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Stochastic Differential

Equations

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Brownian Motion

• Fundamental example of a stochasticprocess.

• Standard Wiener Process.

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Pollen Grain

zn +1 = z

n + ∆ zn

.

∆ z n N (0, 1) ,

E (∆ zn ) = 0 ,Var( ∆ zn ) = E (∆ z 2

n ) − µ2 = E (∆ z 2n ) = 1 .

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Particle Track

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100 Particle Tracks

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Probability Density Function at differenttimes

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Standard deviation as a function of time

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If start time is s then standard

deviation is

√ t

−

s

Change ∆ t = 1 to ∆ t = t − s

∆ z n

√t −sN (0, 1) ,

zn +1 = z

n + ∆ zn

.

∆ z n N (0, 1) ,

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Useful to discretise Brownian motion

• Set δ t = T/N

• Let W (t j ) = W j , t j = j δ t

• W 0 = 0

• W j = W j − 1 + dW j , dW j = √ δ tN (0, 1)

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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1–0.5

0

0.5

1

t

W(t)

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u (W (t )) = e t + 12

W ( t )

Evaluate the function

along 1000 discretized Brownian paths

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N J

N

Trajectory of the differential equation

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Random Di ff erential Equations

It may happen that function is not completely known and contain somerandom parameters. It is also possible that starting condition is random.These equations can be called random di ff erential equations.

Example:

dx = a (ω )xdt + b(t, ω )dt

x (t 0 ) = c(ω )

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Stochastic Di ff erential Equations

The randomness can appear inside the underling function. The solutionhas irregular stochastic behavior. Then we talk about stochasticdiff erential equations.

Symbolically they are written as:

X t = a (t, X t )dt + b(t, X t )dW t

X t 0 = X 0

More precisely we obtain integral representation :

X t = X 0 + t

t 0

a (s, X s )ds + t

t 0

b(s, X s )dW s

d

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Applications

Investment nance

Option Pricing

Seismology

Hydrology...

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Solutions of equations

Stochastic Di ff erential Equation

dX t = a (t, X t )dt + b(t, X t )dW t

X t =?

the solution is a stochastic process

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0 5 10

X*E2

0

2 0

4 0

6 0

Y

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0 5 10

X*E2

- 1 0 0

- 5 0

0

5 0

Y

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Discrete stochastic integrals

We would like an approximate solution to the simplied stochastic di ff erential equation

dx (t ) = W (t )dW (t ) , (1)

where the behavior of W (t ) is governed by the rules of Brownian motion. Integrating bothsides, we obtain

x (t ) = t

0W (s )dW (s ) , (2)

where we have assumed that x (0) = 0 for simplicity.

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The It o integral

In the It o integral, the integral (2) is approximated with the Riemann sum

x (t ) =N − 1

j =0

W j (W j +1 − W j ) , (3)

where we have assumed that t = N ∆ t , W N = W (t ) and W 0 = W (0). This can be written as

x (t ) =12

N − 1

j =0

W 2 j +1

− W 2 j

− W 2 j +1

− 2W j W j +1 + W 2 j ,

= 12

N − 1

j =0

W 2 j +1

− W 2 j

− (W j +1 − W j )2 ,

=12

(W 21

− W 20 ) + ( W 2

2− W 2

1 ) + . . . + ( W 2N − 1

− W 2N − 2 ) + ( W 2

N − W 2

N − 1 )

−N − 1

j =0

(W j +1 − W j )2 ,

= 12

W 2N − W 2

0 − 12

N − 1

j =0

(W j +1 − W j )2 ,

=12

W 2N

− W 20

− 12

N − 1

j =0

∆ W 2 j .

]

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The sum can be written asN − 1

j =0

∆ W 2 j = N 1N

N − 1

j =0

∆ W 2 j ,

the discrete form of the variance of a random variable with zero mean,

Var( ∆ W ) = E (∆ W 2 ) =1N

N − 1

j =0

∆ W 2 j .

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Since we know that ∆ W j = W j +1 − W j is normally distributed with mean 0 and variance∆ t because it governs the jump for Brownian motion, then, as N → ∞ ,

1N

N − 1

j =0

∆ W 2 j = ∆ t , (6)

The approximation of the integral becomes

x (t ) =12

W 2N − W 20 −12

N ∆ t .

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From the denition of Brownian motion, W 0 = W (0) = 0, and after substituting in forW N = W (t ) and N ∆ t = t , we have

x (t ) =12W (t )

2−

12 t . (8)

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The Stratonovich integral

Rather than approximating the integral at the left side of the interval, the Stratonovichintegral approximate the integral (2) with the midpoint rule

x (t ) =N − 1

j =0

W t j + t j +1

2(W j +1 − W j ) . (9)

We will approximate the value of W (t j +1 / 2 ) with

W (t j +1 / 2 ) =12

(W j + W j +1 ) + C j , (10)

C must be determined to satisfy the rules of Brownianmotion

Let Z (t j ) = W (t j +1 / 2 ) satisfy the rules of Brownian motion.

Then∆ Z j = Z (t j + ∆ t ) − Z (t j )

hasE (∆ Z j ) = 0 , V ar (∆ Z j ) = ∆ t

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∆ Z j =12

[(W j +1 + W j +2 ) + 2 C j +1 − (W j + W j +1 ) − 2C j ] ,

=12

(W j +2 − W j ) + ( C j +1 − C j ) .

The increment is given by

SinceW j +1 = W j + ∆ W j

W j +2 = W j +1 + ∆ W j +1

ThenW

j+2 = W

j+ ∆ W

j+ ∆ W

j+1

and

∆ Z j =12

(∆ W j + ∆ W j +1 ) + ∆ C j ,

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E (∆ Z j ) = E 12

(∆ W j + ∆ W j +1 ) + ∆ C j ,

= 12

[E (∆ W j ) + E (∆ W j +1 )] + E (∆ C j ) ,

but since E (∆ W j ) = 0 and E (∆ W j +1 ) = 0, and we require that E (∆ Z j ) = 0, then we knowthat the behavior of C j must satisfy E (∆ C j ) = E (C j +1 ) − E (C j ) = 0. This is most easilysatised by requiring that C j have a zero mean. Taking the variance of ∆ Z j yields

Var( ∆ Z j ) = E (∆ Z 2 j ) − E (∆ Z j )2 ,

but since we require that E (∆ Z j ) = 0, then the variance is given by

Var( ∆ Z j ) = E (∆ Z 2

j ) .

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Substituting in the values from above we have

Var( ∆ Z j ) = E 14

(∆ W j + ∆ W j +1 + 2 ∆ C j )2 ,

= 14

E (∆ W j + ∆ W j +1 )2 + 4 E (∆ C j ∆ W j ) + 4 E (∆ C j ∆ W j +1 ) + 4 E ∆ C 2 j ,

=14

E ∆ W 2 j + 2 E (∆ W j ∆ W j +1 ) + E ∆ W 2 j +1

+4 E (∆ C j ∆ W j ) + 4 E (∆ C j ∆ W j +1 ) + 4 E ∆ C 2 j .

Because ∆ W j , ∆ W j +1 , and ∆ C j are all independent of each other, then we know that

E (∆ W j ∆ W j +1 ) = E (∆ W j )E (∆ W j +1 ) = 0 ,

E (∆ C j ∆ W j +1 ) = E (∆ C j )E (∆ W j +1 ) = 0 ,E (∆ C j ∆ W j ) = E (∆ C j )E (∆ W j ) = 0 ,

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we have

Var( ∆ Z j ) =14

E (∆ W 2 j ) + E (∆ W 2 j +1 ) + E (∆ C 2 j ) .

Because ∆ W j and ∆ W j +1 satisfy the rules for Brownian motion, and we want ∆ Z j to satisfythe rules for Brownian motion, then we must have

Var( ∆ Z j ) = ∆ t ,

E (∆ W 2 j ) = ∆ t ,

E (∆ W 2 j +1 ) = ∆ t ,

we haveE (∆ C 2 j ) =

∆ t2

.

Substituting in for ∆ C j = C j +1 − C j , we have

E (C 2 j +1 ) − 2E (C j +1 C j ) + E (C 2 j ) = ∆ t2

.

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x(t) =12

N − 1

j =0

(W j + W j +1 − 2C j ) (W j +1 − W j ) ,

=12

N − 1

j =0

W 2 j +1 − W 2 j −

N − 1

j =0

C j (W j +1 − W j ) ,

=12 W

21 − W

20 + W

22 − W

21 + . . . + W

2N − 1 − W

2N − 2 + W

2N − W

2N − 1

−

N − 1

j =0

C j (W j +1 − W j ) ,

=12

W 2N − W 20 − N 1N

N − 1

j =0

C j ∆ W j .

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The last term corresponds to an approximation of NE (C j ∆ W j ), which, since C j and ∆ W jare independent, approaches 0 as N → ∞ . Therefore, with W 0 = W (0) = 0 and W N = W (t),

x(t) =12

W (t)2 .

Comparing the two methods, we see that, depending on where W (t) is evaluated whenapproximating

x(t) = t

0W (s)dW (s) ,

the result can be very di ff erent, since

x(t) = 1

2W (t)2

− t

2It o

x(t) = 12 W (t)2 Stratonovich .

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