 said:
Firstly, again thanks a lot for your answers.

Just some points I didn't understant completely.

I think P(0) and P(n) are the stationary distributions, right? Then what can we say about birth and death rates?

And what about the question 1.b, the proportion of time all balls in the urn are white?

Thanks...

The birth and death rate in the unit time [in this case 20 seconds...] are given by...

$\displaystyle R_{b} = R_{d} = \sum_{n=0}^{N} \lambda_{n}\ p_{n} = \sum_{n=0}^{N} \mu_{n}\ p_{n} = 2^{- N}\ \sum_{n=0}^{n} \binom {N-1}{n-1} = 2^{- N}\ 2^{N-1} = \frac{1}{2}\ (1)$

The proportion of time in which alla balls are white is...

$\displaystyle p_{N} = 2^{- N}\ (2)$

Kind regards

$\chi$ $\sigma$
 

FAQ: Some questions about Brownian Motion and Birth-Death in Markov chain

What is Brownian Motion?

Brownian Motion is a type of random motion exhibited by particles in a fluid or gas. It was first observed by botanist Robert Brown in 1827 and was later explained by physicist Albert Einstein in 1905. It is caused by the random collisions of the particles with the molecules of the fluid or gas they are suspended in.

How is Brownian Motion related to Markov chain?

Brownian Motion can be modeled using a Markov chain, which is a mathematical model that describes a sequence of events where the probability of transitioning from one state to another depends only on the current state and not on the previous states. In Brownian Motion, the position of the particle at a given time depends only on its position at the previous time step, making it a suitable application of a Markov chain.

What is Birth-Death process in Markov chain?

Birth-Death process is a type of Markov chain where the states represent the number of individuals in a population and the transitions between states represent births and deaths. It is commonly used to model population growth or decline and can be used to study various biological and social phenomena.

How is Birth-Death process related to Brownian Motion?

Birth-Death process is closely related to Brownian Motion because both models involve a random process and can be described using a Markov chain. In fact, Brownian Motion can be seen as a continuous-time version of a Birth-Death process, where the number of individuals in the population is continuously changing instead of discrete changes in the Birth-Death process.

What are the practical applications of studying Brownian Motion and Birth-Death process?

The study of Brownian Motion and Birth-Death process has many practical applications in various fields. In physics, it is used to study diffusion and thermal conductivity. In biology, it can be used to model population dynamics and the spread of diseases. In finance, it is used to model stock prices and other economic factors. Understanding these processes can help us make predictions and better understand the behavior of complex systems.

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