Quantitative Methods for Ecology and Evolutionary Biology (Cambridge, 2006) - Chapter 7 pptx

7.1, so that the deterministic dynamicscharacterize what the population does ‘‘on average.’’ This identification of the average of the stochastic trajectories with the deterministictraje

The basics of stochastic population dynamics

In this and the next chapter, we turn to questions that require the use ofall of our tools: differential equations, probability, computation, and agood deal of hard thinking about biological implications of the analysis

Do not be dissuaded: the material is accessible However, accessing thismaterial requires new kinds of thinking, because funny things happenwhen we enter the realm of dynamical systems with random compo-nents These are generally called stochastic processes Time can bemeasured either discretely or continuously and the state of the systemcan be measured either continuously or discretely We will encounterall combinations, but will mainly focus on continuous time models.Much of the groundwork for what we will do was laid by physicists

in the twentieth century and adopted in part or wholly by biologists as

we moved into the twentyfirst century (see, for example, May (1974),Ludwig (1975), Voronka and Keller (1975), Costantino and Desharnais(1991), Lande et al (2003)) Thus, as you read the text you may begin tothink that I have physics envy; I don’t, but I do believe that we shouldacknowledge the source of great ideas Both in the text and in Connec-tions, I will point towards biological applications, and the next chapter

is all about them

Thinking along sample paths

To begin, we need to learn to think about dynamic biological systems in

a different way The reason is this: when the dynamics are stochastic,even the simplest dynamics can have more than one possible outcome.(This has profound ‘‘real world’’ applications For example, it means248

that in a management context, we might do everything right and still not

succeed in the goal.)

To illustrate this point, let us reconsider exponential population

growth in discrete time:

Xðt þ 1Þ ¼ ð1 þ lÞX ðtÞ (7:1)

which we know has the solution X(t)¼ (1 þ l)t

X(0) Now suppose that

we wanted to make these dynamics stochastic One possibility would be

to assume that at each time the new population size is determined by the

deterministic component given in Eq (7.1) and a random, stochastic

term Z(t) representing elements of the population that come from

‘‘somewhere else.’’ Instead of Eq (7.1), we would write

Xðt þ 1Þ ¼ ð1 þ lÞX ðtÞ þ ZðtÞ (7:2)

In order to iterate this equation forward in time, we need assumptions

about the properties of Z(t) One assumption is that Z(t), the process

uncertainty, is normally distributed with mean 0 and variance 2 In

that case, there are an infinite number of possibilities for the sequence

{Z(0), Z(1), Z(2), } and in order to understand the dynamics we

should investigate the properties of a variety of the trajectories, or

sample paths, that this equation generates In Figure 7.1, I show ten

such trajectories and the deterministic trajectory

Note that in this particular case, the deterministic trajectory ispredicted to be the same as the average of the stochastic trajectories.

If we take the expectation of Eq (7.2), we have

EfX ðt þ 1Þg ¼ Efð1 þ lÞX ðtÞg þ EfZðtÞg ¼ ð1 þ lÞEfX ðtÞg (7:3)

which is the same as Eq (7.1), so that the deterministic dynamicscharacterize what the population does ‘‘on average.’’ This identification

of the average of the stochastic trajectories with the deterministictrajectory only holds, however, because the underlying dynamics arelinear Were they nonlinear, so that instead of (1þ l)X(t), we had a termg(X(t)) on the right hand side of Eq (7.2), then the averaging as in

Eq (7.3) would not work, since in general E{g(X)}6¼ g(E{X}).The deterministic trajectory shown in Figure7.1accords with ourexperience with exponential growth Since the growth parameter issmall, the trajectory grows exponentially in time, but at a slow rate.How about the stochastic trajectories? Well, some of them are close tothe deterministic one, but others deviate considerably from the deter-ministic one, in both directions Note that the largest value of X(t) in thesimulated trajectories is about 23 and that the smallest value is about

10 If this were a model of a population, for example, we might saythat the population is extinct if it falls below zero, in which case one ofthe ten trajectories leads to extinction Note that the trajectories are just

a little bit bumpy, because of the relatively small value of the variance(try this out for yourself by simulating your own version of Eq (7.2)with different choices of l and 2)

The transition from Eq (7.1) to Eq (7.2), in which we made thedynamics stochastic rather than deterministic, is a key piece of the art ofmodeling We might have done it in a different manner For example,suppose that we assume that the growth rate is composed of a determi-nistic term and a random term, so that we write X(tþ 1) ¼ (1 þ l(t))X(t),where lðtÞ ¼ lþ ZðtÞ, and understand l to be the mean growth rate andZ(t) to be the perturbation in time of that growth rate Now, instead of

Eq (7.2), our stochastic dynamics will be

Xðt þ 1Þ ¼ ð1 þ lÞX ðtÞ þ ZðtÞX ðtÞ (7:4)

Note the difference between Eq (7.4) and Eq (7.2) In Eq (7.4), thestochastic perturbation is proportional to population size This slightmodification, however, changes in a qualitative nature the sample paths(Figure7.2) We can now have very large changes in the trajectory,because the stochastic component, Z(t), is amplified by the current value

of the state, X(t)

Which is the ‘‘right’’ way to convert from deterministic to stochasticdynamics – Eq (7.2) or Eq (7.4)? The answer is ‘‘it depends.’’ It depends

upon your understanding of the biology and on how the random factors

enter into the biological dynamics That is, this is a question of the art of

modeling, at which you are becoming more expert, and which (the

development of models) is a life-long pursuit We will put this question

aside mostly, until the next chapter when it returns with a vengeance,

when new tools obtained in this chapter are used

Brownian motion

In 1828 (Brown1828), Robert Brown, a Scottish botanist, observed that

a grain of pollen in water dispersed into a number of much smaller

particles, each of which moved continuously and randomly (as if with a

‘‘vital force’’) This motion is now called Brownian motion; it was

investigated by a variety of scientists between 1828 and 1905, when

Einstein – in his miraculous year – published an explanation of

Brownian motion (Einstein1956), using the atomic theory of matter

as a guide It is perhaps hard for us to believe today but, at the turn of the

last century, the atomic theory of matter was still just that – considered

to be an unproven theory Fuerth (1956) gives a history of the study of

Brownian motion between its report and Einstein’s publication

Beginning in the 1930s, pure mathematicians got hold of the subject,

and took it away from its biological and physical origins; they tend to

call Brownian motion a Wiener process, after the brilliant Norbert

Wiener who began to mathematize the subject

In compromise, we will use W(t) to denote ‘‘standard Brownianmotion,’’ which is defined by the following four conditions:

(1) W(0)¼ 0;

(2) W(t) is continuous;

(3) W(t) is normally distributed with mean 0 and variance t;

(4) if {t1, t2, t3, t4} represent four different, ordered times with t1<t2<t3<t4(Figure7.3), then W(t2) W(t1) and W(t4) W(t3) are independent randomvariables, no matter how close t3is to t2 The last property is said to be theproperty of independent increments (seeConnectionsfor more details) and

is a key assumption

In Figure7.4, I show five sample trajectories, which in the ness are described as ‘‘realizations of the stochastic process.’’ They allstart at 0 because of property (1) The trajectories are continuous,forced by property (2) Notice, however, that although the trajectoriesare continuous, they are very wiggly (we will come back to thatmomentarily)

busi-For much of what follows, we will work with the ‘‘increment ofBrownian motion’’ (we are going to convert regular differential equa-tions of the sort that we encountered in previous chapters into stochasticdifferential equations using this increment), which is defined as

Now, although Brownian motion and its increment seem verynatural to us (perhaps because have spent so much time working withnormal random variables), a variety of surprising and non-intuitiveresults emerge To begin, let’s ask about the derivative dW/dt SinceW(t) is a random variable, its derivative will be one too Using thedefinition of the derivative

A key assumption of the

process of Brownian motion

is that W(t 2 )  W(t 1 ) and

W(t 4 )  W(t 3 ) are independent

random variables, no matter

how close t 3 is to t 2

and we conclude that the average value of dW/dt is 0 But look what

happens with the variance:

but we had better stop right here, because we know what is going to

happen with the limit – it does not exist In other words, although the

sample paths of Brownian motion are continuous, they are not

differ-entiable, at least in the sense that the variance of the derivative exists

Later in this chapter, in the section on white noise (see p 261), we will

make sense of the derivative of Brownian motion For now, I want to

introduce one more strange property associated with Brownian motion

and then spend some time using it

Suppose that we have a function f (t,W ) which is known and well

understood and can be differentiated to our hearts’ content and for

which we want to find f (tþ dt, w þ dW ) when dt (and thus E{dW2})

Figure 7.4 Five realizations

of standard Brownian motion.

is small and t and W (t ) ¼ w are specified We Taylo r expand in the usualmanner, using a subsc ript to denote a derivativ e

f ðt þ d t ; w þ dW Þ ¼f ðt ; w Þ þ ft dt þ fw dW

þ12

The gamble r’s ruin in a fair game

Many – perha ps all – books on stochasti c proce sses or proba bilityinclude a sectio n on gambling becau se, let’s face it, what is the point

of studying proba bility and stochasti c processe s if you can’ t become abetter gamb ler (see also Dubins and Savage ( 1976 ))? The gam blingproblem als o allow s us to introdu ce some ideas that will flow throughthe rest of this chapt er and the next chapter

Imagi ne that you are playing a fair game in a casino (we will discussreal casinos, which always have the edge, in the next sect ion) and thatyour current holdings are X( t) dollars You are out of the gam e whenX( t) falls to 0 and you break the bank when your holdings X (t ) reach thecasino holdings C If you think that this is a purely mathem aticalproblem and are impatie nt for biology, make the followi ng analogy:X( t) is the size at time t of the popul ation descended from a propa gule ofsize x that reached an island at time t ¼ 0; X( t) ¼ 0 correspo nds toextincti on of the popul ation and X( t) ¼ C correspo nds to succe ssfulcoloniz ation of the island by the descendant s of the propa gule Withthis interp retation, we have one of the models for island biogeogra phy

of MacAr thur and Wilson (1967 ), which will be discusse d in the nextchapter

Since the game is fair, we may assume that the change in your

holdings are determined by a standard Brownian motion; that is, your

holdings at time t and time tþ dt are related by

Xðt þ dtÞ ¼ X ðtÞ þ dW (7:11)

There are many questions that we could ask about your game, but I

want to focus here on a single question: given your initial stake X(0)¼ x,

what is the chance that you break the casino before you go broke?

One way to answer this question would be through simulation of

trajectories satisfying Eq (7.11) We would then follow the

trajec-tories until X(t) crosses 0 or crosses C and the probability of breaking

the casino would be the fraction of trajectories that cross C before

they cross 0 The trajectories that we simulate would look like those

in Figure 7.4 with a starting value of x rather than 0 This method,

while effective, would be hard pressed to give us general intuition and

might require considerable computer time in order for us to obtain

accurate answers So, we will seek another method by thinking along

sample paths

In Figure7.5, I show the t x plane and the initial value of your

holdings X(0)¼ x At at time dt later, your holdings will change to

xþ dW, where dW is normally distributed with mean 0 and variance

dt Suppose that, as in the figure, they have changed to xþ w, where we

can calculate the probability of dW falling around w from the normal

distribution What happens when you start at this new value of

hold-ings? Either you break the bank or you go broke; that is, things start over

exactly as before except with a new level of holdings But what happens

between 0 and dt and after dt are independent of each other because of

the properties of Brownian motion Thus, whatever happens after dt is

determined solely by your holdings at dt And those holdings are

normally distributed

To be more formal about this, let us set

uðxÞ ¼ PrfX ðtÞ hits C before it hits 0jX ð0Þ ¼ xg (7:12)

(which could also be recognized as a colonization probability, using the

metaphor of island biogeography) and recognize that the argument of

the previous paragraph can be summarized as

uðxÞ ¼ EdWfuðx þ dW Þg (7:13)

where EdWmeans to average over dW Now let us Taylor expand the

right hand side of Eq (7.13) around x:

uðxÞ ¼ EdW uðxÞ þ dWuxþ1

to x þ w, where w has a normal distribution with mean 0 and variance dt.

We can thus relate u(x) at this time to the average of u(x þ dW) at a slightly later time (later by dt).

and take the average over dW, remembering that it is normally uted with mean 0 and variance dt:

distrib-uðxÞ ¼ distrib-uðxÞ þ1

2uxxdtþ oðdtÞ (7:14b)

The last two equations share the same number because I want toemphasize their equivalence To finish the derivation, we subtract u(x)from both sides, divide by dt and let dt! 0 to obtain the especiallysimple differential equation

which we now solve by inspection The second derivative is 0, so thefirst derivative of u(x) is a constant ux¼ k1and thus u(x) is a linearfunction of x

uðxÞ ¼ k2þ k1x (7:16)

We will find these constants of integration by thinking about theboundary conditions that u(x) must satisfy

From Eq (7.12), we conclude that u(0) must be 0 and u(C) must

be 1 since if you start with x¼ 0 you have hit 0 before C and if youstart with C you have hit C before 0 Since u(0)¼ 0, from Eq (7.16)

we conclude that k2¼ 0 and to make u(C) ¼ 1 we must have k1¼ 1/C sothat u(x) is

uðxÞ ¼x

What is the typical relationship between your initial holdings andthose of a casino? In general C x, so that u(x)  0 – you are almostalways guaranteed to go broke before hitting the casino limit

But, of course, most of us gamble not to break the bank, but to havesome fun (and perhaps win a little bit) So we might ask how long it will

be before the game ends (i.e., your holdings are either 0 or C) Toanswer this question, set

TðxÞ ¼ average amount of time in the game; given X ð0Þ ¼ x (7:18)

We derive an equation for T(x) using logic similar to that which took

us to Eq (7.15) Starting at X(0)¼ x, after dt the holdings will be

xþ dW and you will have been in the game for dt time units Thus weconclude

TðxÞ ¼ dt þ EdWfT ðx þ dW Þg (7:19)

and we would now proceed as before, Taylor expanding, averaging,dividing by dt and letting dt approach 0 This question is better left as anexercise

Exercise 7.2 (M)

Show that T(x) satisfies the equation1 ¼ (1/2) Txxand that the general solution

of this equation is T(x)¼  x2þ k1xþ k2 Then explain why the boundary

conditions for the equation are T(0)¼ T(C) ¼ 0 and use them to evaluate the

two constants Plot and interpret the final result for T(x)

The gambler’s ruin in a biased game

Most casinos have a slight edge on the gamblers playing there This

means that on average your holdings will decrease (the casino’s edge) at

rate m, as well as change due to the random fluctuations of the game To

capture this idea, we replace Eq (7.11) by

dX¼ X ðt þ dtÞ  X ðtÞ ¼ mdt þ dW (7:20)

Exercise 7.3 (E/M)

Show that dX is normally distributed with mean –mdt and variance dtþ o(dt) by

evaluating E{dX} and E{dX2} using Eq (7.20) and the results of Exercise7.1

As before, we compute u(x), the probability that X(t) hits C

before 0, but now we recognize that the average must be over dX

rather than dW, since the holdings change from x to xþ dX due to

deterministic (mdt) and stochastic (dW) factors The analog of

Eq (7.13) is then

uðxÞ ¼ EdXfuðx þ dX Þg ¼ EdXfuðx  mdt þ dW Þg (7:21)

We now Taylor expand and combine higher powers of dt and dW into a

term that is o(dt)

uðxÞ ¼ EdX uðxÞ þ ðmdt þ dW Þuxþ1

2ðmdt þ dW Þ2uxxþ oðdtÞ

(7:22)

We expand the squared term, recognizing that O(dW2) will be order

dt, take the average over dX, divide by dt and let dt! 0 (you should

write out all of these steps if any one of them is not clear to you) to

obtain

1

2uxx mux¼ 0 (7:23)

which we need to solve with the same boundary conditions as before

u(0)¼ 0, u(C) ¼ 1 There are at least two ways of solving Eq (7.23)

I will demonstrate one; the other uses the same method that we used in

Chapter2to deal with the von Bertalanffy equation for growth

Let us set w¼ ux, so that Eq (7.23) can be rewritten as wx¼ 2mw,for which we immediately recognize the solution w(x)¼ k1e2mx, where

k1is a constant Since w(x) is the derivative of u(x) we integrate again toobtain

uðxÞ ¼ k2e2mxþ k3 (7:24)

where k2and k3are constants and, to be certain that we are on the samepage, try the next exercise

Exercise 7.4 (E)

What is the relationship between k1and k2?

When we apply the boundary condition that u(0)¼ 0, we concludethat k3¼ k2, and when we apply the boundary condition u(C)¼ 1, weconclude that k2¼ 1/(e2mC 1) We thus have the solution for theprobability of reaching the limit of the casino in a biased game:

uðxÞ ¼e2mx 1

x

Figure 7.6 When the game is

biased, the chance of reaching

the limit of the casino before

going broke is vanishingly

small Here I show u(x) given by

Eq ( 7.25 ) for m ¼ 0.1 and

C ¼ 100 Note that if you start

with even 90% of the casino

limit, the situation is not very

good Most of us would start

with x  C and should thus just

enjoy the game (or develop a

system to reduce the value of

m, or even change its sign.)

Exercise 7.5 (M/H)

Derive the equation for T(x), the mean time that you are in the game when

dX is given by Eq (7.20) Solve this equation for the boundary conditions

T(0)¼ T(C) ¼ 0

Exercise 7.6 (E/M)

When m is very small, we expect that the solution of Eq (7.25) should be close

to Eq (7.17) because then the biased game is almost like a fair one Show that

this is indeed the case by Taylor expansion of the exponentials in Eq (7.25) for

m! 0 and show that you obtain our previous result If you have more energy

after this, do the same for the solutions of T(x) from Exercises7.5and7.2

Before moving on, let us do one additional piece of analysis In

general, we expect the casino limit C to be very large, so that 2mC 1

Dividing numerator and denominator of Eq (7.25) by e2mCgives

uðxÞ ¼e

2mðCxÞ e2mC

1 e2mC  e2mðCxÞ (7:26)

with the last approximation coming by assuming that e2mC 1 Now

let us take the logarithm to the base 10 of this approximation to u(x), so

that log10(u(x))¼ 2m(C  x)log10e I have plotted this function in

Figure7.7, for x¼ 10 and C ¼ 50, 500, or 1000 Now, C ¼ 1000, x ¼ 10,

and m¼ 0.01 probably under-represents the relationship of the bank of a

casino to most of us, but note that, even in this case, the chance of

reaching the casino limit before going broke when m¼ 0.01 is about 1 in

on Eq ( 7.26 ) for x ¼ 10 and

C ¼ 50, 500, or 1000, as a function of m.

a billion So go to Vegas, but go for a good time (In spring 1981, myfirst year at UC Davis, I went to a regional meeting of the AmericanMathematical Society, held in Reno, Nevada, to speak in a session onapplied stochastic processes Many famous colleagues were they, andalthough our session was Friday, they had been there since Tuesdaydoing, you guessed it, true work in applied probability All, of course,claimed positive gains in their holdings.)

The transition density and covariance

of Brownian motion

We now return to standard Brownian motion, to learn a little bit moreabout it To do this, consider the interval [0, t] and some intermediatetime s (Figure7.8) Suppose we know that W(s)¼ y, for s < t What can

be said about W(t)? The increment W(t) W(s) ¼ W(t)  y will be ally distributed with mean 0 and variance t s Thus we conclude that

norm-Prfa  W ðtÞ  bg ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1

2pðt  sÞp

ðb aexp ðx  yÞ

22ðt  sÞ

!

dx (7:27)

Note too that we can make this prediction knowing only W(s), and nothaving to know anything about the history between 0 and s A stochasticprocess for which the future depends only upon the current value andnot upon the past that led to the current value is called a Markov process,

so that we now know that Brownian motion is a Markov process.The integrand in Eq (7.27) is an example of a transition densityfunction, which tells us how the process moves from one time and value

to another It depends upon four values: s, y, t, and x, and we shall write it as

qðx; t; y; sÞdx ¼ Prfx  W ðtÞ  x þ dxjW ðsÞ ¼ yg

¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi12pðt  sÞ

p exp ðx  yÞ

22ðt  sÞ

!

dx (7:28)

This equation should remind you of the diffusion equation encountered

in Chapter2, and the discussion that we had there about the strangeproperties of the right hand side as t decreases to s In the next section all

of this will be clarified But before that, a small exercise

Figure 7.8 The time s divides

the interval 0 to t into two

pieces, one from 0 to just before

s (s  ) and one from just after

s (s þ ) to t The increments in

Brownian motion are then

independent random variables.

s–s+

t W(s–) –W(0) and W(t)–W(s+) are independent

random variables

Exercise 7.7 (E/M)

Show that q(x, t, y, s) satisfies the differential equation qt¼ (1/2)qxx What

equation does q(x, t, y, s) satisfy in the variables s and y (think about the

relationship between qtand qsand qxxand qyybefore you start computing)?

Keeping with the ordering of time in Figure7.8, let us compute the

where the last line of Eq (7.29) follows because W(s) W(0) and

W(t) W(s) are independent random variables, with mean 0 Suppose

that we had interchanged the order of t and s Our conclusion would then

be that E{W(t)W(s)}¼ t In other words

EfW ðtÞW ðsÞg ¼ minðt; sÞ (7:30)

and we are now ready to think about the derivative of Brownian motion

Gaussian ‘‘white’’ noise

The derivative of Brownian motion, which we shall denote by

(t)¼ dW/dt, is often called Gaussian white noise It should already be

clear where Gaussian comes from; the origin of white will be understood

at the end of this section, and the use of noise comes from engineers, who

see fluctuations as noise, not as the element of variation that may lead to

selection; Jaynes (2003) has a particularly nice discussion of this point

We have already shown the E{(t)}¼ 0 and that problems arise when we

try to compute E{(t)2} in the usual way because of the variance of

Brownian motion (recall the discussion around Eq (7.8)) So, we are

going to sneak up on this derivative by computing the covariance

EfðtÞðsÞg ¼ q q

qtqsEfW ðtÞW ðsÞg (7:31)

Note that I have exchanged the order of differentiation and integration

in Eq (7.31); we will do this once more in this chapter In general, one

needs to be careful about doing such exchanges; both are okay here (if

you want to know more about this question, consult a good book on

advanced analysis) We know that E{W(t)W(s)}¼ min(t, s) Let us think

about this covariance as a function of t, when s is held fixed, as if it were

just a parameter (Figure7.9)

ðt; sÞ ¼ minðt; sÞ ¼ t if t5 s

s if t s (7:32)

Now the derivative of this function will be discontinuous; since thederivative is 1 if t < s, and is 0 if t > s, there is a jump at t¼ s(Figure 7.9) We are going to deal with this problem by using theapproach of generalized functions described in Chapter2 (and in thecourse of this, learn more about Gaussians).

We will replace the derivative (q/qt)(t, s) by an approximation that issmooth but in the limit has the discontinuity Define a family of functions

nðt; sÞ ¼

ffiffiffinpffiffiffiffiffiffi2ppð1

ðtsÞexp nx

22

p exp nx

22

ðtsÞexp nx

22

dx¼ limn!1

q

qtnðt; sÞ (7:33)

When t¼ s, the lower limit of the integral is 0, so that the integral is 1/2

To understand what happens when t does not equal s, the followingexercise is useful

t s

(d)

t s

(c)

1/2

t s

1 (b)

t s

s

(a) Figure 7.9 (a) The covariance

function (t,s) ¼ E{W(t)W(s)} ¼

min(t,s), thought of as a

function of t with s as a

parameter (b) The derivative

of the covariance function

approximate derivative is the

tail of the cumulative Gaussian

from t ¼ s.

ffiffin p ðtsÞexp y22

The form of the integral in expression (7.34) lets us understand what

will happen when t6¼ s If t < s, the lower limit is negative, so that as

n! 1 the integral will approach 1 If t > s, the lower limit is positive so

that as n increases the integral will approach 0 We have thus

con-structed an approximation to the derivative of the correlation function

Equation (7.31) tells us what we need to do next We have

con-structed an approximation to (q/qt)(t, s), and so to find the covariance

of Gaussian white noise, we now need to differentiate Eq (7.33) with

respect to s Remembering how to take the derivative of an integral with

respect to one of its arguments, we have

q q

qtqsðt; sÞ ¼ limn!1

ffiffiffinpffiffiffiffiffiffi2p

p exp nðt  sÞ

22

!

¼ limn!1nðt  sÞ (7:35)

Now, n(t s) is a Gaussian distribution centered not at 0 but at t ¼ s

with variance 1/n Its integral, over all values of t, is 1 but in the limit

that n! 1 it is 0 everywhere except at t ¼ s, where it is infinite In other

words, the limit of n(t s) is the Dirac delta function that we first

encountered in Chapter2(some n(x) are shown in Figure7.10)

This has been a tough slog, but worth it, because we have shown that

EfðtÞðsÞg ¼ ðt  sÞ (7:36)

We are now in a position to understand the use of the word ‘‘white’’

in the description of this process Historically, engineers have workedinterchangeably between time and frequency domains (Kailath1980)because in the frequency domain tools other the ones that we considerare useful, especially for linear systems (which most biological systemsare not) The connection between the time and frequency (Stratonovich

1963) is the spectrum S(o) defined for the function f(t) by

where the last equality follows because the delta function picks out

t¼ 0, for which the exponential is 1 The spectrum of Eq (7.36) is thusflat (Figure7.11): all frequencies are equally represented in it Well, that

is the description of white light and this is the reason that we call thederivative of Brownian motion white noise In the natural world, thecovariance does not drop off instantaneously and we obtain spectra withcolor (seeConnections)

The Ornstein–Uhlenbeck process and stochastic integrals

In our analyses thus far, the dynamics of the stochastic processhave been independent of the state, depending only upon Brownianmotion We will now begin to move beyond that limitation, but do itappropriately slowly To begin, recall that if X(t) satisfies the dynamics

Spectrum, S( ω)

Frequency, ω

Figure 7.11 The spectrum of

the covariance function given

by Eq ( 7.36 ) is completely

flat so that all frequencies are

equally represented Hence

the spectrum is ‘‘white.’’ In the

natural world, however, the

higher frequencies are less

represented, leading to a

fall-off of the spectrum.

dX/dt¼ f (X) and K is a stable steady state of this system, so that

f (K)¼ 0, and we consider the behavior of deviations from the steady

state Y(t)¼ X(t)  K then, to first order, Y(t) satisfies the linear dynamics

dY / dt¼  | f0(K)|Y, where f0(K) is the derivative of f (X) evaluated at K

We can then define a relaxation parameter ¼ | f0(K)| so that the

dynamics of Y are given by

dY

We call  the relaxation parameter because it measures the rate at which

fluctuations from the steady state return (relax) towards 0 Sometimes

this parameter is called the dissipation parameter

Exercise 7.9 (E)

What is the relaxation parameter if f (X) is the logistic rX(1 (X / K))? If you

have the time, find Levins (1966) and see what he has to say about your result

We fully understand the dynamics of Eq (7.39): it represents return

of deviations to the steady state: which ever way the deviation starts

(above or below K), it becomes smaller However, now let us ask what

happens if in addition to this deterministic attraction back to the steady

state, there is stochastic fluctuation That is, we imagine that in the next

little bit of time, the deviation from the steady state declines because of

the attraction back towards the steady state but at the same time is

perturbed by factors independent of this decline Bjørnstadt and

Grenfell (2001) call this process ‘‘noisy clockwork;’’ Stenseth et al

(1999) apply the ideas we now develop to cod, and Dennis and Otten

(2000) apply them to kit fox

We formulate the dynamics in terms of the increment of Brownian

motion, rather than white noise, by recognizing that in the limit dt! 0,

Eq (7.39) is the same as dY¼  Y dt þ o(dt) and so our stochastic

version will become

where  is allowed to scale the intensity of the fluctuations The

stochastic process generated by Eq (7.40) is called the Ornstein–

Uhlenbeck process (seeConnections) and contains both deterministic

relaxation and stochastic fluctuations (Figure7.12) Our goal is to now

characterize the mixture of relaxation and fluctuation

To do so, we write Eq (7.40) as a differential by using the

integrat-ing factor etso that

dðetYÞ ¼ etdW (7:41)