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A physicist, testing the laws of chance, flips a coin repeatedly until it lands tails. (a) Treating the two states of the physicist ("still flipping" and "done") as states in a Markov process. The current probability vector then is \(\vec{\rho}=\left(\begin{array}{c}\rho_{\text {flipping }} \\ \rho_{\text {done }}\end{array}\right) .\) Write the transition matrix \(\mathcal{P}\), giving the time evolution \(\mathcal{P} \cdot \vec{\rho}_{n}=\vec{\rho}_{n+1}\), assuming that the coin is fair. (b) Find the eigenvalues and right eigenvectors of \(\mathcal{P}\). Which eigenvector is the steady state \(\rho^{*}\) ? Call the other eigenvector \(\tilde{\rho} .\) For convenience, normalize \(\tilde{\rho}\) so that its first component equals one. (c) Assume an arbitrary initial state is written \(\rho_{0}=\) \(A \rho^{*}+B \tilde{\rho} .\) What are the conditions on \(A\) and \(B\) needed to make \rhoo a valid probability distribution? Write \(\rho_{n}\) as a function of \(A\) and \(B, \rho^{*}\) and \(\tilde{\rho}\).

Short Answer

Expert verified
The transition matrix is \( \mathcal{P} = \begin{pmatrix} 0.5 & 0 \\ 0.5 & 1 \end{pmatrix} \). The steady state is \( \rho^{*} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \) and \( \rho_n \) depends on \( A \) and \( B \) such that \( A + B = 1 \).

Step by step solution

01

Define the Transition Matrix

The problem describes a Markov process with two states: "still flipping" and "done." Since the coin is fair, the probability of flipping again ("still flipping") is 0.5, and the probability of being "done" on the first tails flip is also 0.5. The transition matrix \( \mathcal{P} \) can be written as: \[ \mathcal{P} = \begin{pmatrix} 0.5 & 0 \ 0.5 & 1 \end{pmatrix} \] The matrix indicates that with a probability of 0.5, the state stays in "still flipping" or transitions to "done".
02

Find the Eigenvalues

To find the eigenvalues of \( \mathcal{P} \), we solve the characteristic equation \( \det(\mathcal{P} - \lambda I) = 0 \). The matrix subtraction results in: \[ \det \begin{pmatrix} 0.5 - \lambda & 0 \ 0.5 & 1 - \lambda \end{pmatrix} = (0.5 - \lambda)(1 - \lambda) - 0 \ = \lambda^2 - 1.5 \lambda + 0.5 = 0 \] Using the quadratic formula, \( \lambda = \frac{1.5 \pm \sqrt{1.5^2 - 4 \times 0.5}}{2} \), which simplifies to roots \( \lambda_1 = 1 \) and \( \lambda_2 = 0.5 \).
03

Find the Eigenvectors

For \( \lambda_1 = 1 \), solve \( (\mathcal{P} - I)v = 0 \) which gives: \[ \begin{pmatrix} -0.5 & 0 \ 0.5 & 0 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \] This results in: \[ 0.5x = 0 \text{ and } 0.5x = 0 \Rightarrow x = 0 \] Thus, \( y \) can be any number. Choosing \( y = 1 \), the eigenvector is \( \rho^{*} = \begin{pmatrix} 0 \ 1 \end{pmatrix} \).For \( \lambda_2 = 0.5 \), solve \( (\mathcal{P} - 0.5I)v = 0 \):\[ \begin{pmatrix} 0 & 0 \ 0.5 & 0.5 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \] This results in: \( 0.5y = 0 \), therefore \( y = 0 \). Choosing \( x = 1 \), the eigenvector \( \tilde{\rho} = \begin{pmatrix} 1 \ 0 \end{pmatrix} \).
04

Determine Conditions for a Valid Probability Distribution

The initial state \( \rho_0 \) can be written as \( \rho_0 = A \rho^{*} + B \tilde{\rho} \). For this to be a valid probability distribution, the components must sum to 1:\[ A \cdot 0 + B \cdot 1 = B = 1 \] And since \( \rho_0 = A \begin{pmatrix} 0 \ 1 \end{pmatrix} + B \begin{pmatrix} 1 \ 0 \end{pmatrix} \), \[ A + 0 = A = 1 - B \] Thus, the conditions are \( A + B = 1 \).
05

Write \( \rho_n \) as a Function of \( A \) and \( B \)

Under the Markov process, the state evolution is given by: \( \rho_{n} = \mathcal{P}^n \rho_0 \) which simplifies to: \( \rho_n = A \lambda_1^n \rho^{*} + B \lambda_2^n \tilde{\rho} \).Given \( \lambda_1 = 1 \) and \( \lambda_2 = 0.5 \), \( \rho_n = A \begin{pmatrix} 0 \ 1 \end{pmatrix} + B(0.5)^n \begin{pmatrix} 1 \ 0 \end{pmatrix} \).This evolution showcases that over time, the components gravitate towards \( \rho^{*} \).

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Key Concepts

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

Transition Matrix
The transition matrix is a crucial component in understanding Markov processes, which describe systems transitioning from one state to another. In this example, we have two states: "still flipping" and "done." These states represent the physicist's progress while flipping a coin. Since the coin is fair, it lands on heads half the time and tails the other half.
The transition matrix, denoted as \( \mathcal{P} \), is constructed to capture these probabilities transitions. It is written as:
  • The probability of staying in the "still flipping" state is 0.5 because the coin could land heads, prompting another flip.
  • The probability of moving to the "done" state is also 0.5, for the tails result, concluding the flips.
  • Once in the "done" state, it remains so because the process terminates.
Thus, our transition matrix for this Markov process is: \[ \mathcal{P} = \begin{pmatrix} 0.5 & 0 \ 0.5 & 1 \end{pmatrix} \] This matrix helps understand how probabilities transition between states with each coin flip.
Eigenvalues and Eigenvectors
Eigenvalues and eigenvectors of a transition matrix provide insights into the long-term behavior of the Markov process. Calculating these involves solving the characteristic equation \( \det(\mathcal{P} - \lambda I) = 0 \). Here, \( \lambda \) represents the eigenvalues, and the solutions yield insights into the system's stability and steady state.
  • For this problem, solving gives us eigenvalues \( \lambda_1 = 1 \) and \( \lambda_2 = 0.5 \).
  • These values tell us how each state influences the future states. An eigenvalue of 1 indicates a component that remains unchanged over iterations.
  • Corresponding to each eigenvalue is an eigenvector, which dictates the direction of influence in the state space.
For \( \lambda_1 = 1 \), the eigenvector \( \rho^{*} = \begin{pmatrix} 0 \ 1 \end{pmatrix} \) denotes the steady state as the coin flips eventually end in a "done" state. The other eigenvector \( \tilde{\rho} = \begin{pmatrix} 1 \ 0 \end{pmatrix} \) relates to the transitional state of "still flipping." Normalizing the eigenvectors ensures they represent valid physical probabilities when analyzing any arbitrary initial conditions.
Probability Distribution
In modeling this Markov process, the goal is to find a condition on initial probabilities that ensure they sum to one, forming a valid probability distribution. Given any initial state \( \rho_0 = A \rho^{*} + B \tilde{\rho} \), the conditions for \( A \) and \( B \) are essential.
  • We require that the sum \( A + B = 1 \) to ensure the two components represent a complete probability space.
  • \( A \) represents the contribution of the steady state to the initial condition, while \( B \) represents the contribution from the transitional state.
As the process evolves with each coin flip, the state \( \rho_n \) also evolves as follows: \[ \rho_n = A \begin{pmatrix} 0 \ 1 \end{pmatrix} + B(0.5)^n \begin{pmatrix} 1 \ 0 \end{pmatrix} \] This expression shows how the steady state gradually dominates, rendering \( \rho^{*} \) as the focus over time, while the influence of \( \tilde{\rho} \) diminishes due to its smaller eigenvalue \( \lambda_2 = 0.5 \), making this system converge to a state of completion after enough flips.

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Most popular questions from this chapter

Living cells are amazingly complex mixtures of a variety of complex molecules (RNA, DNA, proteins, lipids \(\ldots\) that are constantly undergoing reactions with one another. This complex of reactions has been compared to computation: the cell gets input from external and internal sensors, and through an intricate series of reactions produces an appropriate response. Thus, for example, receptor cells in the retina 'listen' for light and respond by triggering a nerve impulse. The kinetics of chemical reactions are usually described using differential equations for the concentrations of the various chemicals, and rarely are statistical fluctuations considered important. In a cell, the numbers of molecules of a given type can be rather small: indeed, there is (often) only one copy of the relevant part of DNA for a given reaction. It's an important question whether and when we may describe the dynamics inside the cell using continuous concentration variables, even though the actual numbers of molecules are always integers. Consider a dimerization reaction: a molecule \(M\) (called the "monomer") joins up with another monomer and becomes a dimer \(D: 2 M \longleftrightarrow D\). Proteins in cells often form dimers: sometimes (as here) both proteins are the same (homodimers) and sometimes they are different proteins (heterodimers). Suppose the forward reaction rate is \(k_{u}\) and the backward reaction rate is \(k_{d}\). Figure \(8.11\) shows this as a Petri net [39] with each reaction shown as a box with incoming arrows showing species that are consumed by the reaction, and outgoing arrows showing species that are produced by the reaction: the number consumed or produced (the stoichiometry) is given by a label on each arrow. \(^{39}\) There are thus two reactions: the backward unbinding reaction rate per unit volume is \(k_{u}[D]\) (each dimer disassociates with rate \(k_{u}\) ), and the forward binding reaction rate per unit volume is \(k_{b}[M]^{2}\) (since each monomer must wait for a collision with another monomer before binding, the rate is proportional to the monomer concentration squared). \(^{40}\) The brackets [] denote concentrations. We assume, as does reference \([29]\), that the volume per cell is such that one molecule per cell is \(1 \mathrm{nM}\left(10^{-9}\right.\) moles per liter). For convenience, we shall pick nanomoles as our unit of concentration, so \([M]\) is also the number of monomers in the cell. Assume \(k_{b}=1 n M^{-1} s^{-1}\) and \(k_{u}=2 s^{-1}\), and that at \(t=0\) all \(N\) monomers are unbound. (a) Continuum dimerization. Write the differential equation for \(d M /\) dt treating \(M\) and \(D\) as continuous variables. (Hint: remember that two \(M\) molecules are consumed in each reaction.) What are the equilibrium concentrations for \([M]\) and \([D]\) for \(N=2\) molecules in the cell, assuming these continuous equations and the values above for \(k_{b}\) and \(k_{u}\) ? For \(N=90\) and \(N=10100\) molecules? Numerically solve your differential equation for \(M(t)\) for \(N=2\) and \(N=90\), and verify that your solution settles down to the equilbrium values you found. For large numbers of molecules in the cell, we expect that the continuum equations may work well, but for just a few molecules there surely will be relatively large fluctuations. These fluctuations are called shot noise, named in early studies of electrical noise at low currents due to individual electrons in a resistor. We can implement a Monte Carlo algorithm to simulate this shot noise. \({ }^{41}\) Suppose the reactions have rates \(\Gamma_{i}\), with total rate \(\Gamma_{\text {tot }}=\sum_{i} \Gamma_{i}\). The idea is that the expected time to the next reaction is \(1 / \Gamma_{\text {tot }}\), and the probability that the next reaction will be \(j\) is \(\Gamma_{j} / \Gamma_{\text {tot }}\). To simulate until a final time \(t_{f}\), the algorithm runs as follows: (a) Calculate a list of the rates of all reactions in the system. (b) Find the total rate \(\Gamma_{\text {tot }}\). (c) Pick a random time \(t_{\text {wait }}\) with probability distribution \(\rho(t)=\Gamma_{\text {tot }} \exp \left(-\Gamma_{\text {tot }} t\right)\) (d) If the current time \(t\) plus \(t_{\text {wait }}\) is bigger than \(t_{f}\), no further reactions will take place: return. (e) Otherwise, \- Increment \(t\) by \(t_{\text {wait }}\), \- Pick a random number \(r\) uniformly distributed in the range \(\left[0, \Gamma_{\text {tot }}\right)\), \- Pick the reaction \(j\) for which \(\sum_{i

Reading: Reference [29], Michael B. Elowitz and Stanislaw Leibler, "A synthetic oscillator network of transcriptional regulators" Nature \(\mathbf{4 0 3}, 335-338(2000)\). The 'central dogma' of molecular biology is that the flow of information is from DNA to RNA to proteins: DNA is transcribed into RNA, which then is translated into protein. Now that the genome is sequenced, it is thought that we have the parts list for the cell. 'All' that remains is to figure out how they work together. The proteins, RNA, and DNA form a complex network of interacting chemical reactions, which governs metabolism, responses to external stimuli, reproduction (proliferation), differentiation into different cell types, and (when the system perceives itself to be breaking down in dangerous ways) programmed cell death, or apoptosis. Our understanding of the structure of these interacting networks is growing rapidly, but our understanding of the dynamics is still rather primitive. Part of the difficulty is that the cellular networks are not neatly separated into different modules: a given protein may participate in what would seem to be several separate regulatory pathways. In this exercise, we will study a model gene regulatory network, the Repressilator. This experimental system involves three proteins each of which inhibits the formation of the next. They were added to the bacterium \(E .\) coli, with hopefully minimal interactions with the rest of the biological machinery of the cell. We will implement the stochastic model that the authors used to describe their experimental system [29], in order to \- Implement in a tangible system an example both of the central dogma and of transcriptional regulation: the control by proteins of DNA expression into RNA, \- Introduce sophisticated Monte-Carlo techniques for simulations of stochastic reactions, \- Introduce methods for automatically generating continuum descriptions from reaction rates, and \- Illustrate the shot noise fluctuations due to small numbers of molecules and the telegraph noise fluctuations due to finite rates of binding and unbinding of the regulating proteins. Figure \(8.12\) shows the biologist's view of the repressilator network. Three proteins (TetR, \(\lambda\) CI, and LacI) each repress the formation of the next. We shall see that, under appropriate circumstances, this can lead to spontaneous oscillations: each protein peaks in turn, suppressing the suppressor of its suppressor, leading to its own later decrease.

Physical systems usually evolve continuously in time: their laws of motion are differential equations. Computer simulations must approximate these differential equations using discrete time steps. In this exercise, we will introduce some common methods for simulating differential equations using the example of the pendulum: $$ \frac{d^{2} \theta}{d t^{2}}=\ddot{\theta}=-(g / L) \sin (\theta) $$ This equation gives the motion of a pendulum with a point mass at the tip of a massless \(\operatorname{rod}^{47}\) of length \(L\) : rederive it using a free body diagram. Go to our Web site \([108]\) and download the pendulum files for the language you'll be using. The animation should show a pendulum oscillating from an initial condition \(\theta_{0}=2 \pi / 3, \dot{\theta}=0 ;\) the equations being solved have \(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\) and \(L=1 \mathrm{~m}\) There are three independent criteria for picking a good algorithm for solving differential equations: fidelity, \(a c\) curacy, and stability. Fidelity. Notice that in our time step algorithm, we did not do the straightforward choice - using the current \((\theta(t), \dot{\theta}(t))\) to produce \((\theta(t+\delta), \dot{\theta}(t+\delta))\). Rather, we used \(\theta(t)\) to calculate the acceleration and update \(\dot{\theta}\), and then used \(\dot{\theta}(t+\delta)\) to calculate \(\theta(t+\delta)\) $$ \begin{aligned} &\dot{\theta}(t+\delta)=\dot{\theta}(t)+\ddot{\theta}(t) \delta \\ &\theta(t+\delta)=\theta(t)+\dot{\theta}(t+\delta) \delta \end{aligned} $$ Wouldn't it be simpler and make more sense to update \(\theta\) and \(\dot{\theta}\) simultaneously from their current values, so \(\theta(t+\delta)=\theta(t)+\dot{\theta}(t) \delta ?\) (This simplest of all timestepping schemes is called the Euler method, and should not be used for ordinary differential equations (although it is sometimes used in partial differential equations.) (a) Try it. First, see why reversing the order of the updates to \(\theta\) and \(\dot{\theta}\), $$ \begin{aligned} &\theta(t+\delta)=\theta(t)+\dot{\theta}(t) \delta \\ &\dot{\theta}(t+\delta)=\dot{\theta}(t)+\ddot{\theta}(t) \delta \end{aligned} $$ in our loop would give us a simultaneous update. Swap these two lines in the code, and watch the pendulum swing for several turns, until it starts looping the loop. Is the new algorithm as good as the old one? (Make sure you switch the two lines back afterwards.) The simultaneous update scheme is just as accurate as the one we chose, but it is not as faithful to the physics of the problem: its fidelity is not as good. For subtle reasons we won't explain here, updating first \(\dot{\theta}\) and then \(\theta\) allows our algorithm to exactly conserve an approximation to the energy: it's called a symplectic algorithm. \(^{48}\) Improved versions of this algorithm - like the Verlet algorithms below - are often used to simulate systems that conserve energy (like molecular dynamics) because they exactly \(^{49}\) simulate the dynamics for an approximation to the Hamiltonian - preserving important physical features not kept by just approximately solving the dynamics. Accuracy. Most computational methods for solving differential equations (and many other continuum problems like integrating functions) involve a step size \(\delta\), and become more accurate as \(\delta\) gets smaller. What is most important is not the error in each time step, but the accuracy of the answer after a fixed time \(T\), which is the accumulated error after \(T / \delta\) time steps. If this accumulated error varies as \(\delta^{n}\), we say that the algorithm has \(n^{\text {th }}\) order cumulative accuracy. Our algorithm is not very high order! (b) Plot the pendulum trajectory \(\theta(t)\) for time steps \(\delta=\) \(0.1,0.01\), and \(0.001\). Zoom in on the curve at one of the coarse points (say, \(t=1)\) and compare the values from the three time steps. Does it appear that this time is converging \(^{50}\) as \(\delta \rightarrow 0\) ? From your measurement, what order accuracy is our method? We can write higher-order symplectic algorithms. The approximation to the second derivative $$ \ddot{\theta} \approx(\theta(t+\delta)-2 \theta(t)+\theta(t-\delta)) / \delta^{2} $$ (which you can verify with a Taylor expansion is correct to \(\left.O\left(\delta^{4}\right)\right)\) motivates the Verlet Algorithm $$ \theta(t+\delta)=2 \theta(t)-\theta(t-\delta)+\ddot{\theta} \delta^{2} $$ This algorithm is a bit awkward to start up since you need to initialize \(^{51} \theta(t-\delta)\); it's also often convenient to know the velocities as well as the positions. The Velocity Verlet algorithm fixes both of these problems; it is motivated by the constant acceleration formula \(x(t)=x_{0}+v_{0} t+1 / 2 a t^{2}\) : $$ \begin{array}{r} \theta(t+\delta)=\theta(t)+\dot{\theta}(t) \delta+1 / 2 \ddot{\theta}(t) \delta^{2} \\ \dot{\theta}(t+\delta / 2)=\dot{\theta}(t)+{ }^{1} \ddot{2} \ddot{\theta}(t) \delta \\ \dot{\theta}(t+\delta)=\dot{\theta}(t+\delta / 2)+1 / 2 \ddot{\theta}(t+\delta) \delta \end{array} $$ The trick that makes this algorithm so good is to cleverly split the velocity increment into two pieces, half for the acceleration at the old position and half for the new position. \(^{52}\) (You'll want to initialize \(\ddot{\theta}\) once before starting the loop.) (c) Pick one of the Verlet algorithms, implement it, and plot the trajectory for time steps \(\delta=0.1,0.01\), and \(0.001\). You should see a dramatic improvement in convergence. What cumulative order accuracy does Verlet have? \(^{253}\) Stability. In many cases high accuracy is not crucial. What prevents us from taking enormous time steps? In a given problem, there is usually a typical fastest time scale: a vibration or oscillation period (as in our exercise) or a growth or decay rate. When our time step becomes a substantial fraction of this fastest time scale, algorithms like ours usually become unstable: the first few time steps may be fairly accurate, but small errors build up until the errors become unacceptable (indeed, often ones first warning of problems are machine overflows). (d) Plot the pendulum trajectory \(\theta(t)\) for time steps \(\delta=\) \(0.1,0.2, \ldots, 0.8\), using a small amplitude oscillation \(\theta_{0}=0.01, \dot{\theta}_{0}=0.0\), up to \(t_{\max }=10 .\) At about what \(\delta_{c}\) does it go unstable? Looking at the first few points of the trajectory, does it seem like sampling the curve at steps much larger than \(\delta_{c}\) would miss the oscillations? At \(\delta_{c} / 2\), how accurate is the amplitude of the oscillation? (You'll) need to observe several periods in order to estimate the maximum amplitude of the solution.) In solving the properties of large, nonlinear systems (e.g., partial differential equations (PDEs) and molecular dynamics) stability tends to be the key difficulty. The maximum stepsize depends on the local configuration, so highly nonlinear regions can send the system unstable before one might expect. The maximum safe stable stepsize often has accuracy far higher than needed; indeed, some algorithms become less stable if the stepsize is decreased \(!^{54}\) ODE packages: higher order, variable stepsize, stiff systems ... The Verlet algorithms are not hard to code, and we use higher-order symplectic algorithms in Hamiltonian systems mostly in unusual applications (planetary motion) where high accuracy is demanded, because they are typically significantly less stable. In systems of differential equations where there is no conserved energy or Hamiltonian, or even in Hamiltonian systems (like high- energy collisions) where accuracy at short times is more crucial than fidelity at long times, we use general purpose methods. The general-purpose solvers come in a variety of basic algorithms (Runge- Kutta, predictor-corrector, ...), and methods for maintaining and enhancing accuracy (variable step size, Richardson extrapolation). There are also implicit methods for stiff systems. A system is stiff if there is a large separation between the slowest and fastest relevant time scales: implicit methods often allow one to take time steps much larger than the fastest time scale (unlike the explicit Verlet methods you studied in part (d), which go unstable). Large, sophisticated packages have been developed over many years for solving differential equations - switching between algorithms and varying the time steps to most efficiently maintain a given level of accuracy. They solve \(d \mathbf{y} / d t=\operatorname{dyd} \mathbf{t}(\mathbf{y}, t)\), where for us \(\mathbf{y}=[\theta, \dot{\theta}]\) and \(\mathbf{d y d} \mathbf{t}=[\dot{\theta}, \ddot{\theta}] .\) They typically come in the form of subroutines or functions, which need as arguments \- Initial conditions \(\mathbf{y}_{0}\) \- The right-hand side dydt, a function of the vector \(\mathbf{y}\) and time \(t\), which returns a vector giving the current rate of change of \(\mathbf{y}\), and \- The initial and final times, and perhaps intermediate times, at which the trajectory \(\mathbf{y}(t)\) is desired. They often have options that \- Ask for desired accuracy goals, typically a relative (fractional) accuracy and an absolute accuracy, sometimes set separately for each component of \(\mathbf{y}\), \- Ask for and return derivative and time step information from the end of the last step (to allow efficient restarts after intermediate points), \- Ask for a routine that computes the derivatives of dydt with respect to the current components of \(\mathbf{y}\) (for use by the stiff integrator), and \- Return information about the methods, time steps, and performance of the algorithm. You will be supplied with one of these general-purpose packages, and instructions on how to use it. (e) Write the function dydt, and use the general purpose solver to solve for the motion of the pendulum as in parts (a)- \((c)\), and informally check that the trajectory is accurate.

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