FYTN03HT09

Exercises: Stochastic processes

1. Calculate the first three cumulants Xnc (n = 1, 2, 3) for a normally distributed randomnumber X with mean and variance 2. Use the fact that the characteristic function isthe Fourier-transform of the probability distribution and that the cumulants are definedin terms of the Taylor expansion of the logarithm of the characteristic function.

2. Consider the probability distribution for the 2-distribution with n degrees of freedom:

p(x) =

{

xn/21ex/2/[2n/2(n/2)] if x 00 otherwise

where (x) =

0 ettx1dt is the gamma function. What is the characteristic function?

What are the first two cumulants? What are the first two moments of p(x)?

3. Consider the (discrete) Poisson distribution:

pn = e

n

n!n = 0, 1, 2, 3, .. ( > 0)

Show that this distribution is normalized:

n=0 pn = 1. What is the first moments ?What is the variance 2?

4. First-passage time problems can be solved by introducing an absorbing boundary into theequations of motion. Alternatively, renewal theory relates the first passage time densityf(t) (from x = x0 to the absorbing point x = c) to the probability distribution, P (x, t|x0),in the absence of the boundary, according to:

f(s) =P (c, s|x0)P (c, s|c)

where a hat denotes the Laplace-transform: A(s) =

0 estA(t). Show by explicit calcu-

lation that the result above agrees with the result in the lecture notes for a one-dimensionalrandom walker satisfying the diffusion equation: P (x, t|x0)/t = D2P (x, t|x0)/x2.

5. Show how a random number with frequency function

p(x) =

{

(1 + a)xa if 0 < x < 10 otherwise

(a > 1)

can be obtained from a rectangularly distributed random number.

6. Show how a random number with frequency function

p(x) =

{

4/[(1 + x2)] if 0 < x < 10 otherwise

can be obtained from rectangularly distributed random numbers by (a) the transformationmethod, and (b) the accept/reject method.

Solutions

1. Calculate the first three cumulants Xnc (n = 1, 2, 3) for a normally distributed randomnumber X with mean and variance 2.

Solution: The characteristic function is the Fourier-transform of the probability distribu-tion.

(k) =1

22

dxeikxe(x)2/22

we do the variable transformation y = (x )/ which gives us

(k) =12

dyeik(y+)ey2/2 = eikk

22/2 12

dye(iky)2/2

where the integral is simply

2. Now we see that the logarithm of the characteristicfunction is simply given by

ln (k) = ik k22/2Identifying the powers of k gives Xc = , X2c = 2 and Xnc = 0 for n > 2.

2. Consider the probability distribution for the 2-distribution with n degrees of freedom:

p(x) =

{

xn/21ex/2/[2n/2(n/2)] if x 00 otherwise

where (x) =

0 ettx1dt is the gamma function. What is the characteristic function?

What are the first two cumulants? What are the first two moments of p(x)?

Solution: Taking the Fourier-transform of p(x) gives:

(k) =

eikxp(x)dx =1

2n/2(n/2)

0ex(ik1/2)xn/21dx.

Making the variable substitution s = (1/2 ik)x we obtain:

(k) =1

2n/2(1/2 ik)n/2

0 essn/21ds

(n/2)= (1 2ik)n/2

where we in the last step used the definition of the gamma function. In order to determinethe cumulants we expand:

ln (k) = ln[(1 2ik)n/2] = n2

ln(1 2ik) = n(ik) + n(ik)2 + O(k3)

from which we identifyXc = n X2c = 2n

Using the relation between cumulants and moments: Xc = X and X2c = X2X2we obtain the moments:

X = n X2 = n(n + 2)

3. Consider the (discrete) Poisson distribution:

pn = e

n

n!n = 0, 1, 2, 3, .. ( > 0)

Show that this distribution is normalized:

n=0 pn = 1. What is the first moments ?What is the variance 2?

Solution: From the fact that the Taylor-series of e is

n=0 n/n! it follows immediately

that

n=0 pn = 1. The first moment is:

=

n=0

npn =

n=1

npn = e

n=1

n

(n 1)! = e

n=1

n1

(n 1)!

e

=

The variance is:

2 =

n=0

(n )2pn =

n=0

( n2

n(n1)+n

2n + 2)pn

=

n=0

n(n 1)pn + (1 2)

n=0

npn

+2

n=0

pn

1

= e2

n=2

n2

(n 2)!

e

+ 2 =

i.e., for the Poisson distribution = 2 = .

4. First-passage time problems can be solved by introducing an absorbing boundary into theequations of motion. Alternatively, renewal theory relates the first passage time densityf(t) (from x = x0 to the absorbing point x = c) to the probability distribution, P (x, t|x0),in the absence of the boundary, according to:

f(s) =P (c, s|x0)P (c, s|c)

where a hat denotes the Laplace-transform: A(s) =

0 estA(t). Show by explicit calcu-

lation that the result above agrees with the result in the lecture notes for a one-dimensionalrandom walker satisfying the diffusion equation: P (x, t|x0)/t = D2P (x, t|x0)/x2.Solution: The solution to the 1d diffusion equation with initial condition P (x, t = 0|x0) =(x x0) is:

P (x, t|x0) =1

(4Dt)1/2exp

(

(x x0)2

4Dt

)

Taking the Laplace-transform gives:

P (x, s|x0) =1

2

s/Dexp

(

|x x0|

s/D

)

We hence get:

f(s) =P (c, s|x0)P (c, s|c)

= exp

(

|c x0|

s/D

)

so that by inverse Laplace-transform we have:

f(t) =|c x0|4Dt3

exp

(

(c x0)2

4Dt

)

.

in agreement with the lecture notes.

5. Show how a random number with frequency function

p(x) =

{

(1 + a)xa if 0 < x < 10 otherwise

(a > 1)

can be obtained from a rectangularly distributed random number.

Solution: The cumulative distribution is

C(x) =

x

0(1 + a)(x)adx = xa+1.

If R denotes a uniform random number [0, 1] then we set

C(X) = R Xa+1 = R X = R1

a+1

6. Show how a random number with frequency function

p(x) =

{

4/[(1 + x2)] if 0 < x < 10 otherwise

can be obtained from rectangularly distributed random numbers by (a) the transformationmethod, and (b) the accept/reject method.

Solution: (a) The cumulative distribution is

C(x) =4

x

0

dx

1 + (x)2=

4

tan1(x)

so that by introducing a uniform random number R [0, 1] and setting C(X) = R we getX = tan(R/4).

(b) The simplest overestimating function is f0(x) = 4/. Thus,

p0(x) =f0(x) 10 f0(x)

= 1, if 0 < x < 1

and 0 otherwise. So, we first generate a random number Xtrial from the p0(x) (uniform)distribution. We then draw a new uniform random number R and accept Xtrial withprobability p(Xtrial)/f0(Xtrial) =

11+X2

trial

, i.e., if

R