A new derivation of the randomness parameter

JOURNAL OF MATHEMATICAL PHYSICS 48, 103301 共2007兲

A new derivation of the randomness parameter Hongyun Wanga兲 Department of Applied Mathematics and Statistics, University of California, Santa Cruz, California 95064, USA 共Received 18 April 2007; accepted 15 September 2007; published online 10 October 2007兲

For a stochastic stepper that can only step forward, there are two randomnesses: 共1兲 the randomness in the cycle time and 共2兲 the randomness in the number of steps 共cycles兲 over long time. The equivalence between these two randomnesses was previously established using the approach of Laplace transform 关M. J. Schnitzer and S. M. Block, “Statistical kinetics of processive enzymes,” Cold Spring Harbor Symp. Quant. Biol. 60, 793 共1995兲兴. In this study, we first discuss the problems of this approach when the cycle time distribution has a discrete component, and then present a new derivation based on the framework of semi-Markov processes with age structure. We also show that the equivalence between the two randomnesses depends on the existence of the first moment of the waiting time for completing the first cycle, which is strongly affected by the initial age distribution. Therefore, any derivation that concludes the equivalence categorically regardless of the initial age distribution is mathematically questionable. © 2007 American Institute of Physics. 关DOI: 10.1063/1.2795215兴

I. INTRODUCTION

Many molecular motors move in steps. For example, a kinesin dimer walks on a microtubule filament in 8 nm steps.1–4 For a kinesin dimer, each motor step is coupled to one adenosine triphosphate 共ATP兲 hydrolysis cycle.1,5 Assuming at the end of each reaction cycle, the kinesin dimer returns to the same statistical configuration as the one at the beginning of the cycle with the roles of two heads switched; in other words, each reaction cycle starts with the same statistical configuration, then each reaction cycle is independent of any other reaction cycle. This allows us to model the kinesin dimer as a stochastic stepper, a special case of semi-Markov processes,6 in which the cycle time of each motor step has the same distribution and is statistically independent of that of any other step.7 The distribution of the cycle time, however, can be arbitrary and is not necessarily exponential. The framework of stochastic steppers is a very general one. It can accommodate backward motor steps and variable motor step sizes.8 In this study, we focus on the simple case where the stochastic stepper can only move forward and the step size is uniform. As a first step in peeking into the chemical reaction inside the motor and deciphering the motor mechanism, one would like to obtain the statistics of the cycle time. A key statistical quantity of the motor that one would like to know is the randomness in the cycle time, which is defined as the variance of cycle time divided by the square of the average cycle time.9,7 Unfortunately, with the current single molecule experimental technologies, it is still not possible to observe/record the position of motor directly and accurately. If the motor position could be measured directly and accurately, then it would be straightforward to collect a large number of samples of the cycle time. As shown in Fig. 1, in many single molecule motor experiments, the position of motor is inferred by following the position of a large cargo 共a latex bead兲 that is elastically linked to the motor.1,10,2,11 Although certain behaviors of the motor can still be observed by following the cargo position, it is difficult to extract accurate sample values of cycle time from the time series of

a兲

Electronic mail: [email protected]

0022-2488/2007/48共10兲/103301/18/$23.00

48, 103301-1

© 2007 American Institute of Physics

Downloaded 19 Oct 2007 to 128.114.49.85. Redistribution subject to AIP license or copyright, see http://jmp.aip.org/jmp/copyright.jsp

103301-2

Hongyun Wang

J. Math. Phys. 48, 103301 共2007兲

FIG. 1. Schematic diagram of a motor-cargo system in single molecule experiments.1,10,2,11 The motor moves on a polymer track in steps. Each motor step corresponds to one reaction cycle and is statistically independent of any other steps. Mathematically, the motor is modeled as a stochastic stepper. The motor is elastically linked to a cargo that is much larger than the motor, and can be controlled and monitored by a laser trap. The motor behaviors are inferred from the measurements of the cargo.

cargo positions. On the positive side, the motor and the cargo are linked. Consequently, over long time, they must have asymptotically the same average displacement and the same variance in displacement. Thus, the average velocity and the effective diffusion coefficient of the motor can be calculated from the measured time series of cargo positions. Since the statistics of motor displacement over long time can be measured reliably, we can consider the randomness in the number of motor steps over long time, which is defined as the variance of number of motor steps divided by the average number of steps. In Ref. 7, it was established that the randomness in the number of motor steps is the same as the randomness in the cycle time. This result is very significant in that it relates what we can measure externally from the time series of cargo positions to what we want to know about the internal motor operation. The equivalence between the two randomnesses was derived using the approach of Laplace transform.7 As we will show below, when the cycle time distribution has a discrete component the approach of Laplace transform breaks down but amazingly the equivalence between the two randomnesses still holds. In this study, we present a new derivation for the equivalence between the two randomnesses. The new derivation is based on the framework of semi-Markov processes with age structure.12,13 More precisely, we prove the equivalence between these two randomnesses for the general case where the cycle time of regular cycles has the first two moments and the waiting time for completing the first cycle has the first moment. We also give an example to show that the two randomnesses are different if the first moment of the waiting time for completing the first cycle is not finite. The waiting time for completing the first cycle is strongly affected by the initial age distribution. Therefore, any serious attempt on proving the equivalence between these two randomnesses must take into consideration the initial age distribution. In other words, any derivation that concludes the equivalence categorically regardless of the initial age distribution is doomed mathematically not rigorous, and consequently, such a derivation should not be viewed as a rigorous substitution for the derivation presented here. The rest of the paper is organized as follows. In Sec. II, we first describe the renewal model for molecular motors and review the equivalence proof using the approach of Laplace transform. Then we present a counter example to show how the approach of Laplace transform breaks down. In Sec. III, we present a new derivation for the equivalence between the two randomnesses. To facilitate the new proof, the renewal model introduced in Sec. II is described again in the framework of semi-Markov processes with age structure. The new proof is based on this framework. In Sec. IV, we give a special example to demonstrate that the initial age distribution does affect the validity of the equivalence. II. RANDOMNESS PARAMETER

As shown in Fig. 1, we consider a motor-cargo system in which 共a兲 the motor goes forward stochastically in steps of uniform size, 共b兲 each motor step corresponds to one reaction cycle and

Downloaded 19 Oct 2007 to 128.114.49.85. Redistribution subject to AIP license or copyright, see http://jmp.aip.org/jmp/copyright.jsp

103301-3

A new derivation of the randomness parameter

J. Math. Phys. 48, 103301 共2007兲

is independent of any other steps, 共c兲 the cycle time of every step has the same probability density function p共t兲, and 共d兲 the motor and the cargo are elastically linked whereas only the cargo behaviors are measurable in experiments. Let L be the motor step size, and T be the random cycle time of a motor step, i.e., the time it takes to complete a motor step. Let p共t兲 denote the probability density of the random cycle time T. Mathematically, that is, p共t兲 = lim

⌬t→0

Pr关t 艋 T ⬍ t + ⌬t兴 . ⌬t

共1兲

The randomness in the cycle time is defined as RT ⬅

var共T兲 , 具T典2

共2兲

where 具T典 denotes the mean and var共T兲 = 具T2典 − 具T典2 denotes the variance of random cycle time T. As we pointed out above, RT informs us about the internal motor operation but RT is not directly measurable in experiments. A quantity that can be measured in experiments is the randomness in the number of steps over long time. Let N共t兲 be the stochastic number of steps the motor has finished by time t, and Y共t兲 be the stochastic position of the cargo at time t. The stochastic position of the motor at time t is X共t兲 = N共t兲L. Since the motor and the cargo are linked, over long time, X共t兲 and Y共t兲 must have asymptotically the same mean and the same variance. The randomness in the number of steps is defined below and can be calculated from Y共t兲:9,14 RN ⬅ lim

t→⬁

var共N共t兲兲 var共X共t兲兲 var共Y共t兲兲 = lim = lim . 具N共t兲典 t→⬁ 具X共t兲典L t→⬁ 具Y共t兲典L

共3兲

If each reaction cycle is simply a Markov step with transition rate r, then the cycle time has the exponential distribution, p共t兲 = re−rt, and the number of steps finished by time t has the Poisson distribution, Pr关N共t兲 = n兴 = ␭ne−␭ / n!, where ␭ = rt. It follows that 具T典2 = var共T兲 = 1 / r2, and 具N共t兲典 = var共N共t兲兲 = ␭. Thus, we have RT = RN for this special case. In Ref. 7, RT = RN was established for the general case using the approach of Laplace transform. Below we first review the derivation of RT = RN in Ref. 7. Then we discuss the problems of this approach. In the next section, we present a new derivation of RT = RN. Suppose the motor has just arrived at step 0 at time 0. This specification is necessary because the advance from step 0 to step 1 may not be a Markov step and the motor system does remember how long it has been waiting at step 0. Let P共n , t兲 ⬅ Pr关N共t兲 = n兴. For n 艌 0, P共n , t兲 satisfies P共0,t兲 =

冕

⬁

p共␶兲d␶ ,

t

P共n + 1,t兲 =

冕

t

P共n,t − ␶兲p共␶兲d␶ .

共4兲

0

Let ˜f 共s兲 denote the Laplace transform of f共t兲: ˜f 共s兲 ⬅ L关f共t兲兴 ⬅ 兰⬁0 e−st f共t兲dt. Taking the Laplace transform of Eq. 共4兲 gives us ˜P共0,s兲 = 1 − ˜p共s兲 , s ˜P共n + 1,s兲 = ˜P共n,s兲p ˜ 共s兲.

共5兲

Applying Eq. 共5兲 recursively, we obtain an explicit expression for ˜P共n , s兲:

Downloaded 19 Oct 2007 to 128.114.49.85. Redistribution subject to AIP license or copyright, see http://jmp.aip.org/jmp/copyright.jsp

103301-4

J. Math. Phys. 48, 103301 共2007兲

Hongyun Wang

˜P共n,s兲 = 1 − ˜p共s兲 ˜pn共s兲. s

共6兲

Expressing 具N共t兲典 and 具N2共t兲典 in terms of P共n , t兲, we have ⬁

具N共t兲典 = 兺 nP共n,t兲, n=0

⬁

具N 共t兲典 = 兺 n2 P共n,t兲.

共7兲

2

n=0

Taking the Laplace transform of 具N共t兲典 and 具N2共t兲典 yields ⬁

⬁

˜p共s兲 1 − ˜p共s兲 ˜ n共s兲 = , np 兺 s s共1 − ˜p共s兲兲 n=0

˜ 共n,s兲 = L关具N共t兲典兴共s兲 = 兺 nP n=0

⬁

⬁

˜p共s兲 + ˜p2共s兲 1 − ˜p共s兲 n2˜pn共s兲 = . L关具N 共t兲典兴共s兲 = 兺 n P共n,s兲 = 兺 s s共1 − ˜p共s兲兲2 n=0 n=0 2˜

2

共8兲

共9兲

In deriving Eqs. 共8兲 and 共9兲 we have used the summation formulas listed below: ⬁

1

, xn = 兺 1−x n=0 ⬁

d nx = x 兺 dx n=0 n

⬁

兺 n 2x n = x

n=0

d dx

冉兺 冊 冉兺 冊 ⬁

xn =

n=0

x , 共1 − x兲2

⬁

nxn =

n=0

x + x2 . 共1 − x兲3

To find the behaviors of L关具N共t兲典兴共s兲 and L关具N2共t兲典兴共s兲 for small s, we expand ˜p共s兲:

冏 冏 冏冕 dk˜p共s兲 dsk

⬁

=

s=0

共− t兲ke−st p共t兲dt

0

˜p共s兲 = 1 − 具T典s +

冏

= 共− 1兲k具Tk典, s=0

具T2典 2 s + O共s3兲. 2

Substituting the expansion into L关具N共t兲典兴共s兲 and L关具N2共t兲典兴共s兲 yields L关具N共t兲典兴 =

L关具N共t兲2典兴 =

冉

冊

共10兲

冉

冊 冉冊

共11兲

1 1 2具T典2 − 具T2典 1 + O共1兲, 2 − 具T典 s 2具T典2 s

2 1 1 3具T典2 − 2具T2典 1 − . 2 3 3 2 +O 具T典 s 具T典 s s

In Ref. 7, it was assumed that 具N共t兲典 and 具N2共t兲典 have Laurent expansions as t → ⬁:

Downloaded 19 Oct 2007 to 128.114.49.85. Redistribution subject to AIP license or copyright, see http://jmp.aip.org/jmp/copyright.jsp

103301-5

J. Math. Phys. 48, 103301 共2007兲

A new derivation of the randomness parameter ⬁

Assumption :具N共t兲典 = គ

兺

⬁

k c共1兲 k t ,

具N 共t兲典 = 2

k=−⬁

兺

k c共2兲 k t .

共12兲

k=−⬁

Under this assumption, the coefficients in the Laurent expansions as t → ⬁ are calculated from the Laurent expansions as s → 0 of their Laplace transforms 共10兲 and 共11兲. 具N共t兲典 =

具N共t兲2典 =

冉

冊 冉冊

冉

冊

1 2具T典2 − 具T2典 1 t− +O , 2 具T典 2具T典 t

共13兲

1 2 3具T典2 − 2具T2典 t − t + O共1兲, 具T典2 具T典3

共14兲

which leads to lim t→⬁

具N2共t兲典 − 具N共t兲典2 具T2典 − 具T典2 = . 具N共t兲典 具T典2

共15兲

In other words, the two randomnesses are the same: RN = RT. Assumption 共12兲 is the key in the derivation in Ref. 7. Note, in particular, that the existence of Laurent expansion as s → 0 for the Laplace transform does not imply the existence of Laurent expansion as t → ⬁ for the original function. For example, f共t兲 = t + sin t does not have a Laurent expansion as t → ⬁ while its Laplace transform ˜f 共s兲 = 共1 / s2兲 + 关1 / 共s2 + 1兲兴 has a Laurent expansion as s → 0: ˜f 共s兲 = 共1 / s2兲 + 0共1 / s兲 + O共1兲. When the cycle time has a discrete distribution, assumption 共12兲 breaks down but the conclusion RN = RT still holds. To illustrate this, we consider the cycle time distribution given below:

冦

0.5, t = 1

Pr关T = t兴 = 0.5, t = 2 0 otherwise.

冧

共16兲

The first two moments of T and the randomness in T are given by 具T典 = 23 ,

具T2典 = 25 ,

var共T兲 = 41 ,

RT = 91 .

共17兲

For time t in 关k , k + 1兲 where k is an integer, the first two moments of N共t兲 have the expressions 共see Appendix for the derivation兲

冋 冉 冊册 冋 冉 冊册 冋 冉 冊册

1 −1 2 1− 具N共t兲典 = k − 3 9 2 4 2 −1 具N2共t兲典 = k2 − k 1 − 9 27 2

k

+

k

,

k 艋 t ⬍ k + 1,

1 −1 1− 9 2

共18兲

k

,

k 艋 t ⬍ k + 1.

共19兲

It is clear that Laurent expansions 共13兲 and 共14兲, predicted from the Laplace transform, are invalid for this discrete cycle time distribution. The equivalence between RN and RT, however, still holds. The randomness in N共t兲 as t → ⬁ is RN = 1 / 9, the same as RT. III. A NEW DERIVATION

The motor is modeled as a stochastic stepper, a special case of semi-Markov processes.15 In a semi-Markov process, the system does not remember how it got to the current state but it does remember its age. The age of the system is defined as the time elapsed since the latest arrival at the current state. At the moment when the system jumps to another state, the age is reset to zero. After that, the age increases with the time until the next jump. Let A共t兲 be the age of the system at time

Downloaded 19 Oct 2007 to 128.114.49.85. Redistribution subject to AIP license or copyright, see http://jmp.aip.org/jmp/copyright.jsp

103301-6

J. Math. Phys. 48, 103301 共2007兲

Hongyun Wang

t. Let ␳共n , ␶ , t兲 be the probability density that the system has age ␶ in state n at time t. Here n refers to the number of motor steps completed. Mathematically, ␳共n , ␶ , t兲 is defined as

␳共n, ␶,t兲 ⬅ lim

␦␶→0

Pr关N共t兲 = n, ␶ 艋 A共t兲 ⬍ ␶ + ␦␶兴 . ␦␶

共20兲

For our motor system, ␳共n , ␶ , t兲 is governed by the conservation of probability:

⳵␳共n, ␶,t兲 ⳵␳共n, ␶,t兲 =− − ␳共n, ␶,t兲␤共␶兲, ⳵t ⳵␶

␤共␶兲 ⬅

冕

p共␶兲

共21兲

,

⬁

p共s兲ds

␶

where ␤共␶兲 is called the hazard function and is the conditional rate of completing the current cycle at age ␶ given that the cycle is not completed in age 关0 , ␶兲. In Eq. 共21兲, the first term on the right hand side corresponds to the fact that if no transition occurs then the age of the system increases with the time. The second term on the right hand side of Eq. 共21兲 reflects that a transition out of a state reduces the probability of that state. From its definition, it follows that the hazard function ␤共␶兲 satisfies −

冕

␶

冉冕

␤共s兲ds = log

0

⬁

冊

共22兲

p共s兲ds .

␶

This relation between ␤共s兲 and p共s兲 will be very useful in the analysis below. Since all new arrivals have age 0, the boundary condition for ␳共n , ␶ , t兲 is

␳共0,0,t兲 = 0,

␳共n,0,t兲 =

冕

t ⬎ 0, ⬁

␳共n − 1, ␶,t兲␤共␶兲d␶,

t ⬎ 0,

n ⬎ 0.

共23兲

0

Here ␳共0 , 0 , t兲 = 0 reflects the fact that the motor system starts in state 0 at time 0. Because we assume that the motor goes only forward there is no transition into state 0. To determine the evolution of ␳共n , ␶ , t兲, an initial state-age distribution needs to be specified. A straightforward initial state-age distribution is initial distribution 0:

␳共0, ␶,0兲 = ␦共␶兲, ␳共n, ␶,0兲 = 0,

n ⬎ 0,

共24兲

which corresponds to that all motors in the ensemble start with age 0 in state 0 at time 0. It turns out that this seemingly simple initial condition is neither experimentally easy to implement nor mathematically convenient. In experiments, to implement the initial condition ␳共0 , ␶ , 0兲 = ␦共␶兲, we need to start a new cycle with age 0 at the beginning of the experiment, or equivalently we need to detect accurately the time at which a cycle is completed and the next cycle is started. In experiments, a more realistic situation is that the system has a random age at time 0. This is especially true if in the statistical calculation the ensemble is created by cutting one or a few long time series into many time series.9,1,14 Thus, it is more consistent with the experiments if we use the stationary age distribution as the initial condition. Mathematically, the age distribution 共regardless of the state兲 at time t is

Downloaded 19 Oct 2007 to 128.114.49.85. Redistribution subject to AIP license or copyright, see http://jmp.aip.org/jmp/copyright.jsp

103301-7

J. Math. Phys. 48, 103301 共2007兲

A new derivation of the randomness parameter ⬁

␳共␶,t兲 ⬅ 兺 ␳共n, ␶,t兲.

共25兲

n=0

The governing equation and the boundary condition for the age distribution are obtained by summing Eqs. 共21兲 and 共23兲 over all states:

⳵␳共␶,t兲 ⳵␳共␶,t兲 − ␳共␶,t兲␤共␶兲, =− ⳵t ⳵␶

冕

␳共0,t兲 =

⬁

␳共␶,t兲␤共␶兲d␶ .

共26兲

0

Let ␳共S兲共␶兲 be the stationary age distribution. Setting the left side of Eq. 共26兲 to zero, solving for ␳共S兲共␶兲, and using the relation 共22兲 yield

冉冕 冊 冕 冕 冉冕 冊 冕冕 ␶

exp −

共S兲

␳ 共␶兲 =

⬁

␤共s兲ds

0

␶

⬁

exp −

0

=

␤共s兲ds d␶

0

p共s兲ds

␶

⬁

⬁

␶

0

= p共s兲dsd␶

冕 冕

⬁

␶ ⬁

p共s兲ds =

␶ p共␶兲d␶

冕

⬁

p共s兲ds

␶

具T典

.

共27兲

0

A more reasonable 共more consistent with experiments兲 initial state-age distribution is to start the ensemble of motors at state 0 with the stationary age distribution ␳共S兲共␶兲:

␳共0, ␶,0兲 = ␳共S兲共␶兲,

initial distribution 1:

␳共n, ␶,0兲 = 0,

n ⬎ 0.

共28兲

In the rest of this section, we will first prove RN = RT for initial state-age distribution 共28兲. Then we will extend the conclusion RN = RT to initial state-age distribution 共24兲. Finally, we will extend the conclusion RN = RT to the general case where the first moment of the waiting time for completing the first cycle is finite. To demonstrate the importance of the first moment of the waiting time for completing the first cycle, in the next section, we will present an example to show RN ⫽ RT if the first moment of the waiting time for completing the first cycle is not finite. A. The derivation in the case of initial condition „28… ⬁ With the initial condition 共28兲, the age distribution will remain stationary: 兺n=0 ␳共n , ␶ , t兲 = ␳ 共␶兲. To calculate 具N共t兲典, we first express 具N共t兲典 in terms of ␳共n , ␶ , t兲: 共S兲

⬁

具N共t兲典 = 兺 n n=0

冕

⬁

␳共n, ␶,t兲d␶ .

共29兲

0

Differentiating with respect to t, using differential equation 共21兲 and boundary condition 共23兲, we obtain ⬁

d具N共t兲典 =−兺n dt n=0 ⬁

冕冉 ⬁

0

= 兺 ␳共n,0,t兲 = ␳共S兲共0兲 = n=0

冊

⬁

⳵␳共n, ␶,t兲 + ␳共n, ␶,t兲␤共␶兲 d␶ = 兺 n共␳共n,0,t兲 − ␳共n + 1,0,t兲兲 ⳵␶ n=0 1 . 具T典

共30兲

Similarly, for the second moment, we have

Downloaded 19 Oct 2007 to 128.114.49.85. Redistribution subject to AIP license or copyright, see http://jmp.aip.org/jmp/copyright.jsp

103301-8

J. Math. Phys. 48, 103301 共2007兲

Hongyun Wang ⬁

具N 共t兲典 = 兺 n2 2

n=0

冕

⬁

␳共n, ␶,t兲d␶ .

共31兲

0

Differentiating with respect to t, using differential equation 共21兲 and boundary condition 共23兲, yields ⬁

d具N2共t兲典 = − 兺 n2 dt n=0

冕冉 ⬁

0

冊

⬁

⳵␳共n, ␶,t兲 + ␳共n, ␶,t兲␤共␶兲 d␶ = 兺 n2共␳共n,0,t兲 − ␳共n + 1,0,t兲兲 ⳵␶ n=0

⬁

= 兺 共2n − 1兲␳共n,0,t兲 ⬅ n=0

1 2t 2 q共0,t兲 + , 2 − 具T典 具T典 具T典

共32兲

where function q共␶ , t兲 is defined as ⬁

q共␶,t兲 ⬅ 具T典 兺 n␳共n, ␶,t兲 + 共␶ − t兲␳共S兲共␶兲.

共33兲

n=0

The initial condition 共28兲 implies 具N共0兲典 = 0 and 具N2共0兲典 = 0. Integrating Eqs. 共30兲 and 共32兲 with respect to t yields 具N共t兲典 =

具N2共t兲典 =

2 具T典

冕

t , 具T典

共34兲

t

q共0,s兲ds +

0

t2 t . − 具T典2 具T典

共35兲

Here we should point out that result 共34兲 corresponds to the renewal theorem in Ref. 16. Using these results to calculate RN, we obtain RN ⬅ lim

t→⬁

具N2共t兲典 − 具N共t兲典2 2 = lim 具N共t兲典 t→⬁ t

冕

冉 冕

t

q共0,s兲ds − 1 = 2 lim

t→⬁

0

1 t

t

0

q共0,s兲ds −

冊

具T2典 + RT . 2具T典2 共36兲

For the limit enclosed in the parentheses in Eq. 共36兲, we have the theorem below. Theorem 1: Function q共␶ , t兲 defined in Eq. 共33兲 has the property lim t→⬁

1 t

冕

t

q共0,s兲ds =

0

具T2典 . 2具T典2

共37兲

Remark: Combining the result of Theorem 1 and Eq. 共36兲 leads immediately to the desired conclusion RN = RT. To facilitate the proof of Theorem 1, we introduce a lemma. Lemma 1: Consider a function q共␶ , t兲 关not necessarily the one defined in Eq. 共33兲兴. Suppose q共␶ , t兲 satisfies the differential equation

⳵q共␶,t兲 ⳵q共␶,t兲 =− − q共␶,t兲␤共␶兲. ⳵t ⳵␶

共38兲

␳共S兲共␶兲 . ␳共S兲共␶ − c兲

共39兲

Then we have q共␶,t兲 = q共␶ − c,t − c兲

Proof of Lemma 1: We rewrite differential equation 共38兲 as

Downloaded 19 Oct 2007 to 128.114.49.85. Redistribution subject to AIP license or copyright, see http://jmp.aip.org/jmp/copyright.jsp

103301-9

J. Math. Phys. 48, 103301 共2007兲

A new derivation of the randomness parameter

d q共␶ + s,t + s兲 = − q共␶ + s,t + s兲␤共␶ + s兲. ds It follows that

冉冕

q共␶,t兲 = q共␶ − c,t − c兲exp −

␶

␶−c

冊

␤共s兲ds .

Using property 共22兲 of ␤共␶兲 and the expression of ␳共S兲共␶兲 given in Eq. 共27兲, we obtain

冉冕

␶

exp −

␶−c

冊

␤共s兲ds =

␳共S兲共␶兲 , ␳共S兲共␶ − c兲

which leads directly to Eq. 共39兲. Proof of Theorem 1: Using Eqs. 共21兲, 共23兲, and 共27兲, it is straightforward to verify that q共␶ , t兲 defined in Eq. 共33兲 satisfies differential equation 共38兲 and boundary condition

冕

q共0,t兲 =

⬁

q共␶,t兲␤共␶兲d␶ .

共40兲

0

Setting t = 0 in Eq. 共33兲, we see that q共␶ , t兲 satisfies the initial condition q共␶ , 0兲 = ␶␳共S兲共␶兲. Using differential equation 共38兲, boundary condition 共40兲, and Lemma 1, one can show the following. 共i兲 共ii兲 共iii兲 共iv兲 共v兲 共vi兲

q共␶ , t兲 is non-negative. 兰⬁0 ␶␳共S兲共␶兲d␶ = 兰⬁0 ␶兰␶⬁ p共s兲dsd␶ / 具T典 = 兰⬁0 ␶2 p共␶兲d␶ / 2具T典 = 具T2典 / 2具T典 ⬅ ␣. 兰⬁0 q共␶ , t兲d␶ = 兰⬁0 q共␶ , 0兲d␶ = ␣. It follows from 共iii兲 that 兰t0兰⬁0 q共␶ , s兲d␶ds = ␣t. It follows from Lemma 1 that q共␶ , t兲 = 共␶ − t兲␳共S兲共␶兲 for ␶ ⬎ t. It follows from Lemma 1 that q共␶ , t兲 = q共0 , t − ␶兲␳共S兲共␶兲具T典 for ␶ ⬍ t.

For the convenience of discussion below, let us introduce h共t兲 ⬅

冕

t

共41兲

q共0,s兲ds.

0

We divide the integration region in 共iv兲 into two parts: the region of ␶ ⬎ t and the region of ␶ ⬍ t. For the region of ␶ ⬎ t, using result 共v兲 listed above and then applying integration by parts, we have

冕冕 t

0

⬁

冕冕 t

q共␶,s兲d␶ds =

s

0

⬁

␶␳共S兲共␶ + s兲d␶ds =

0

1 2具T典

冕冕 t

⬁

␶2 p共␶ + s兲d␶ds.

共42兲

␶2 p共␶兲d␶ .

共43兲

0

0

The inner layer integral on the right hand side of Eq. 共42兲 satisfies 0⬍

冕

⬁

␶2 p共␶ + t兲d␶ ⬍

0

冕

⬁

共␶ + t兲2 p共␶ + t兲d␶ =

0

冕

⬁

t

Since the second moment of T exists, we have limt→⬁ 兰⬁t ␶2 p共␶兲d␶ = 0. Combining this result with inequality 共43兲 yields limt→⬁ 兰⬁0 ␶2 p共␶ + t兲d␶ = 0. Dividing Eq. 共42兲 by t, taking the limit as t → ⬁, and applying L’Hospital’s rule, we obtain lim t→⬁

1 t

冕冕 t

0

⬁

s

q共␶,s兲d␶ds = lim

t→⬁

1 2具T典

冕

⬁

␶2 p共␶ + t兲d␶ = 0.

共44兲

0

Thus, using result 共iv兲 listed above and result 共44兲, we arrive at

Downloaded 19 Oct 2007 to 128.114.49.85. Redistribution subject to AIP license or copyright, see http://jmp.aip.org/jmp/copyright.jsp

103301-10

J. Math. Phys. 48, 103301 共2007兲

Hongyun Wang

lim t→⬁

1 t

冕冕 t

0

s

q共␶,s兲d␶ds = lim

t→⬁

0

冉 冕冕 t

1 ␣t − t

⬁

冊

q共␶,s兲d␶ds = ␣ .

s

0

共45兲

On the other hand, for the region of ␶ ⬍ t, using result 共vi兲 listed above, we have

冕冕 t

0

s

冕冕 冕 冉冕 t

q共␶,s兲d␶ds =

0

t

␶

0

t

= 具T典

0

q共␶,s兲dsd␶ = 具T典 t−␶

冊

冕冕 t

t

0

␶

q共0,s − ␶兲␳共S兲共␶兲dsd␶

q共0,s兲ds ␳共S兲共␶兲d␶ = 具T典

0

冕

t

h共t − ␶兲␳共S兲共␶兲d␶ .

共46兲

0

Combining results 共45兲 and 共46兲 for the region of ␶ ⬍ t yields lim t→⬁

g共t兲 ␣ = , t 具T典

共47兲

where g共t兲 ⬅ 兰t0h共t − s兲␳共S兲共s兲ds. Since ␳共S兲共s兲 is a probability density, for any ⑀ ⬎ 0 there exists M such that 兰0M ␳共S兲共s兲ds 艌 共1 − ⑀兲. Recall that h共t兲 is non-negative and nondecreasing. It follows that g共t兲 =

冕

t

h共t − s兲␳共S兲共s兲ds 艋 h共t兲

0

g共t + M兲 =

冕

t+M

冕

t

␳共S兲共s兲ds 艋 h共t兲,

0

h共t + M − s兲␳共S兲共s兲ds 艌

0

冕

M

h共t + M − s兲␳共S兲共s兲ds 艌 h共t兲

0

冕

M

␳共S兲共s兲ds

0

艌 h共t兲共1 − ⑀兲. Combining these two inequalities, we obtain 1 t + M g共t + M兲 g共t兲 h共t兲 . 艋 艋 t+M t t 1−⑀ t Taking the limit as t → ⬁, using Eq. 共47兲, and then taking the limit as ⑀ → 0, we arrive at lim t→⬁

h共t兲 ␣ 具T2典 = , = t 具T典 2具T典2

which is equivalent to property 共37兲. This completes the proof of Theorem 1. B. Extending the conclusion to the general case

In the previous subsection, we proved the equivalence of the two randomnesses RN = RT for initial condition 共28兲. In single molecule motor experiments, initial condition 共28兲 is the only realistic and relevant initial condition. Thus, in studies of single molecule motors, we need RN = RT only for initial condition 共28兲. For mathematical completeness, in this subsection, we first extend the conclusion RN = RT to the case of initial condition 共24兲. The case of initial condition 共24兲 can be regarded as the canonical case where all motors start with age 0 at state 0 at time 0. Finally, based on the results for the canonical case, we extend the conclusion RN = RT to the general case where the first moment of the waiting time for completing the first cycle is finite. In the discussion below, we need to refer to both of these two cases. To clearly distinguish them, for the canonical initial condition 共24兲 we shall use ␳0共n , ␶ , t兲 and N0共t兲 to denote the probability density and the number of motor steps; for initial condition 共28兲 we shall use the notations ␳1共n , ␶ , t兲 and N1共t兲. In the previous subsection, we have proved

Downloaded 19 Oct 2007 to 128.114.49.85. Redistribution subject to AIP license or copyright, see http://jmp.aip.org/jmp/copyright.jsp

103301-11

J. Math. Phys. 48, 103301 共2007兲

A new derivation of the randomness parameter

具N1共t兲典 =

t , 具T典

lim t→⬁

var共N1共t兲兲 RT = . t 具T典

共48兲

Below we will first extend the conclusion to the canonical case 共Theorem 2兲 lim t→⬁

1 具N0共t兲典 = , t 具T典

lim t→⬁

var共N0共t兲兲 RT = . t 具T典

共49兲

Then we will extend the conclusion to the general case 共Theorem 3兲 lim t→⬁

1 具N1*共t兲典 = , t 具T典

lim t→⬁

var共N1*共t兲兲 RT = . t 具T典

共50兲

We shall proceed step by step and present the derivation as a sequence of lemmas leading to the main theorems. Lemma 2: 共i兲

具N0共t兲典 is non-negative, is a nondecreasing function of t, and satisfies 0 艋 具N0共t1 + t2兲典 − 具N0共t1兲典 艋 c0 + c0t2

共ii兲

for t1 艌 0,

t2 艌 0.

共51兲

具N20共t兲典 is non-negative, is a nondecreasing function of t, and satisfies 0 艋 具N20共t1 + t2兲典1/2 − 具N20共t1兲典1/2 艋 c1 + c1t2

for t1 艌 0,

t2 艌 0,

共52兲

where constants c0 ⬎ 1 and c1 ⬎ 1 are independent of t1 and t2. Remark: Note that 具N0共t兲典 and 具N20共t兲典 may be discontinuous. Result 共i兲 of Lemma 2 corresponds to a modified form of the renewal theorem in Ref. 16. Proof of Lemma 2: Since each realization N0共t兲 is non-negative and is a nondecreasing function of t, it follows that both 具N0共t兲典 and 具N20共t兲典 are non-negative and are nondecreasing functions of t. Each realization of N0共t1 + t2兲 can be written as N0共t1 + t2兲 =

再

N0共t1兲 + 1 + N0共t2 − tc兲, 0 ⬍ tc 艋 t2 N0共t1兲,

tc ⬎ t2 ,

冎

共53兲

where t1 + tc is the time at which the first motor step after time t1 is completed. Taking average on both sides of Eq. 共53兲 yields 具N0共t1 + t2兲典 艋 具N0共t1兲典 + 1 + 具N0共t2兲典.

共54兲

Setting t2 = 1 in Eq. 共54兲 and applying Eq. 共54兲 repeatedly, we obtain that for positive integer k, 具N0共t1 + k兲典 艋 具N0共t1兲典 + c0k,

c0 = 1 + 具N0共1兲典.

共55兲

For k − 1 ⬍ t2 艋 k, we have 具N0共t1 + t2兲典 艋 具N0共t1兲典 + c0 + c0t2 .

共56兲

To derive 共ii兲, we set t2 = 1 in Eq. 共53兲, square both sides, and then take average 具N20共t1 + 1兲典 艋 具共N0共t1兲 + 1 + N0共1兲兲2典. Expanding the average on the right hand side, using the Cauchy-Schwarz inequality, and then taking square root of both sides, we obtain

Downloaded 19 Oct 2007 to 128.114.49.85. Redistribution subject to AIP license or copyright, see http://jmp.aip.org/jmp/copyright.jsp

103301-12

J. Math. Phys. 48, 103301 共2007兲

Hongyun Wang

具N20共t1 + 1兲典1/2 艋 共具N20共t1兲典 + 2具N0共t1兲共1 + N0共1兲兲典 + 具共1 + N0共1兲兲2典兲1/2 艋 共具N20共t1兲典 + 2具N20共t1兲典1/2具共1 + N0共1兲兲2典1/2 + 具共1 + N0共1兲兲2典兲1/2 = 具N20共t1兲典1/2 + 具共1 + N0共1兲兲2典1/2 .

共57兲

Applying Eq. 共57兲 repeatedly, we obtain that for positive integer k, 具N20共t1 + k兲典1/2 艋 具N20共t1兲典1/2 + c1k,

c1 = 具共1 + N0共1兲兲2典1/2 .

共58兲

For k − 1 ⬍ t2 艋 k, we have 具N20共t1 + t2兲典1/2 艋 具N20共t1兲典1/2 + c1 + c1t2 .

共59兲

This completes the proof of Lemma 2. Let us consider ␳1共0 , ␶ , t兲, the probability density of age ␶ at state 0 at time t for initial condition 共28兲. ␳1共0 , ␶ , t兲 satisfies differential equation 共38兲. Lemma 1 tells us that

␳1共0, ␶,t兲 =

冦

␳1共0, ␶ − t,0兲 0

␳共S兲共␶兲 = ␳共S兲共␶兲 for ␶ 艌 t ␳共S兲共␶ − t兲 for ␶ ⬍ t.

冧

共60兲

兰⬁0 ␳1共0 , ␶ , t兲d␶ is the probability of having not completed the first cycle by time t. The probability density of the waiting time for completing the first cycle is p共1兲 1 共t兲 = −

d dt

冕

⬁

␳1共0, ␶,t兲d␶ = ␳共S兲共t兲.

共61兲

0

Here the superscript 共1兲 refers to the first cycle. Below we shall use the notation p共1兲 1 共t兲 instead of ␳共S兲共t兲 although they are the same. Furthermore, we shall use only the property that 兰⬁0 tp共1兲 1 共t兲dt is finite. The benefit of doing so is that all lemmas developed below will still be valid in the general case where the first moment of the waiting time for completing the first cycle is finite. At the moment the first cycle is completed, the system is renewed and continues with the canonical initial condition 共24兲.17 Thus, 具N1共t兲典 and 具N0共s兲典 are related by the convolution 具N1共t兲典 =

冕

t

具1 + N0共t − ␶兲典p共1兲 1 共␶兲d␶ .

共62兲

0

It should be pointed out that this convolution relation was used in Ref. 16 to relate a delayed renewal process to an ordinary renewal process. Lemma 3: 具N0共t兲典 is bounded in terms of 具N1共t兲典 by 具N1共t兲典 − 1 艋 具N0共t兲典 艋 具N1共t兲典 + c2 ,

共63兲

where constant c2 ⬎ 1 is independent of t. Proof of Lemma 3: Combining 具N0共t − ␶兲典 艋 具N0共t兲典 with Eq. 共62兲, we have 具N1共t兲典 艋 1 + 具N0共t兲典, which leads to the first half of Eq. 共63兲. Subtracting 具N0共t兲典 from both sides of Eq. 共62兲 and then using the result of Lemma 2 yields

Downloaded 19 Oct 2007 to 128.114.49.85. Redistribution subject to AIP license or copyright, see http://jmp.aip.org/jmp/copyright.jsp

103301-13

J. Math. Phys. 48, 103301 共2007兲

A new derivation of the randomness parameter

具N0共t兲典 − 具N1共t兲典 艋

冕 冕

t

共具N0共t兲典 − 具N0共t − ␶兲典兲p共1兲 1 共␶兲d␶ +

0

艋

⬁

共c0 + c0␶兲p共1兲 1 共␶兲d␶ = c0 + c0

0

冕

冕

⬁

具N0共t兲典p共1兲 1 共␶兲d␶

t

⬁

␶ p共1兲 1 共␶兲d␶ ,

0

which leads to the second half of Eq. 共63兲. Here we have used that 兰⬁0 ␶ p共1兲 1 共␶兲d␶ is finite. This completes the proof of Lemma 3. The significance of Lemma 3 is that Eq. 共63兲 leads to 0 艋 具共N1共t兲 − 具N0共t兲典兲2典 − var共N1共t兲兲 = 具共N1共t兲 − 具N1共t兲典 + 具N1共t兲典 − 具N0共t兲典兲2典 − var共N1共t兲兲 = 共具N1共t兲典 − 具N0共t兲典兲2 艋 c22 .

共64兲

Thus, we can replace var共N1共t兲兲 in the limit of Eq. 共48兲 with 具共N1共t兲 − 具N0共t兲典兲2典. Furthermore 具共N1共t兲 − 具N0共t兲典兲2典 is expressed in a convolution form similar to Eq. 共62兲: 具共N1共t兲 − 具N0共t兲典兲2典 =

冕

t

具共1 + N0共t − ␶兲 − 具N0共t兲典兲2典p共1兲 1 共␶兲d␶ .

共65兲

0

The two lemmas below relate var共N1共t兲兲 to var共N0共t兲兲 through Eq. 共65兲. Lemma 4: Let P1共t兲 ⬅ 兰t0 p共1兲 1 共␶兲d␶ and let Q共t兲 ⬅ 兩var共N1共t兲兲 − var共N0共t兲兲P1共t兲兩.

共66兲

Q共t兲 艋 c3 + c21␴共t兲 + 2c4冑var共N0共t兲兲,

共67兲

We have

where all constants are independent of t and ␴共t兲 satisfies limt→⬁␴共t兲 / t = 0. Proof of Lemma 4: We first relate the right hand side of Eq. 共65兲 to var共N0共t兲兲. Let I共t, ␶兲 ⬅ 兩具共1 + N0共t − ␶兲 − 具N0共t兲典兲2典 − var共N0共t兲兲兩.

共68兲

Step 1: We show that for 0 艋 ␶ 艋 t, I共t , ␶兲 is bounded by I共t, ␶兲 艋 共c1 + c1␶兲2 + 2共c1 + c1␶兲冑var共N0共t兲兲.

共69兲

Each realization of N0共t兲 can be written as N0共t兲 = N0共t − ␶兲 + 1 + V共t − ␶, ␶兲, where random variable V共t − ␶ , ␶兲 is given by V共t − ␶, ␶兲 =

再

N0共␶ − tc兲, 0 ⬍ tc 艋 ␶ − 1,

tc ⬎ ␶ ,

共70兲

冎

共71兲

and t − ␶ + tc is the time at which the first motor step after time t − ␶ is completed. Let pc ⬅ Pr关tc ⬎ ␶兴. Using Eq. 共71兲 directly to calculate 具V2共t − ␶ , ␶兲典, using the result of Lemma 2, and using the fact that c1 ⬎ 0, we obtain 具V2共t − ␶, ␶兲典 艋 具N20共␶兲典共1 − pc兲 + pc 艋 共c1 + c1␶兲2 .

共72兲

Using Eq. 共70兲 to expand Eq. 共68兲 yields I共t, ␶兲 = 兩具共N0共t兲 − 具N0共t兲典 − V共t − ␶, ␶兲兲2典 − var共N0共t兲兲兩 艋 具V2共t − ␶, ␶兲典 + 2兩具V共t − ␶, ␶兲共N0共t兲 − 具N0共t兲典兲典兩.

共73兲

Applying the Cauchy-Schwarz inequality and using inequality 共72兲 lead directly to Eq. 共69兲.

Downloaded 19 Oct 2007 to 128.114.49.85. Redistribution subject to AIP license or copyright, see http://jmp.aip.org/jmp/copyright.jsp

103301-14

J. Math. Phys. 48, 103301 共2007兲

Hongyun Wang

Step 2: We show that Q共t兲 is bounded by inequality 共67兲. Combining inequality 共64兲, Eq. 共65兲, notation 共68兲, and inequality 共69兲 yields Q共t兲 艋 c22 +

冕

t

2 2 I共t, ␶兲p共1兲 1 共␶兲d␶ 艋 c2 + c1

0

艋 c3 +

冕

t

共1 + ␶兲2 p共1兲 1 共␶兲d␶ + 2c1

0

c21␴共t兲

冕

t

冑 共1 + ␶兲p共1兲 1 共␶兲d␶ var共N0共t兲兲

0

+ 2c4冑var共N0共t兲兲,

共74兲

where ␴共t兲 ⬅ 兰t0␶2 p共1兲 1 共␶兲d␶, and constants c3 and c4 are given by

冉 冕

⬁

c3 = c22 + c21 1 + 2

冊

␶ p共1兲 1 共␶兲d␶ ,

0

冉 冕

冊

⬁

␶ p共1兲 1 共␶兲d␶ .

c4 = c1 1 +

0

共75兲

We now show limt→⬁␴共t兲 / t = 0. Using integration by parts, we write ␴共t兲 / t as

␴共t兲 = t

冕

t

0

t

冕 冉冕 t

␶2 p共1兲 1 共␶兲d␶

␶d −

=

0

⬁

␶

sp共1兲 1 共s兲ds

冊

t

冕 冉冕 t

=−

冕

⬁

sp共1兲 1 共s兲ds

+

⬁

␶

0

冊

sp共1兲 1 共s兲ds d␶ t

t

. 共76兲

⬁ 共1兲 Since 兰⬁0 sp共1兲 1 共s兲ds is finite, we have limt→⬁兰t sp1 共s兲ds = 0. Applying L’Hospital’s rule yields limt→⬁␴共t兲 / t = 0. This completes the proof of Lemma 4. Lemma 5: var共N0共t兲兲 is bounded in terms of var共N1共t兲兲 by

冑

冉 冊

var共N1共t兲兲 − c3 − c21␴共t兲 c4 + P1共t兲 P1共t兲

2

−

c4 艋 冑var共N0共t兲兲 P1共t兲 艋

冑

冉 冊

var共N1共t兲兲 + c3 + c21␴共t兲 c4 + P1共t兲 P1共t兲

2

+

c4 . P1共t兲 共77兲

Proof of Lemma 5: Dividing both sides of Eq. 共67兲 by P1共t兲 yields c3 + c21␴共t兲 2c4 var共N1共t兲兲 冑var共N0共t兲兲. − var共N0共t兲兲 艋 + P1共t兲 P1共t兲 P1共t兲 Moving var共N0共t兲兲 to the right side and completing the square, we have

冉 冊 冉冑

var共N1共t兲兲 − c3 − c21␴共t兲 c4 + P1共t兲 P1共t兲

2

艋

var共N0共t兲兲 +

c4 P1共t兲

冊

2

.

Taking the square root leads to the first half of Eq. 共77兲. The second half of Eq. 共77兲 is proved in a similar way. This completes the proof of Lemma 5. Now we extend the conclusion to the canonical case. Theorem 2: N0共t兲 satisfies lim t→⬁

1 具N0共t兲典 = , t 具T典

lim t→⬁

var共N0共t兲兲 RT = . t 具T典

共78兲

Proof of Theorem 2: Dividing inequality 共77兲 by 冑t, taking the limit as t → ⬁, and using limt→⬁ P1共t兲 = 1, limt→⬁␴共t兲 / t = 0, and Eq. 共48兲, we obtain

Downloaded 19 Oct 2007 to 128.114.49.85. Redistribution subject to AIP license or copyright, see http://jmp.aip.org/jmp/copyright.jsp

103301-15

J. Math. Phys. 48, 103301 共2007兲

A new derivation of the randomness parameter

lim t→⬁

冑

var共N0共t兲兲 = t

冑

RT . 具T典

共79兲

Using the result of Lemma 3 and Eq. 共48兲, we arrive at lim t→⬁

1 具N0共t兲典 = . t 具T典

共80兲

This completes the proof of Theorem 2. Finally, we extend the conclusion to the general case. Theorem 3: Consider the general case where the first moment of the waiting time for completing the first cycle is finite. 共Of course, by using 具T典 and 具T2典, we have assumed that the first two moments of the cycle time of regular cycles are finite.兲 Let N1*共t兲 denote the number of steps completed by time t. Then N1*共t兲 satisfies lim t→⬁

var共N1*共t兲兲 = RT . 具N1*共t兲典

共81兲

Proof of Theorem 3: In deriving Lemmas 2–5, we only used that 兰⬁0 ␶ p共1兲 1 共␶兲d␶ is finite. As a result, Lemmas 2–5 are also valid for N1*共t兲. Using Lemma 3 and Theorem 2, we have lim t→⬁

1 具N1*共t兲典 具N1*共t兲典 具N0共t兲典 = lim = . t t 具T典 t→⬁ 具N0共t兲典

共82兲

Using the result of Lemma 4, we obtain var共N0共t兲兲P1共t兲 − c3 − c21␴共t兲 − 2c4冑var共N0共t兲兲 艋 var共N1*共t兲兲 艋 var共N0共t兲兲P1共t兲 + c3 + c21␴共t兲 + 2c4冑var共N0共t兲兲.

共83兲

Dividing inequality 共83兲 by t, taking the limit as t → ⬁, and using limt→⬁ P1共t兲 = 1, limt→⬁␴共t兲 / t = 0, and Theorem 2, we arrive at lim t→⬁

var共N1*共t兲兲 RT = . t 具T典

共84兲

Combining Eqs. 共82兲 and 共84兲 leads immediately to the desired conclusion. IV. AN EXAMPLE OF RN Å RT AND DISCUSSION

In this section, we present an example to demonstrate that the initial age distribution does affect the validity of RN = RT. This example indicates that any serious attempt on proving RN = RT must take into consideration the initial age distribution. In other words, all proofs that conclude RN = RT categorically without considering the initial age distribution are, at least, mathematically questionable. We consider a stochastic stepper with cycle time probability density p共t兲 = 共2 + a兲共1 + t兲−共3+a兲,

t 艌 0,

0 ⬍ a ⬍ 1.

共85兲

The first two moments of the cycle time and the randomness in the cycle time are 具T典 =

1 , 1+a

具T2典 =

2 , a共1 + a兲

var共T兲 =

2+a , a共1 + a兲2

RT =

2+a . a

共86兲

The stationary age distribution is

Downloaded 19 Oct 2007 to 128.114.49.85. Redistribution subject to AIP license or copyright, see http://jmp.aip.org/jmp/copyright.jsp

103301-16

J. Math. Phys. 48, 103301 共2007兲

Hongyun Wang

␳共S兲共␶兲 =

兰␶⬁ p共s兲ds = 共1 + a兲共1 + ␶兲−共2+a兲 . 具T典

共87兲

Let us start with the initial age distribution, initial distribution 2:

␳2共0, ␶,0兲 = a共1 + ␶兲−共1+a兲 , ␳2共n, ␶,0兲 = 0,

n ⬎ 0.

共88兲

We shall use subscript 2 to denote quantities corresponding to initial condition 共88兲. ␳2共0 , ␶ , t兲 satisfies differential equation 共38兲. Using the result of Lemma 1, we have

␳2共0, ␶,t兲 =

冦

␳2共0, ␶ − t,0兲 0

␳共S兲共␶兲 = a共1 + ␶ − t兲共1 + ␶兲−共2+a兲 for ␶ 艌 t ␳共S兲共␶ − t兲 for ␶ ⬍ t.

冧

共89兲

兰⬁0 ␳2共0 , ␶ , t兲d␶

is the probability of having not completed the first cycle by time t. The probability density of the waiting time for completing the first cycle is p共1兲 2 共t兲 = −

d dt

冕

⬁

␳2共0, ␶,t兲d␶ = a共1 + t兲−共2+a兲 +

0

a 共1 + t兲−共1+a兲 , 1+a

共90兲

where the superscript 共1兲 refers to the first cycle. It is important to notice that the first moment of p共1兲 2 共t兲 diverges. As we will see below, the nonexistence of the first moment is the cause of RN ⫽ R T. Let N2共t兲 denote the number of steps completed by time t for initial condition 共88兲. At the moment the first cycle is completed, the system is renewed and continues with the canonical initial condition 共24兲. Thus, we have 具N2共t兲典 =

冕

t

具1 + N0共t − ␶兲典p共1兲 2 共␶兲d␶ ,

0

具N22共t兲典 =

冕

t

具共1 + N0共t − ␶兲兲2典p共1兲 2 共␶兲d␶ .

共91兲

0

The variance of N2共t兲 satisfies var共N2共t兲兲 = 具N22共t兲典 − 具N2共t兲典2 = 艌

冕

冕

t

var共N0共t − ␶兲兲p共1兲 2 共␶兲d␶ +

t

2 具1 + N0共t − ␶兲典2 p共1兲 2 共␶兲d␶ − 具N2共t兲典

0

0

t

冕

2 2 具1 + N0共t − ␶兲典2 p共1兲 2 共␶兲d␶ − 具N2共t兲典 ⬅ I2 − 具N2共t兲典 .

共92兲

0

Now we use the result of Lemma 3 and 具N1共t兲典 = t / 具T典 = 共1 + a兲t to estimate I2 and 具N2共t兲典. I2 =

冕

t

具1 + N0共t − ␶兲典2 p共1兲 2 共␶兲d␶ 艌

0

= 共1 + a兲2

冕

冋

t

0

冋

冕

t

具N1共t − ␶兲典2 p共1兲 2 共␶兲d␶

0

共t − ␶兲2 a共1 + ␶兲−共2+a兲 +

= 共1 + a兲2 共1 + t兲2 −

册

a 共1 + ␶兲−共1+a兲 d␶ 1+a

册

2 共1 + t兲共2−a兲 + O共t兲 + ¯ , 共1 − a2兲共2 − a兲

共93兲

Downloaded 19 Oct 2007 to 128.114.49.85. Redistribution subject to AIP license or copyright, see http://jmp.aip.org/jmp/copyright.jsp

103301-17

J. Math. Phys. 48, 103301 共2007兲

A new derivation of the randomness parameter

具N2共t兲典 =

冕

t

具1 + N0共t − ␶兲典p共1兲 2 共␶兲d␶ 艋

0

= 共1 + a兲

冕

冋

t

冕

t

0

冋

共t − ␶兲 a共1 + ␶兲−共2+a兲 +

0

= 共1 + a兲 共1 + t兲 −

具N1共t − ␶兲典2 p共1兲 2 共␶兲d␶ + 共1 + c2兲

册

a 共1 + ␶兲−共1+a兲 d␶ + O共1兲 1+a

册

1 共1 + t兲共1−a兲 + O共1兲 + ¯ . 共1 − a2兲

共94兲

Combining Eqs. 共92兲–共94兲 yields var共N2共t兲兲 艌 I2 − 具N2共t兲典2 艌 共1 + a兲2

冋

册

2 共1 + t兲共2−a兲 + O共t兲 + ¯ , 共1 + a兲共2 − a兲

共95兲

which leads to 2 共1 + t兲共1−a兲 + O共1兲 + ¯ var共N2共t兲兲 共2 − a兲 艌 lim = ⬁. RN = lim 1 + O共t−a兲 + ¯ t→⬁ 具N2共t兲典 t→⬁

共96兲

Therefore, for initial condition 共88兲, we have RN ⫽ RT. This example demonstrates that besides the existence of 具T典 and 具T2典, the validity of RN = RT also depends on the existence of the first moment of p共1兲共t兲, the waiting time distribution for completing the first cycle. If the first moment of p共1兲共t兲 diverges, then we have RN = ⬁. If the first moment of p共1兲共t兲 is finite, then we have RN = RT 共Theorem 3 in the previous section兲. In conclusion, we discussed the randomness in the cycle time and the randomness in the number of cycles over long time. We showed the equivalence between these two randomnesses for the general case where the cycle time of regular cycles has the first two moments and the waiting time for completing the first cycle has the first moment. In general, if the tails of the density of the waiting time for completing the first cycle are no worse than those of subsequent cycles, then the two randomnesses are equivalent. The waiting time for completing the first cycle is strongly affected by the initial age distribution. Therefore, any serious attempt on proving the equivalence between these two randomnesses must take into consideration the initial age distribution. In other words, any derivation that concludes the equivalence categorically regardless of the initial age distribution is mathematically not rigorous, and consequently, such a derivation should not be viewed as a rigorous substitution for the derivation presented here.

ACKNOWLEDGMENTS

The author would like to thank Steven M. Block and Mark J. Schnitzer for encouraging him to pursue a more general derivation of the randomness parameter and for many helpful discussions during the project. This work was partially supported by the US National Science Foundation.

APPENDIX: DERIVATION OF EQUATIONS „18… and „19…

The cycle time is either 1 or 2 with equal probabilities. It follows that 具N共t兲典 can change only at an integer time. For k ⬍ 2, we have 具N共0兲典 = 0 and 具N共1兲典 = 0.5. For k 艌 2, we have the recursive relation 具N共k兲典 = 21 关1 + 具N共k − 1兲典兴 + 21 关1 + 具N共k − 2兲典兴.

共A1兲

To solve this recursive equation, we write 具N共k兲典 = 共2 / 3兲k + ak. The term 共2 / 3兲k is chosen such that 兵ak其 satisfies a homogeneous recursive equation, which is

Downloaded 19 Oct 2007 to 128.114.49.85. Redistribution subject to AIP license or copyright, see http://jmp.aip.org/jmp/copyright.jsp

103301-18

J. Math. Phys. 48, 103301 共2007兲

Hongyun Wang

1 1 ak = ak−1 + ak−2, 2 2

a0 = 0,

a1 =

−1 . 6

共A2兲

Substituting ak = ␭k into Eq. 共A2兲 to get a quadratic equation for ␭, calculating the two roots of the quadratic equation, and using linear combination ak = c1␭k1 + c2␭k2 to satisfy the two initial conditions, we arrive at ak = 共−1 / 9兲关1 − 共−1 / 2兲k兴, which leads immediately to Eq. 共18兲. For 具N2共t兲典, we have 具N2共0兲典 = 0, 具N2共1兲典 = 0.5, and the recursive relation

具N2共k兲典 = 21 具共1 + N共k − 1兲兲2典 + 21 具共1 + N共k − 2兲兲2典 = 1 + 具N共k − 1兲典 + 具N共k − 2兲典 + 21 具N2共k − 1兲典 + 21 具N2共k − 2兲典.

共A3兲

To solve this recursive equation, we write 具N2共k兲典 = 共2 / 3兲k具N共k兲典 + bk. The term 共2 / 3兲k具N共k兲典 is chosen such that 兵bk其 satisfies a homogeneous recursive equation, which is bk = 21 bk−1 + 21 bk−2,

b0 = 0,

b1 = 61 .

共A4兲

Using the method described above for solving Eq. 共A2兲, we obtain bk = 共1 / 9兲关1 − 共−1 / 2兲k兴, which leads directly to Eq. 共19兲. M. J. Schnitzer and S. M. Block, Nature 共London兲 388, 386 共1997兲. A. J. Hunt, F. Gittes, and J. Howard, Biophys. J. 67, 766 共1994兲. 3 K. Svoboda, C. F. Schmidt, B. J. Schnapp, and S. M. Block, Nature 共London兲 365, 721 共1993兲. 4 J. Howard, A. J. Hudspeth, and R. D. Vale, Nature 共London兲 342, 154 共1989兲. 5 D. L. Coy, M. Wagenbach, and J. Howard, J. Biol. Chem. 274, 3667 共1999兲. 6 E. Cinlar, Introduction to Stochastic Processes 共Prentice-Hall, Englewood Cliffs, NJ, 1975兲. 7 M. J. Schnitzer and S. M. Block, “Statistical kinetics of processive enzymes,” Cold Spring Harbor Symp. Quant. Biol. 60, 793 共1995兲. 8 J. W. Shaevitz, S. M. Block, and M. J. Schnitzer, Biophys. J. 89, 2277 共2005兲. 9 K. Svoboda, P. P. Mitra, and S. M. Block, Proc. Natl. Acad. Sci. U.S.A. 91, 11782 共1994兲. 10 C. Coppin, D. Pierce, L. Hsu, and R. Vale, Proc. Natl. Acad. Sci. U.S.A. 94, 8539 共1997兲. 11 S. M. Block, C. L. Asbury, J. W. Shaevitz, and M. J. Lang, Proc. Natl. Acad. Sci. U.S.A. 100, 2351 共2003兲. 12 J. D. Murray, Mathematical Biology 共Springer-Verlag, New York, 2002兲. 13 H. Qian and H. Wang, Europhys. Lett. 76, 15 共2006兲. 14 K. Visscher, M. Schnitzer, and S. Block, Nature 共London兲 400, 184 共1999兲. 15 S. I. Resnick, Adventures in Stochastic Processes 共Birkhauser, Boston, MA, 1992兲. 16 S. Karlin and H. M. Taylor, A First Course in Stochastic Processes 共Academic, New York, NY, 1975兲. 17 A. Budhiraja and J. Fricks, Discrete Contin. Dyn. Syst., Ser. B 6, 711 共2003兲. 1 2

Downloaded 19 Oct 2007 to 128.114.49.85. Redistribution subject to AIP license or copyright, see http://jmp.aip.org/jmp/copyright.jsp