Stochastic processes

STOCHASTIC PROCESSES

• this is a theoretical course
• applied examples
• books: Sacerdote, Kolokoltsov
• exercises to do during the semester and written part in June and lectures with exercises

Schilling, Partzsch, Karlin and Taylor (Towards applications)

Brownian motion

In 1888, a biologist observed the movement of pollen plant in the river ⇒ the goal is to understand how they move They observed:

• irregular movement; translations and rotations
• moves stops
• each particle moves independently from the others
• motion is more active the less viscous is the fluid

In 1905, Einstein arrived to put together these observations ⇒ he understood that atoms are moving and bombarding macroparticles, they are moving accordingly to the movement of atoms.

❗ We can have many different microscopic behavior that give arise to the same macroscopic behavior

White noise

when a signal have all possible frequencies

Properties of brownian motion

• the movement start at x=0
• change position only at discrete times kΔt with T (= T fixed) and k ∈ ℕ
• maximum movement Δx (unit fixed) to the left or to the right with probability 1/2
• Δx does not depend on any previous position nor on the present position but on time t = kΔt
• as Δt → 0 we also have Δx → 0 due the fact that motion is continuous

We consider X_t the position of the particle at time t ∈ [0,T] with T=NΔt and N = ⌊t/Δt⌋ (the biggest integer part of t/Δt)

• The left/right movement is independent and identically distributed
• N is the number of changing position

Introducing i.i.d Bernoulli random variables {E_k}_k such that P(E_k=1) = P(E_k=0) = 1/2

The number of movement toward right is

and the number of movement toward left is given by N-S_N

So we get

Considering λ = mΔt and T = NΔt we get

We have a broken time

and we are adding points - and we are getting our new broken time -

as we have a denoty quantity of points => we don’t have any point in which we have differentiability!

⁴ Let B(t) a Brownian motion in 1 dimension on [0,1) and let B(t) be

B(t) - B(t/2) - B(t/2)

where

B(1/2) = B(t) - B(t/2)

B(t) - B(1/2) = B(t/2)

and we can observe that B(1/2) || B(t/2) and in particular both are normal with zero mean and variance 1/2

⁵ Let A and B be independent N(0,σ²) random variables.

Then A+B and A-B are two independent N(0,2σ²) random variables

_DIM\ Cov(A+B,A-B) = E([(A+B)(A-B)] = E[A²-B²] = E[A²] - E[B²] = 0

of course they are orthogonal ■

⁹ B(1/2), B(t/2) and B(t/2), B(1/2), so ...? are independent

- B(1/2) - 1/2 B(t/2) = 1/2 B(t/2) - B(t/2)

_DIM\ B(t/2) = B(t/2) + 1/2 B(t) + 1/2 B(t/2) + 1/2 B(t)

the COVARIANCE MATRIX OF Y

STANDARD GAUSSIAN RANDOM VECTORS

• independent components that are all gaussian with zero mean and variance 1. (N(0,1))

A vector of random variables Y: Ω → ℝ^m is GAUSSIAN MULTIVARIATE if there exists

1. a standard gaussian vector X in ℝ^m, m=m,
2. an m×m matrix A
3. an m-dimensional vector b such that y = Ax + b

BOX and MILLER method gives us the chance to pass from cartesian coordinates to polar coordinates (x,y) as independent ⇒ we are able to generate (r,θ) as random variables

! If we apply a linear transformation on a standard gaussian vector, we are getting a multivariate gaussian vector which is not anymore standard (in general).

! In a gaussian multivariate, we have

• cov(Y) = E[(Y-EY)(Y-EY)ᵀ] = AAT

(This equation can be proved) and from cov(Y) = AAT we can deduce that AAT should be symmetric and positively semidefined

* If AAT = I then we say that A is ORTHOGONAL.

LEMMA

If Θ is an m×m orthogonal matrix and X is an m-dimensional standard gaussian vector then ΘX is an m-dimensional standard gaussian vector.

some α > 0. Then (W_t), W_t := B_t+α - B_t is again a b.m.

3) Markov property of a Brownian motion

P [B_s ∈ A | B_t] = P [B_s ∈ A | B_ta] with t ≤ s

that is equivalent to say,

E [f (B_t+h) | B_s] = E [f (B_t+h) | B_t] with t ≤ s

for every bounded function f

4)

We say that B_s and B_t are independent if we know B_s with t ≥ s ≥ 0, 2t and we have that covariance matrix is diagonal.

(X₁,...,X_m) ~ N(0, ξ) and (X_t+1,...,X_m)

Conditioned to (X_k = x_k,...,X_l = x_l), is

N(0, Σ_2|1) where Σ_2|1 = Σ₂₂ - Σ₂₁ Σ₁₁^-1 Σ₁₂

and in our case (with t ≥ s ≥ 0, 2t) we have

ξ = _{(σ  σ  σ)}

     (σ  σ  σ)

     (σ  σ  σ)

Σ_2|1 = Σ₂₂ - Σ₂₁ Σ₁₁^-1 Σ₁₂ =

= _{(σ  σ)    (σ  σ)   ^{-1 (σ  0) =}}

    (σ  σ)      ( σ  σ)      σ²

     (σ   σ)     σ

     (σ²/σ )    

     σ

     0

5) Time inversion

Let (B_t), t ≥ 0 be a Brownian motion and fix some α > 0 (time). Then W_t := B_αt - B_αt

with λ ∈ [0, α], is again a b.m.

6) Scaling property

For all c > 0 and t > 0 we have B_ct ∼ c^1/2 B_t

and in general is too small for applications, we can enlarge it using

(i.e. it is a stopping time with respect to)

LEMMA

Let be a d-dimensional stochastic process with right-continuous sample paths and be an open set.

The first hitting time satisfies

and

EXAMPLE

For a 1-dimensional Brownian motion consider the FIRST PASSAGE TIME, the entry time into the closed set {b}.

Observe that

where the random variables are i.i.d. standard normal random variables.

Then

By the i.i.d. property of the, we get that and

Since, we conclude that almost surely.

The same argument applies to the minimum and we get

cum t → 0 p_t0,t(x,t) = δ(x-x₀) = { 1 if x = x₀ }

0 if x ≠ x₀

Now we show that in dimension 1 (∂/∂t = 1/2 ∂²/∂x²) the process BM with σ² = 2 satisfies ∂p/∂t = ∂²p/∂x².

we need the FOURIER TRANSFORM ϕ(λ,t) = 1/√(2π) ∫^∞_-∞ p_x0,t(x,t) e^-iλx dx

and does the INVERSE FOURIER TRANSFORM will be

p_x0,t(x,t) = ∫^∞_-∞ 1/√(2π) ϕ(λ,t) e^iλx dλ

To prove it, first let us suppose that p is a solution of ∂p/∂t = ∂²p/∂x² and apply the Fourier transform

we have ∂ϕ/∂t(λ,t) = ∫^∞_-∞ e^-iλx ∂p/∂t dx = ∫^∞_-∞ e^-iλx ∂²p/∂x² dx = e^-iλx ∂p/∂x_-∞^∞ - ∫^∞_-∞ e^-iλx ∂p/∂x dx

= -λ² ϕ(λ,t)

↑ we integrate by parts

Then we have ∂/∂t ϕ(λ,t) = -λ² ϕ(λ,t), that is an

ORDINARY DIFFERENTIAL EQUATION

⇒ ϕ(λ,t) = e^-λ2t const ϕ(λ)

and worry lim_{t → 0} ϕ(λ,t) = lim_{t → 0} 1/√(2π) ∫^∞_-∞ p(x,t) e^-iλx dx = = 1/√(2π) ∫^∞_-∞ e^-iλx δ(x-x₀) dx = e^-iλx₀

ϕ(λ,0) = e^-iλx₀ => ϕ(λ,t₀) = e^{-iλx₀+λ2t₀}

Now apply the inverse Fourier transform and we get

p_x0,t0(x,t) = 1/√(2π) ∫^∞_-∞ e^iλxe^{λ2(t + t₀) + iλx₀} dλ

= 1/√(2π) ∫^∞_-∞ e ^{-[λ2(t-t₀) + iλ(x-x₀)]} dλ

= 1/√(2π) e ^{-(x-x₀)2/4(t-t₀) ∫∞_-∞ e -(λN=ξ₀ - i (x-x₀)/2√(t-t₀))2 dλ}