Lectures 1–2 : Repetition. Stochastic Processes. The Kolmogorov Construction. Stochastically Equivalent Processes. Processes with Independent Increments and Independent Stationary Increments. Stochastically Continuous Processes. The Wiener Process, Brownian Motion. Construction and Simulation of the Brownian Motion. The Poisson Process. Construction and Simulation of the Poisson Process.

Literature.

1. Ash, R.B., Gardner, M.L. Topics in Stochastic Processes, Academic Press, 1975.

2. Cox, D.R., Miller, H.D. The Theory of Stochastic Processes, Methuen, 1970.

3. Cramer, H., Leadbetter, M.R. Stationary and Related Stochastic Processes, John Wiley, 1968.

4. Doob, J.L. Stochastic Processes, John Wiley, 1967.

5. Hoel, P.G., Port, S.C., Stone, C.J. Introduction to Stochastic Processes, Houghton Mifflin, 1972.

6. Todorovic, P. An Introduction to Stochastic Processes and Their Applications, Springer-Verlag, 1992.

Repetition

Definition 1. The set of all possible outcomes of an experiment is called the sample space and it is denoted by Ω.

Definition 2. A collection F of subsets of Ω is called a σ-field if it satisfies the following conditions:

(a) ∅, Ω ∈ F;

(b) if A ∈ F then A^{c} ∈ F;

(c) if A_{1}, A_{2}, … ∈ F then ∪_{k=1}^{∞}A_{k} ∈ F.

Definition 3. A Borel σ-field of subsets of R is the smallest σ-field containing all open sets. It is denoted by B(R).

Definition 4. A Borel σ-field of subsets of R^{n} is the smallest σ-field containing all open sets. It is denoted by B(R^{n}).

Definition 5. A pair (S, G), where S is a set and G is a σ-field of subsets of S is called a measurable space.

Definition 6. A probability P on (Ω, F) is a function P : F → [0, 1] satisfying the conditions:

(a) P (A) ≥ 0 for any A ∈ F;

(b) P (Ω) = 1;

(c) P (∪_{k=1}^{∞}A_{k}) = ∑_{k=1}^{∞} P (A_{k}) whenever A_{i} ∩ A_{j} = ∅ for any i, j, i ≠ j.

Definition 7. A triple (Ω, F, P) comprising a measurable space (Ω, F) with a probability P on it is called a probability space.

Definition 8. A function X : Ω → S is said to be measurable iff {ω : X(ω) ∈ A} ∈ F for each A ∈ G. It is denoted by X : (Ω, F) → (S, G).

Definition 9. A random variable X is a measurable function X : (Ω, F) → (R, B(R)).

Definition 10. An n-dimensional random vector is a measurable function X : (Ω, F) → (R^{n}, B(R^{n})). The random vector X may be regarded as an n-tuple (X_{1}, …, X_{n}) of random variables.

The next several definitions are formulated for n-dimensional random vectors. The analogous definitions for random variables can be obtained from them taking n = 1.

Definition 11. An n-dimensional random vector X is called discrete if it takes at most countable number of values {(x_{11}, …, x_{1n}), …, (x_{k1}, …, x_{kn}), …}.

Definition 12. An n-dimensional random vector X is called continuous if it can take any value from an n-dimensional subset of R^{n}.

Definition 13. The distribution of a discrete random vector X = (X_{1}, …, X_{n}) is the collection of probabilities

P (X_{1} = x_{1}, …, X_{n} = x_{n}) for all possible (x_{1}, …, x_{n}).

Definition 14. The distribution function of a continuous random vector X = (X_{1}, …, X_{n}) is the function F_{X} : R^{n} → [0, 1] given by

F_{X}(x_{1}, …, x_{n}) = P (X_{1} ≤ x_{1}, …, X_{n} ≤ x_{n}) (x_{1}, …, x_{n}) ∈ R^{n}.

Definition 15. An n-dimensional continuous random vector X is called absolutely continuous if its distribution function can be expressed as

F_{X}(x_{1}, …, x_{n}) = ∫_{−∞}^{x_{1}} … ∫_{−∞}^{x_{n}} f_{X}(u_{1}, …, u_{n})du_{1} … du_{n} ∀(x_{1}, … x_{n}) ∈ R^{n},

for some integrable function f_{X} : R^{n} → [0, ∞). The function f_{X} is called the density function of X.

Fact 1. If n-dimensional random vector X is absolutely continuous then

P (X ∈ A) = ∫ … ∫_{A} f_{X}(u_{1}, …, u_{n})du_{1} … du_{n} ∀A ∈ B(R^{n}).

Definition 16. Continuous random variables X_{1}, …, X_{n} are called independent iff

P (X_{1} ≤ x_{1}, …, X_{n} ≤ x_{n}) = P (X_{1} ≤ x_{1}) … P (X_{n} ≤ x_{n}) ∀(x_{1}, …, x_{n}) ∈ R^{n}

or equivalently, in the case when they are absolutely continuous,

f_{X}(u_{1}, … u_{n}) = f_{X_{1}}(u_{1}) … f_{X_{n}}(u_{n}) ∀(u_{1}, …, u_{n}) ∈ R^{n},

where X = (X_{1}, …, X_{n}). If they are discrete then they are independent iff

P (X_{1} = x_{1}, …, X_{n} = x_{n}) = P (X_{1} = x_{1}) … P (X_{n} = x_{n}) for all possible (x_{1}, …, x_{n}).

Definition 17. The conditional density function of a random variable Y given a random variable X, written f_{Y |X}, is defined to be

f_{Y |X}(x, y) = \frac{f_{(X,Y)}(x, y)}{f_{X}(x)}

for every y ∈ R and every x ∈ R such that f_{X}(x) > 0.

Definition 18. The conditional distribution function of a random variable Y given a random variable X, written F_{Y |X} or P (Y ≤ y|X = x), is defined to be

F_{Y |X}(x, y) = ∫_{−∞}^{y} \frac{f_{(X,Y)}(x, u)}{f_{X}(x)}du

for every y ∈ R and every x ∈ R such that f_{X}(x) > 0.

Definition 19. The conditional expectation of a random variable Y given a random variable X, written E(Y |X = x), is defined to be

E(Y |X = x) = ∫_{−∞}^{∞} yf_{Y |X}(x, y)dy.

It is a function of x only.

Fact 2. For any random variables X and Y and any measurable function g : (R, B(R)) → ((0, ∞), B((0, ∞)) the equality E(g(X)Y |X = x) = g(x)E(Y |X = x) holds.

Fact 3. If random variables X and Y are independent then E(g(Y)|X = x) = Eg(Y) for any measurable function g : (R, B(R)) → (R, B(R)).

Fact 4. Let X be absolutely continuous n-dimensional random vector with the density function f_{X}(x_{1}, …, x_{n}) and let Y = g(X), where g is a measurable function g : (R^{n}, B(R^{n})) → (R^{n}, B(R^{n})) admitting the inverse function h = (h_{1}, …, h_{n}). Suppose the matrix A = ∥\frac{∂h_{i}}{∂u_{k}} ∥ to be non-degenerate and denote |J| = det A. Then the density function of Y has the form

f_{Y} (u_{1}, …, u_{n}) = |J|f_{X}(h_{1}(u_{1}, …, u_{n}), …, h_{n}(u_{1}, …, u_{n})).

Definition 20. We say that a sequence of functions {f_{n}}_{n≥1}, f_{n} : R → R, uniformly converges to a limit function f : R → R as n → ∞ iff

∀ε > 0 ∃N = N(ε) : ∀n ≥ N |f_{n}(x) − f(x)| < ε

simultaneously for every x ∈ R.

Fact 5. Uniform limit of continuous functions is a continuous function.

Fact 6. If a random vector X = (X_{1}, …, X_{n}) has a multivariate normal distribution and A is an m × n-matrix then the random vector AX also has a multivariate normal distribution, i.e. a linear transformation of a multivariate normally distributed random vector is normally distributed.

Definition 21. The covariance of any two random variables X and Y is given by

Cov (X, Y) = E(X − EX)(Y − EY)

if all mathematical expectations exist.

Definition 22. Any two random variables X and Y are called uncorrelated iff E(XY) exists and equal to zero.

Fact 7. (Cauchy-Schwartz inequality) For any two random variables X and Y

(E(XY))^{2} ≤ E(X^{2})E(Y^{2})

if all mathematical expectations exist.

Fact 8. If a random vector X = (X_{1}, …, X_{n}) has a multivariate normal distribution and its components are uncorrelated then they are independent.

Fact 9. ((strong) law of large numbers) If X_{k}, k ≥ 1, are independent identically distributed random variables and EX_{1} exists then \frac{S_{n}}{n} → EX_{1} in probability (and even with probability 1) as n → ∞, i.e. P (|\frac{S_{n}}{n} − EX_{1}| > ε) → 0 as n → ∞ for every ε > 0

(P ( lim_{n→∞} \frac{S_{n}}{n} = EX_{1}) = 1). Here S_{n} = X_{1} + ⋯ + X_{n}.

Fact 10. (central limit theorem) If X_{k}, k ≥ 1, are independent identically distributed random variables with zero means and variances equal to 1 then as n → ∞

P (\frac{S_{n}}{\sqrt{n}} ≤ x) → \frac{1}{\sqrt{2π}} ∫_{−∞}^{x} e^{−u^{2}/2}du

uniformly in x ∈ R. Here S_{n} = X_{1} + ⋯ + X_{n}.

Stochastic Processes

Let (Ω, F, P) be a probability space, (S, G) be a measurable space and T be a set.

Definition 23. A stochastic process on (Ω, F, P) with state space (S, G) and index set T is a family of measurable functions X(t) : (Ω, F) → (S, G), t ∈ T. It is denoted by {X(t), t ∈ T }.

If T is a set of positive integers and (S, G) = (R, B(R)) then stochastic process is simply a sequence of random variables {X(1), …, X(n), …}. Similarly, we can obtain an n-dimensional random vector (T = {1, …, n}, (S, G) = (R, B(R))) and even a single random variable (n = 1).

Usually T is interpreted as time. The most popular cases:

(a) T is a set of all integers or a set of all nonnegative integers; then it is said about stochastic process with discrete time;

(b) T = [0, a] or T = [0, ∞) or T = R; then it is said about stochastic process with continuous time.

In fact, stochastic process should be considered as a function of \underline{two variables} (t, ω), t ∈ T, ω ∈ Ω.

Further on, we mainly consider the case (S, G) = (R, B(R)).

For fixed t we have a random variable X(t, ·) defined on Ω. The family of random variables {X(t, ω), t ∈ T } is called a stochastic process.

For fixed ω we have a function X(·, ω) defined on T. It is called a realization, trajectory, sample function or sample path of the stochastic process. Varying ω ∈ Ω we get {X(t, ω), ω ∈ Ω} – a set of all sample paths of the stochastic process.

Examples.

1. Let ξ and η be two independent random variables with distribution functions F and G, respectively. Define a stochastic process {X(t), t ∈ R} as X(t, ω) = 0 for t < ξ(ω) and X(t, ω) = η(ω) for t ≥ ξ(ω). The process represents a step function with random starting time and random amplitude.

2. Let ϕ be a random variable uniformly distributed in [0, 2π]. Define a stochastic process {X(t), t ∈ R} as X(t, ω) = sin(at + ϕ(ω)), where a ∈ R is fixed. The process represents a sine-wave with a given frequence and random phase angle.

3. Define a stochastic process {X(t), t ∈ R} as X(t, ω) = ∑_{k=1}^{n} α_{k}(ω) cos(a_{k}t + ϕ_{k}(ω)), where a_{k} ∈ R are fixed but α_{k} and ϕ_{k} are random. The process represents a sum of cosine-waves with given frequences but random amplitudes and random phase angles.

The Kolmogorov Construction

As it is known, a random vector X = (X_{1}, …, X_{n}) is completely characterized by its distribution function F_{X}(x_{1}, …, x_{n}) = P (X_{1} ≤ x_{1}, …, X_{n} ≤ x_{n}). What about stochastic processes ?

Let {X(t), t ∈ T } be a stochastic process and define for given t_{1}, …, t_{n} ∈ T, t_{1} < … < t_{n}, an n-dimensional random vector (X(t_{1}), …, X(t_{n})). The distribution of such a vector is called the finite-dimensional distribution of the process. It is obvious that having stochastic process we can compute all its finite-dimensional distributions.

Examples.

1. Let us calculate P (X(2) ≤ 1, X(5) ≤ 4) under the conditions of Example 1 (see page 6). We have

P (X(2) ≤ 1, X(5) ≤ 4) = 1 − P (X(5) > 4) − P (X(2) > 1, X(5) ≤ 4) =

1 − P (ξ ≤ 5, η > 4) − P (ξ ≤ 2, 1 < η ≤ 4) =
1 − F (5)(1 − G(4)) − F (2)(G(4) − G(1)).

2. Let us calculate P (X(1) ≤ 0, X(2) ≤ 1/2) under the conditions of Example 2 (see page 6) with a = π/2. We have

P (X(1) ≤ 0, X(2) ≤ 1/2) = P (sin(π/2 + ϕ) ≤ 0, sin(π + ϕ) ≤ 1/2) =

P (π/2 ≤ ϕ ≤ 3π/2, 0 ≤ ϕ ≤ 4π/3) = \frac{4π/3 − π/2}{2π} = \frac{5}{12}.

It turns out that the converse statement is also true ! Namely, if we specify the distribution of (X(t_{1}), …, X(t_{n})) for all finite subsets {t_{1}, …, t_{n}} of T then it is possible to construct a stochastic process with prescribed finite-dimensional distributions in a unique way. This is a consequence of the so-called Kolmogorov extension theorem (which is not going to be proved here). In other words, specification of all finitedimensional distributions determines the distribution of entire process uniquely.

Stochastically Equivalent Processes

However, it is interesting to note that specification of all finite-dimensional distributions does not determine the character of the sample paths of the process !

Indeed, let {X_{0}(t), t ∈ [0, ∞)} be a stochastic process and define another process {X_{1}(t), t ∈ [0, ∞)} as X_{1}(t, ω) = X_{0}(t, ω) + f(t + ξ(ω)), where f(t) = t for t ∈ Q – the set of all rational numbers, and zero otherwise, while ξ is an absolutely continuous random variable (e.g. uniformly distributed in [0, 1]).

For any t we have

P ({ω : X_{0}(t, ω) ≠ X_{1}(t, ω)}) =
P ({ω : f(t + ξ(ω)) ≠ 0}) = P ({ω : ξ(ω) ∈ Q − t}) = 0

since the set Q − t is countable. Hence,

P ({ω : X_{0}(t, ω) = X_{1}(t, ω)}) = 1 ∀t ∈ T.

In such a case we say that the processes X_{0} and X_{1} are stochastically equivalent.

Obviously, both processes have the same finite-dimensional distributions. Indeed, let us denote

A = {ω : X_{1}(t_{1}) ≤ x_{1}, …, X_{1}(t_{n}) ≤ x_{n}},
B = {ω : X_{0}(t_{1}) ≤ x_{1}, …, X_{0}(t_{n}) ≤ x_{n}}.

Since A ⊂ B, we have

P (B) − P (A) = P (B \ A) ≤ ∑_{k=1}^{n} P ({ω : ξ(ω) ∈ Q − t_{k}}) = 0.

Hence, due to the Kolmogorov theorem the distributions of both processes are the same. But if, for example, we take X_{0} having all sample paths to be continuous, we can state that all sample paths of X_{1} are discontinuous !

Processes with Independent Increments

Definition 24. We say that a process {X(t), t ∈ T } has independent increments if for any n and any t_{0} < t_{1} < … < t_{n}, t_{i} ∈ T, the random variables X(t_{0}), X(t_{1}) − X(t_{0}), …, X(t_{n}) − X(t_{n−1}) are independent, i.e. the increments of the process for disjoint time intervals are mutually independent.

If T = [0, a] or T = [0, ∞) then t_{0} = 0 is usually taken. The random variable X(0) is called the initial value of the process, while its distribution is called the initial distribution of the process.

Example.

Let X_{1}, …, X_{n}, … be independent random variables. Define the process in such a way : T is a set of all positive integers, X(n) = X_{1} + ⋯ + X_{n}. Then {X(t), t ∈ T } is a process with independent increments.

Let {X(t), t ∈ [0, ∞)} be a stochastic process with independent increments and such that P (X(0) = 0) = 1. Define a family of the distribution functions {F_{t,s}, t > s, t, s ∈ T } given by

F_{t,s}(x) = P (X(t) − X(s) ≤ x) ∀x ∈ R.

This is a family of the distribution functions of all possible increments of the process.

All finite-dimensional distributions of such a stochastic process are completely determined by this two-parametric family. Indeed (for simplicity suppose that all distribution functions F_{t,s} possess densities f_{t,s}),

P (X(t_{1}) ≤ x_{1}, …, X(t_{n}) ≤ x_{n}) = P (X(t_{1}) ≤ x_{1}, X(t_{1}) + (X(t_{2}) − X(t_{1})) ≤ x_{2},

…, X(t_{1}) + (X(t_{2}) − X(t_{1})) + ⋯ + (X(t_{n}) − X(t_{n−1})) ≤ x_{n}) =
P (X(t_{1}) − X(0), X(t_{2}) − X(t_{1}), …, X(t_{n}) − X(t_{n−1}) ∈ A) =

∫ … ∫_{A} f_{t_{1},0}(y_{1})f_{t_{2},t_{1}}(y_{2}) … f_{t_{n},t_{n−1}}(y_{n})dy_{1}dy_{2} … dy_{n},

where A = {(y_{1}, …, y_{n}) ∈ R^{n} : y_{1} ≤ x_{1}, y_{1} + y_{2} ≤ x_{2}, …, y_{1} + ⋯ + y_{n} ≤ x_{n}}.

”Lack of Memory”

Let T = [0, ∞) and P (X(0) = 0) = 1. Lemma 1. If {X(t), t ∈ [0, ∞)} is a process with independent increments then

P (X(t_{n}) ≤ x_{n}|X(t_{1}) = x_{1}, …, X(t_{n−1}) = x_{n−1}) = P (X(t_{n}) ≤ x_{n}|X(t_{n−1}) = x_{n−1}) for any n any 0 ≤ t_{1} < … < t_{n} and any x_{n} ∈ R. Proof. The value X(t_{n}) can be represented as a sum of independent random variables since (we set t_{0} = 0)

X(t_{n}) = ∑_{k=1}^{n}(X(t_{k}) − X(t_{k−1})).

Denote for simplicity X(t_{n}) = X_{n}, X(t_{k}) − X(t_{k−1}) = Y_{k}. Then X_{n} = ∑_{k=1}^{n} Y_{k}. Our goal is to show that

P (X_{n} ≤ x_{n}|X_{1} = x_{1}, …, X_{n−1} = x_{n−1}) = P (X_{n} ≤ x_{n}|X_{n−1} = x_{n−1}). First of all, it is worth to note that P (X_{n} ≤ x_{n}|X_{1} = x_{1}, …, X_{n−1} = x_{n−1}) = P (X_{n} ≤ x_{n}|Y_{1} = x_{1}, …, Y_{n−1} = x_{n−1}−x_{n−2}). Then

P (X_{n−1} ≤ x_{n−1}, Y_{n} ≤ y|Y_{1} = x_{1}, …, Y_{n−1} = x_{n−1} − x_{n−2}) =
E(1_{(−∞,x_{n−1}]}(X_{n−1})1_{(−∞,y]}(Y_{n})|Y_{1} = x_{1}, …, Y_{n−1} = x_{n−1} − x_{n−2}) =

1_{(−∞,x_{n−1}]}(x_{n−1})E(1_{(−∞,y]}(Y_{n})|Y_{1} = x_{1}, …, Y_{n−1} = x_{n−1} − x_{n−2}) = E(1_{(−∞,y]}(Y_{n})). Similarly,

P (X_{n−1} ≤ x_{n−1}, Y_{n} ≤ y|X_{n−1} = x_{n−1}) = E(1_{(−∞,y]}(Y_{n})).

Hence, P (X_{n−1} ≤ x_{n−1}, Y_{n} ≤ y|Y_{1} = x_{1}, …, Y_{n−1} = x_{n−1}−x_{n−2}) = P (X_{n−1} ≤ x_{n−1}, Y_{n} ≤ y|X_{n−1} = x_{n−1}) for all x_{n−1}, y ∈ R. As a consequence,

P ((X_{n−1}, Y_{n}) ∈ A|Y_{1}=x_{1}, …, Y_{n−1}=x_{n−1} − x_{n−2})=P ((X_{n−1}, Y_{n}) ∈ A|X_{n−1}=x_{n−1}) (1)

for any A ∈ B([0, ∞)) × B([0, ∞)). Define

B = {(x, y) : x + y ≤ x_{n}} ∈ B([0, ∞]) × B([0, ∞)).

For B the equality (1) holds and it remains to note that (X_{n−1}, Y_{n}) ∈ B ⟺ X_{n−1} + Y_{n} ≤ x_{n} ⟺ X_{n} ≤ x_{n}.

Processes with Independent Stationary Increments

Definition 25. We call a stochastic process {X(t), t ∈ T } with independent increments a stochastic process with independent stationary (homogeneous) increments if the distributions of X(t)−X(s) and X(t+h)−X(s+h) are the same for any t, s ∈ T, t > s, and any h such that t + h, s + h ∈ T.

In other words, if T is the set of all nonnegative integers or T = [0, ∞) then the distribution of X(t + h) − X(h) does not depend on h !

Example.

Let X_{1}, …, X_{n}, … be independent identically distributed random variables. Define the process in such a way : T is a set of all nonnegative integers, X(0) = 0 and X(n) = X_{1} + ⋯ + X_{n}. Then {X(t), t ∈ T } is a process with independent stationary increments.

Let {X(t), t ∈ [0, ∞)} be a stochastic process with independent stationary increments and P (X(0) = 0) = 1. Then all its finite-dimensional distributions are determined by the one-parametric family of the distribution functions of all possible increments {F_{t}, t ≥ 0} given by

F_{t}(x) = P (X(t + u) − X(u) ≤ x) ∀x ∈ R.

Now let us divide the interval [0, t] onto n intervals of the same length. So, 0 = t_{0} < t_{1} < … < t_{n} = t, where t_{k} = \frac{tk}{n}. We can write

X(t) = ∑_{k=1}^{n} (X (\frac{tk}{n} ) − X (\frac{t(k − 1)}{n})) .

Clearly, each summand in the sum has the same distribution since

X (\frac{tk}{n} ) − X (\frac{t(k − 1)}{n}) = X (\frac{t(k − 1) + t}{n}) − X (\frac{t(k − 1)}{n}) =^{d}X (\frac{t}{n}) .

Therefore, for any given t the random variable X(t) for any n can be represented as a sum of n independent identically distributed random variables. The distribution having such a property is called the infinite-divisible distribution.

Stochastically Continuous Processes

Quite often stochastic processes with independent stationary increments are also stochastically continuous. We call a stochastic process stochastically continuous if for any t ∈ T and any h → 0 such that t + h ∈ T we have X(t + h) − X(t) → 0 in probability, that is

P (|X(t + h) − X(t)| > ε) → 0 ∀ε > 0.

Obviously, if the stochastic process has all sample paths to be continuous then it is stochastically continuous. The converse statement does not hold.

Example.

Define a stochastic process {X(t), t ∈ [0, 1]} as

X(t, ω) = \{ \begin{matrix}0, & t < ξ(ω) \\ 1, & t ≥ ξ(ω), \end{matrix}

where ξ is a random variable having uniform distribution in [0, 1]. All sample paths of this stochastic process are discontinuous. On the same time, the process is stochastically continuous. Indeed,

P ({ω : |X(t + h, ω) − X(t, ω)| > ε}) = P ({ω : X(t + h, ω) = 1, X(t, ω) = 0}) =

P ({ω : t+h ≥ ξ(ω), t < ξ(ω)}) = P ({ω : t < ξ(ω) ≤ t+h}) = F (t+h)−F (t) = h → 0,

where F (t) = t is a distribution function of ξ.

The Wiener Process, Brownian Motion

Definition 26. A stochastic process {W (t), t ≥ 0} with independent stationary increments is called the Wiener process if P (W (0) = 0) = 1 and the distribution of any increment W (t + u) − W (u), u ≥ 0, t ≥ 0, is normal N(θt, σ^{2}t) distribution with some θ ∈ R and σ^{2} > 0. Also it is often assumed that the process {W (t), t ≥ 0} has all sample paths to be continuous.

Thus, W (t) − W (s) ∼ N(θ(t − s), σ^{2}(t − s)) and W (t) ∼ N(θt, σ^{2}t).

It was N.Wiener who first described such a process in 1923 and credited his discovery of its construction to previous conversations with P.Levy.

Definition 27. The Wiener process {W (t), t ≥ 0} with θ = 0 and σ^{2} = 1 is called the Brownian motion.

It was the English botanist R.Brown who first noticed in 1827 that microscopic particles settling in a liquid made chaotic movements. And only in 1905 the famous physicist A.Einshtein was managed to explain why they did such movements: because of undergoing molecular bombardment. So, if we imagine that we observe the movement of the particle then we can see that it looks as a broken line with chaotic directions. Such a movement was named after R.Brown. It turned out that the Wiener process can approximately describe that movement quite satisfactory.

Indeed, let us assume that the particle starts at the origin and denote by X(t) its distance from the origin at the moment t. Of course, for fixed t we can suppose that X(t) is a random variable. Each moment the particle undergoes a huge amount of bombardment. Hence, the increment of the distance X(t) − X(s) during the time interval (s, t) can be considered as a sum of very large number of very small increments. In such a situation, the central limit theorem can be applied and approximately we can assume that the distribution of the increment X(t) − X(s) is normal. Moreover, it is reasonable to assume that variance of such an increment increasing proportionally with increasing of the time interval. If the liquid is homogeneous and in a balance, we can suppose that distributions of X(t) − X(s) and X(t + h) − X(s + h) for all h > 0 are the same and all directions are equivalent, that is the mean of the increment X(t)−X(s) is zero. And at last intuitively it is clear that increments on disjoint time intervals could be approximately assumed to be independent.

Some Properties of the Brownian Motion

1. EW (t) = 0, EW (t)W (s) = min{t, s}.

Indeed, let first 0 ≤ s ≤ t. Then

EW (t)W (s) = E(W (t)−W (s))W (s)+EW^{2}(s) = E(W (t)−W (s))EW (s)+s = s.

In the same way it can be shown that for 0 ≤ t ≤ s we have EW (t)W (s) = t.

2. Density function of the random vector (W (t_{1}), …, W (t_{n})), 0 ≤ t_{1} < … < t_{n}. First, consider the random vector (W (t_{1}), W (t_{2}) − W (t_{1}), …, W (t_{n}) − W (t_{n−1})). Clearly, its density function has the form

f(x_{1}, x_{2}, …, x_{n}) = \frac{1}{(2π)^{n/2}\sqrt{t_{1}(t_{2} − t_{1}) … (t_{n} − t_{n−1})}} exp \{−\frac{1}{2} ∑_{k=1}^{n} \frac{x_{k}^{2}}{t_{k} − t_{k−1}} \} .

After the transformation

\{ \begin{matrix}u_{1} & = & x_{1} \\ u_{2} & = & x_{1} + x_{2} \\ … & … & . . . . . . . . . \\ u_{n} & = & x_{1} + … + x_{n} \end{matrix}

we obtain the desired random vector and its density looks as (we set t_{0} = u_{0} = 0)

g(u_{1}, u_{2}, …, u_{n}) = \frac{1}{(2π)^{n/2}\sqrt{t_{1}(t_{2} − t_{1}) … (t_{n} − t_{n−1})}} exp \{−\frac{1}{2} ∑_{k=1}^{n} \frac{(u_{k} − u_{k−1})^{2}}{t_{k} − t_{k−1}} \} .

3. Conditional distribution of W (t) given W (t_{1}) = x_{1}, W (t_{2}) = x_{2}, where 0 ≤ t_{1} < t < t_{2}. First of all, let us write the density function of the random vector (W (t_{1}), W (t), W (t_{2})). It has the form

g(u_{1}, u, u_{2}) = \frac{1}{(2π)^{3/2}\sqrt{t_{1}(t − t_{1})(t_{2} − t)}} exp \{−\frac{1}{2} (\frac{u_{1}^{2}}{t_{1}} + \frac{(u − u_{1})^{2}}{t − t_{1}} + \frac{(u_{2} − u)^{2}}{t_{2} − t} )\}

= \frac{1}{2π\sqrt{t_{1}(t_{2} − t_{1})}} exp \{−\frac{1}{2} (\frac{u_{1}^{2}}{t_{1}} + \frac{(u_{2} − u_{1})^{2}}{t_{2} − t_{1}} )\} \frac{1}{(2π)^{1/2}σ} exp \{−\frac{(u − μ)^{2}}{2σ^{2}} \} ,

where

μ = μ(t) = x_{1} \frac{t_{2} − t}{t_{2} − t_{1}} + x_{2} \frac{t − t_{1}}{t_{2} − t_{1}} , σ^{2} = σ^{2}(t) = \frac{(t_{2} − t)(t − t_{1})}{t_{2} − t_{1}}.

The conditional density of W (t) given W (t_{1}) = x_{1}, W (t_{2}) = x_{2}, has the form

f(u) = \frac{g(x_{1}, u, x_{2})}{∫ g(x_{1}, u, x_{2})du} = \frac{1}{(2π)^{1/2}σ} exp \{−\frac{(u − μ)^{2}}{2σ^{2}} \} .

Therefore, the distribution of W (t)|_{W (t_{1})=x_{1},W (t_{2})=x_{2}} is normal N(μ, σ^{2}) distribution.

It is interesting to note that if we write the equation of a line passing through the points (t_{1}, x_{1}) and (t_{2}, x_{2}) we get

x = x(t) = x_{1} \frac{t_{2} − t}{t_{2} − t_{1}} + x_{2} \frac{t − t_{1}}{t_{2} − t_{1}} .

It means that if we know that the Brownian motion is at point x_{1} at moment t_{1} and at point x_{2} at moment t_{2} then at moment t_{1} < t < t_{2} it is, in average, on the line passing through (t_{1}, x_{1}) and (t_{2}, x_{2}) since

W (t)|_{W (t_{1})=x_{1},W (t_{2})=x_{2}} =^{d} μ(t) + σ(t)ξ_{t},

where ξ_{t} ∼ N(0, 1).

As a corollary, taking t = \frac{t_{1}+t_{2}}{2} we get

W (\frac{t_{1} + t_{2}}{2} ) |_{W (t_{1})=x_{1},W (t_{2})=x_{2}} =^{d} \frac{x_{1} + x_{2}}{2} + \sqrt{\frac{t_{2} − t_{1}}{4}} ξ_{\frac{t_{1}+t_{2}}{2}} .

Construction of the Brownian Motion

The fundamental step in the investigation of the Brownian motion was to prove the existence of a process with such properties. There exist several different constructions of the Brownian motion. Each of them gives access to different aspects of the process itself. We consider the construction of the Brownian motion by P.Levy and start with the interval [0, 1].

Let {ξ_{k}^{(n)}, k ∈ I(n), n−nonnegative integer} be a family of independent identically distributed random variables with normal N(0, 1) distribution, where I(n) = {0 ≤ k ≤ 2^{n}, k − odd numbers}. For instance, I(0) = {1}, I(1) = {1}, I(2) = {1, 3}, I(3) = {1, 3, 5, 7}, etc.

Define a sequence of processes {W_{n}(t), t ∈ [0, 1]}, n ≥ 0, in the following way:

W_{0}(t) : \{ \begin{matrix}W_{0}(0) & = & 0 \\ W_{0}(1) & = & ξ_{1}^{(0)} \\ W_{0}(t) & = & tξ_{1}^{(0)}, 0 < t < 1 \end{matrix} ,

W_{1}(t) : \{ \begin{matrix}W_{1}(0) & = & W_{0}(0) \\ W_{1}(1) & = & W_{0}(1) \\ W_{1} (\frac{1}{2}) & = & W_{0} (\frac{1}{2}) + \frac{1}{2}ξ_{1}^{(1)} \\ W_{1}(t) & = & 2tW_{1} (\frac{1}{2}) , 0 < t < \frac{1}{2} \\ W_{1}(t) & = & 2(1 − t)W_{1} (\frac{1}{2}) + (2t − 1)W_{1}(1), \frac{1}{2} < t < 1 \end{matrix} ,

and so on.

So, the process {W_{n}(t), t ∈ [0, 1]} is defined basing on {W_{n−1}(t), t ∈ [0, 1]} in the following way:

(a) at the points of the form \frac{k}{2^{n}} , k – even (”old points”)

W_{n} (\frac{k}{2^{n}} ) = W_{n−1} (\frac{k}{2^{n}} );

(b) at the points of the form \frac{k}{2^{n}} , k – odd (”new points”)

W_{n} (\frac{k}{2^{n}} ) = W_{n−1} (\frac{k}{2^{n}} ) + \frac{1}{2^{(n+1)/2}} ξ_{k}^{(n)}

(why \frac{1}{2^{(n+1)/2}} ? because (see page 15) σ^{2} = \frac{t−s}{4} = \frac{1}{4·2^{n−1}} = \frac{1}{2^{n+1}} ⟹ σ = \frac{1}{2^{(n+1)/2}} ); (c) and apply simple linear interpolation between those points, i.e.

W_{n}(t) = (m + 1 − 2^{n}t)W_{n} (\frac{m}{2^{n}} ) + (2^{n}t − m)W_{n} (\frac{m + 1}{2^{n}})

for \frac{m}{2^{n}} < t < \frac{m+1}{2^{n}}.

It is possible to show that:

(i) for fixed ω the sequence {W_{n}(t, ω)} as n → ∞ uniformly converges to a continuous limit function W (t, ω); (ii) this limit function satisfies all conditions to be the Brownian motion.

(i) Define the so-called Schauder functions {S_{k}^{(n)}(t), t ∈ [0, 1], k ∈ I(n), n ≥ 0} as S_{k}^{(n)}(t) = ∫_{0}^{t} H_{k}^{(n)}(u)du, where H_{1}^{(0)}(t) = 1 for all t ∈ [0, 1] and

H_{k}^{(n)}(t) = \{ \begin{matrix}2^{(n−1)/2}, & t ∈ [\frac{k−1}{2^{n}}, \frac{k}{2^{n}}) \\ −2^{(n−1)/2}, & t ∈ [\frac{k}{2^{n}} , \frac{k+1}{2^{n}}) \\ 0, & otherwise \end{matrix} ; t ∈ [0, 1], k ∈ I(n), n ≥ 0.

It is easy to note that S_{k}^{(n)}(t) ≠ 0 only for t ∈ (\frac{k−1}{2^{n}}, \frac{k+1}{2^{n}}). For the same n and different k the supports of the Schauder functions are disjoint. Moreover, max_{t} S_{k}^{(n)}(t) = 2^{−(n+1)/2}, arg max_{t} S_{k}^{(n)}(t) = \frac{k}{2^{n}} .

It is clear that W_{0}(t, ω) = ξ_{1}^{(0)}(ω)S_{1}^{(0)}(t). By induction, it can be checked that the following important represesntation takes place

W_{n}(t, ω) = ∑_{m=0}^{n} ∑_{k∈I(m)} ξ_{k}^{(m)}(ω)S_{k}^{(m)}(t), t ∈ [0, 1], n ≥ 0.

As n → ∞,

W_{n}(t, ω) → W (t, ω) = ∑_{m=0}^{∞} ∑_{k∈I(m)} ξ_{k}^{(m)}(ω)S_{k}^{(m)}(t) ∀t ∈ [0, 1] ∀ω ∈ Ω

and it is possible to show that the convergence of the last series is uniform for fixed ω. Then the limit function W (t, ω) is continuous as the uniform limit of continuous functions.

(ii) Indeed, the process {W (t), t ∈ [0, 1]} satisfies all conditions to be the Brownian motion.

First, for fixed t the random variable W (t) has a normal distribution since it could be represented as a linear transformation of a multivariate normally distributed random vector.

Second, from the same representation it follows that EW (t) = 0 and

EW (t)W (s) = E ( ∑_{m=0}^{∞} ∑_{k∈I(m)} ξ_{k}^{(m)}S_{k}^{(m)}(t) ∑_{l=0}^{∞} ∑_{i∈I(l)} ξ_{i}^{(l)}S_{i}^{(l)}(s)) =

∑_{m=0}^{∞} ∑_{k∈I(m)} ∑_{l=0}^{∞} ∑_{i∈I(l)} E (ξ_{k}^{(m)}ξ_{i}^{(l)}) S_{k}^{(m)}(t)S_{i}^{(l)}(s) = ∑_{m=0}^{∞} ∑_{k∈I(m)} S_{k}^{(m)}(t)S_{k}^{(m)}(s) = min{t, s}.

As a particular case, EW^{2}(0) = 0, and therefore P (W (0) = 0) = 1.

Third, for 0 ≤ t_{1} < t_{2} ≤ t_{3} < t_{4} we have

E(W (t_{2}) − W (t_{1}))(W (t_{4}) − W (t_{3})) = t_{2} − t_{1} − t_{2} + t_{1} = 0.

Hence, the random vector (W (t_{1}), W (t_{2}) − W (t_{1}), …, W (t_{n}) − W (t_{n−1})) for any 0 < t_{1} < … < t_{n} and any n has uncorrelated components. Since it has a multivariate normal distribution, all its components are independent. So, the process {W (t), t ∈ [0, 1]} has independent increments.

Fourth, W (t + h) − W (h) is normally distributed with mean zero and variance

E(W (t + h) − W (h))^{2} = t + h − h − h + h = t.

Thus, the distribution of W (t + h) − W (h) does not depend on h, that is, the process has stationary increments.

Therefore, we have completed the construction of the Brownian motion on the interval [0, 1]. It is very simple now to extend it on the whole half-line. Indeed, let us construct the sequence of such processes {W^{(n)}(t), t ∈ [0, 1]}, n ≥ 1. Define now the process {W (t), t ≥ 0} as follows: W (t) = W^{(1)}(t) for t ∈ [0, 1]; W (t) = W (1) + W^{(2)}(t − 1) for t ∈ [1, 2]; W (t) = W (2) + W^{(3)}(t − 2) for t ∈ [2, 3] and so on. Clearly, such a process has all sample paths to be continuous and is a Brownian motion on [0, ∞).

Simulation of the Brownian Motion

First of all, let us notice how to simulate, in general, the stochastic process {X(t), t ≥ 0} with independent stationary increments and a property P (X(0) = 0) = 1 at given points 0 < t_{1} < … < t_{n}. The next algorithm is quite clear.

1. Set X(0) = 0.

2. Let F_{t} be a distribution function of the increment X(t + u) − X(u) (see page 11). Simulate a random variable Y_{1} with the distribution F_{t_{1}} and set X(t_{1}) = Y_{1}.

3. Simulate a random variable Y_{2} with the distribution F_{t_{2}−t_{1}} and set X(t_{2}) = Y_{1}+Y_{2} and so on.

Of course, it is not possible to simulate a trajectory of the process by that method.

Consider two methods for simulation of a trajectory of the Brownian motion. Without loss of generality we consider only the case of the interval [0, 1].

Method based on Levy’s construction.

We approximate a trajectory of {W (t), t ∈ [0, 1]} by that of {W_{n}(t), t ∈ [0, 1]} which looks as a broken line (see pages 16–17). Choosing sufficiently large n we can achieve quite good precision of approximation. We construct a trajectory of {W_{n}(t), t ∈ [0, 1]} step by step starting from that of {W_{0}(t), t ∈ [0, 1]}. Clearly, on each step we only need to simulate the suitable number of N(0, 1) random variables. For instance. having W_{n−1}(t) we need to simulate 2^{n−1} of them for producing W_{n}(t).

Method based on infinite divisibility.

We have (see page 11)

W (t) = ∑_{k=1}^{n} (W (\frac{tk}{n} ) − W (\frac{t(k − 1)}{n})) .

Therefore, for every n

W (t) = ξ_{1,n} + ⋯ + ξ_{n,n}, ξ_{k,n}=^{d}W (\frac{t}{n}) ∼ N (0, \frac{t}{n}) .

It means that we can represent W (t) as

W (t) = \sqrt{\frac{t}{n}}X_{1} + ⋯ + \sqrt{\frac{t}{n}}X_{n},

where X_{1}, …, X_{n} are independent identically distributed with normal N(0, 1) distribution. Hence,

W (\frac{n}{m}) = \sqrt{\frac{1}{m}}X_{1} + ⋯ + \sqrt{\frac{1}{m}}X_{n} = \frac{S_{n}}{\sqrt{m}},

where S_{n} = X_{1} + ⋯ + X_{n}.

\underline{The idea is:} increasing m we can approximate a trajectory of the Brownian motion by the broken line passing through points (\frac{n}{m}, \frac{S_{n}}{\sqrt{m}}) , 0 ≤ n ≤ m. Choosing sufficiently large m we can hope to achieve quite good precision of approximation. The next algorithm represents a realization of this idea.

1. Simulate X_{1}^{(1)} ∼ N(0, 1) and take as W_{1}(t) a line passing through the points

(0, 0), (1, X_{1}^{(1)}), i.e.

W_{1}(t) = tX_{1}^{(1)}, t ∈ [0, 1].

2. Simulate X_{1}^{(2)}, X_{2}^{(2)} ∼ N(0, 1) and take as W_{2}(t) a broken line passing through

the points (0, 0), (\frac{1}{2}, \frac{X_{1}^{(2)}}{\sqrt{2}} ), (1, \frac{X_{1}^{(2)}+X_{2}^{(2)}}{\sqrt{2}} ), i.e.

W_{2}(t) = 2t\frac{X_{1}^{(2)}}{\sqrt{2}} , if t ∈ [0, \frac{1}{2}],

W_{2}(t) = \frac{X_{1}^{(2)}}{\sqrt{2}} + (2t − 1)\frac{X_{2}^{(2)}}{\sqrt{2}} , if t ∈ [\frac{1}{2}, 1],

and so on.

3. On nth step we simulate X_{1}^{(n)}, …, X_{n}^{(n)} ∼ N(0, 1) and take

W_{n}(t) = \frac{S_{k−1}^{(n)}}{\sqrt{n}} + \frac{t − \frac{k−1}{n}}{\frac{1}{n}} \frac{X_{k}^{(n)}}{\sqrt{n}} =

\frac{\frac{k}{n} − t}{\frac{1}{n}}W_{n} (\frac{k − 1}{n}) + \frac{t − \frac{k−1}{n}}{\frac{1}{n}} W_{n} (\frac{k}{n}) , if t ∈ [\frac{k − 1}{n}, \frac{k}{n}],

or equivalently, since for t ∈ [\frac{k−1}{n}, \frac{k}{n}] we have k − 1 = [nt],

W_{n}(t) = \frac{S_{[nt]}^{(n)}}{\sqrt{n}} + (nt − [nt])\frac{X_{[nt]+1}^{(n)}}{\sqrt{n}} , t ∈ [0, 1].

In practice, according to the invariance principle we can base a simulation of a trajectory of the Brownian motion not on the simulation of N(0, 1) random variables but on that of random variables from any distribution having mean zero and variance equal to 1 ! For instance, it is possible to take a discrete distribution P (X = ±1) = 1/2.

The Poisson Process

Definition 28. A stochastic process {Y (t), t ≥ 0} with independent stationary increments taking values on the nonnegative integers is called the Poisson process if P (Y (0) = 0) = 1 and the distribution of any increment Y (t + u) − Y (u), u ≥ 0, t ≥ 0, is Poissonian P(μt) with some μ > 0.

Thus, Y (t) − Y (s) ∼ P(μ(t − s)) and Y (t) ∼ P(μt), i.e.

P (Y (t) = n) = \frac{(μt)^{n}}{n!} e^{−μt}, n ≥ 0.

Clearly, all sample paths of the Poisson process are discontinuous.

The Poisson process can describe quite satisfactory a lot of processes in the real life. In general, as the value Y (t) we could take the number of events of the same kind occuring in the period of time [0, t]. It could be: the number of particles emitted by a substance under the radioactive decay, the number of telephone calls from the given district to the central station, the number of accidents at a given road intersection, the number of customers to be served anywhere, etc.

Some Properties of the Poisson Process.

1. EY (t) = μt, E(Y (t) − EY (t))(Y (s) − EY (s)) = μ min{t, s}. Indeed, let first 0 ≤ s ≤ t. Then

E(Y (t) − EY (t))(Y (s) − EY (s)) = EY (t)Y (s) − EY (t)EY (s) =

E(Y (t) − Y (s))Y (s) + EY^{2}(s) − μtμs = μ(t − s)μs + μ^{2}s^{2} + μs − μ^{2}ts = μs.

In the same way it can be shown that for 0 ≤ t ≤ s we have

E(Y (t) − EY (t))(Y (s) − EY (s)) = μt.

2. Distribution of the random vector (Y (t_{1}), …, Y (t_{n})), 0 ≤ t_{1} < … < t_{n}. First, consider the random vector (Y (t_{1}), Y (t_{2}) − Y (t_{1}), …, Y (t_{n}) − Y (t_{n−1})). Clearly, its distribution has the form

P (Y (t_{1}) = k_{1}, Y (t_{2}) − Y (t_{1}) = k_{2}, …, Y (t_{n}) − Y (t_{n−1}) = k_{n}) =

\frac{(μt_{1})^{k_{1}}}{k_{1}!} e^{−μt_{1}} \frac{(μ(t_{2} − t_{1}))^{k_{2}}}{k_{2}!} e^{−μ(t_{2}−t_{1})} … \frac{(μ(t_{n} − t_{n−1}))^{k_{n}}}{k_{n}!} e^{−μ(t_{n}−t_{n−1})} =

\frac{μ^{k_{1}+k_{2}+···+k_{n}}t_{1}^{k_{1}}(t_{2} − t_{1})^{k_{2}} … (t_{n} − t_{n−1})^{k_{n}}}{k_{1}!k_{2}! … k_{n}!} e^{−μt_{n}}.

On the other hand,

P (Y (t_{1}) = k_{1}, Y (t_{2}) − Y (t_{1}) = k_{2}, …, Y (t_{n}) − Y (t_{n−1}) = k_{n}) =
P (Y (t_{1}) = k_{1}, Y (t_{2}) = k_{1} + k_{2}, …, Y (t_{n}) = k_{1} + k_{2} + ⋯ + k_{n}).

Therefore,

P (Y (t_{1}) = j_{1}, Y (t_{2}) = j_{2}, …, Y (t_{n}) = j_{n}) = \frac{μ^{j_{n}}t_{1}^{j_{1}}(t_{2} − t_{1})^{j_{2}−j_{1}} … (t_{n} − t_{n−1})^{j_{n}−j_{n−1}}}{j_{1}!(j_{2} − j_{1})! … (j_{n} − j_{n−1})!} e^{−μt_{n}}.

3. Consider the interval [0, t] and its partition by the points 0 < t_{1} < … < t_{n} < t onto n+1 intervals [0, t_{1}), [t_{1}, t_{2}), …, [t_{n−1}, t_{n}), [t_{n}, t]. Denote Δ_{k} = t_{k} −t_{k−1}, k = 1, 2, …, n + 1, t_{0} = 0, t_{n+1} = t. What happens when n → ∞ and max_{k} Δ_{k} → 0 ? Using Taylor’s expansion we get the following relations for the distributions of increments:

P (Y (t_{k}) − Y (t_{k−1}) = 0) = P (Y (Δ_{k}) = 0) = e^{−μΔ_{k}} = 1 − μΔ_{k} + O(Δ_{k}^{2}),
P (Y (t_{k}) − Y (t_{k−1}) = 1) = P (Y (Δ_{k}) = 1) = μΔ_{k}e^{−μΔ_{k}} = μΔ_{k} + O(Δ_{k}^{2}),
P (Y (t_{k}) − Y (t_{k−1}) ≥ 2) = P (Y (Δ_{k}) ≥ 2) = O(Δ_{k}^{2}).

It follows from those relations that:

(a) probability of having a jump at given point t tends to zero;

(b) probability of the event that the process has a jump more than on 1 is negligible with respect to that of a jump equal to 1;

(c) it can be shown that the length of all intervals without jumps tends to the length of the whole interval [0, t].

Moreover, we have P (Y (t + h) − Y (t) = 0) → 1 as h → 0 for any t ≥ 0. It means that the Poisson process is stochastically continuous. Let us remind, that all the sample paths are discontinuous. Usually the sample paths of the Poisson process are considered to be continuous from the right.

So, the Poisson process is a jump process with jumps equal to 1. The value Y (t) is equal to the number of jumps on the interval [0, t]. The points of jumps are random variables a_{k}, k ≥ 1.

Characterization of the Poisson Process

Lemma 2. Let {Y (t), t ≥ 0} be the Poisson process and a_{k}, k ≥ 1, be the points of its jumps. Then the random variables X_{1} = a_{1}, X_{2} = a_{2} − a_{1}, …, X_{n} = a_{n} − a_{n−1}, … are independent identically distributed with exponential E(μ) distribution, i.e.

P (X_{i} ≤ x) = 1 − e^{−μx}, x > 0,

and zero otherwise.

Proof. First, let us note that

{ω : X_{1}(ω) > t} = {ω : Y (t, ω) = 0}.

Hence,

P (X_{1} > t) = P (Y (t) = 0) = e^{−μt}.

That means that X_{1} has exponential E(μ) distribution.

Second, let us calculate P (s_{1} < a_{1} ≤ t_{1}, s_{2} < a_{2} ≤ t_{2}) for 0 ≤ s_{1} < t_{1} < s_{2} < t_{2}. We get

P (s_{1} < a_{1} ≤ t_{1}, s_{2} < a_{2} ≤ t_{2}) =

P (Y (s_{1}) = 0, Y (t_{1}) − Y (s_{1}) = 1, Y (s_{2}) − Y (t_{1}) = 0, Y (t_{2}) − Y (s_{2}) ≥ 1) =

e^{−μs_{1}}μ(t_{1} − s_{1})e^{−μ(t_{1}−s_{1})}e^{−μ(s_{2}−t_{1})} (1 − e^{−μ(t_{2}−s_{2})}) = μ(t_{1} − s_{1}) (e^{−μs_{2}} − e^{−μt_{2}}) =

∫_{s_{2}}^{t_{2}} ∫_{s_{1}}^{t_{1}} μ^{2}e^{−μy_{2}}dy_{1}dy_{2}.

Therefore, the random vector (X_{1}, X_{1} + X_{2}) has the density function

f(y_{1}, y_{2}) = μ^{2}e^{−μy_{2}}, 0 < y_{1} < y_{2},

and zero otherwise. It means that X_{1} and X_{2} are independent and identically distributed with exponential E(μ) distribution (see Exercise 14).

Third, in the same way we can find that the random vector (X_{1}, X_{1} + X_{2}, X_{1} + X_{2} + ⋯ + X_{n}) has the density function

f(y_{1}, y_{2}, …, y_{n}) = μ^{n}e^{−μy_{n}}, 0 < y_{1} < y_{2} < … < y_{n},

and zero otherwise. Therefore, the random variables X_{1}, X_{2}, …, X_{n} are independent and identically distributed with exponential E(μ) distribution.

Construction of the Poisson Process

Construction of the Poisson process is based on Lemma 2.

Let X_{k}, k ≥ 1, be independent identically distributed random variables with exponential E(μ) distribution. Denote S_{n} = X_{1} + ⋯ + X_{n} (S_{0} = 0). As it is known, the random variable S_{n} has gamma Γ(n, 1/μ) distribution with shape parameter n and scale parameter 1/μ, i.e. its density function has the form

f_{S_{n}}(x) = \frac{μ^{n}}{(n − 1)!}x^{n−1}e^{−μx}, x > 0

and zero otherwise.

Define a stochastic process {Y (t), t ≥ 0} as

Y (t, ω) = \{ \begin{matrix}max\{n ≥ 0, S_{n}(ω) ≤ t\} \\ ∞ if S_{n}(ω) ≤ t ∀n. \end{matrix}

It is worth to note that P (Y (t, ω) = ∞) = P (S_{n}(ω) ≤ t, n ≥ 0) = 0. Indeed, basing on the strong law of large numbers we have \frac{S_{n}}{n} → EX_{1} = 1/μ with probability 1 as n → ∞ that implies P (S_{n} > c) = P (\frac{S_{n}}{n} > \frac{c}{n}) → 1 for every c > 0 as n → ∞.

Moreover, it is easy to see that for fixed ω the function Y (·, ω) is discontinuous but continuous from the right and takes only nonnegative integers.

It is possible to show that {Y (t), t ≥ 0} is the Poisson process.

First of all, it is useful to note that the following relations hold :

{ω : Y (t, ω) = n} = {ω : S_{n}(ω) ≤ t < S_{n+1}(ω)},
{ω : Y (t, ω) ≤ n} = {ω : S_{n+1}(ω) > t},
{ω : Y (t, ω) ≥ n} = {ω : S_{n}(ω) ≤ t}.

It follows from those relations that P (Y (0) = 0) = 1 since

P (Y (0) > 0) = P (Y (0) ≥ 1) = P (S_{1} ≤ 0) = P (X_{1} ≤ 0) = 0.

Now let us find a distribution of Y (t). We have

P (Y (t) ≤ n) = P (S_{n+1} > t) = ∫_{t}^{∞} μ\frac{(μx)^{n}}{n!} e^{−μx}dx = \frac{1}{n!} ∫_{μt}^{∞} u^{n}e^{−u}du =

−\frac{1}{n!}u^{n}e^{−u}|{∞ \atop μt} + \frac{1}{(n − 1)!} ∫_{μt}^{∞} u^{n−1}e^{−u}du = \frac{(μt)^{n}}{n!} e^{−μt} + \frac{1}{(n − 1)!} ∫_{μt}^{∞} u^{n−1}e^{−u}du =

∑_{k=0}^{n} \frac{(μt)^{k}}{k!} e^{−μt}.

Hence (see Exercise 15),

P (Y (t) = n) = P (Y (t) ≤ n) − P (Y (t) ≤ n − 1) = \frac{(μt)^{n}}{n!} e^{−μt}.

Further on, it is possible to prove that for 0 < t_{1} < … < t_{n} the increments Y (t_{1}), Y (t_{2}) − Y (t_{1}), …, Y (t_{n}) − Y (t_{n−1}) are independent with Poissonian distribution

P (Y (t_{k}) − Y (t_{k−1}) = m) = \frac{(μ(t_{k} − t_{k−1}))^{m}}{m!} e^{−μ(t_{k}−t_{k−1})}, m ≥ 0.

Simulation of the Poisson Process

Algorithm for simulation of the Poisson process is quite clear.

1. Set Y (0) = 0.

2. Simulate X_{1} ∼ E(μ) and set Y (t) = 0 for t < X_{1} and Y (X_{1}) = 1.

3. Simulate X_{2} ∼ E(μ) and set Y (t) = 1 for X_{1} < t < X_{1} + X_{2} and Y (X_{1} + X_{2}) = 2, and so on.

4. On nth step we simulate X_{n} ∼ E(μ) and set Y (t) = n − 1 for X_{1} + ⋯ + X_{n−1} < t < X_{1} + ⋯ + X_{n} and Y (X_{1} + ⋯ + X_{n}) = n.

Questions and Exercises

1. Define a stochastic process {X(t), t ≥ 0} as X(t, ω) = 1_{{t=0}}ξ(ω) + at, where a > 0 is fixed and ξ is a random variable. What does the process represent ? Could you describe the set of all its sample paths ?

2. Define a stochastic process {X(t), t ∈ [0, 1]} as X(t, ω) = 1 if ξ(ω) = t and zero otherwise, where ξ is a uniformly distributed in [0, 1] random variable. Calculate P (X(0.1) ≤ 1, X(0.5) ≤ 1/2, X(0.9) ≤ 1).

3. Define a stochastic process {X(t), t ∈ [0, 1]} as X(t, ω) = tξ(ω) + η(ω), where ξ and η are independent discrete random variables with distributions: P (ξ = ±1) = 1/2, P (η = 2) = 1/3, P (η = 0) = 2/3. What is the set of all sample paths of the process ? Calculate P (X(1/3) ≤ 1/2, X(1) ≤ 1).

4. Let {X(t), t ∈ [0, 1]} be a stochastic process defined in Exercise 2 and {Y (t), t ∈ [0, 1]} be another stochastic process such that Y (t) = 0 everywhere. What can you say about finite-dimensional distributions and sample paths of both processes ? Calculate and compare

P (max_{t} X(t) ≤ \frac{1}{2}) , P (max_{t} Y (t) ≤ \frac{1}{2}) .

5. Let {X(t), t ∈ T } and {Y (t), t ∈ T } be two stochastic processes with independent increments. Is it true that for given a > 0, b > 0, the stochastic process {aX(t) + bY (t), t ∈ T } is also with independent increments ?

6. Consider the stochastic process from Example 1 (see page 6). Under what conditions on ξ and η it is with independent increments ?

7. Let {X(t), t ∈ T } and {Y (t), t ∈ T } be two stochastic processes with independent stationary increments. What should we also assume about them in order to have the stochastic process {aX(t) + bY (t), t ∈ T } for given a > 0, b > 0, to be with independent stationary increments ?

8. Consider the stochastic process from Example 1 (see page 6). Is it stochastically continuous ? Are all sample paths continuous ?

9. Consider the stochastic process from Example 2 (see page 6). Is it stochastically continuous ? Are all sample paths continuous ?

10. Let {W (t), t ≥ 0} be Brownian motion. Determine the joint conditional density function of W (t_{1}) and W (t_{2}) given W (1) = 0, where 0 < t_{1} < t_{2} < 1.

11. If {W (t), t ≥ 0} is a Brownian motion, show that the stochastic process {X(t), t ≥ 0} defined as X(t) = tW (t^{−1}) for t > 0 and X(0) = 0 is also a Brownian motion.

12. Check that under the construction of the Brownian motion (see page 17),

W_{1}(t, ω) = ξ_{1}^{(0)}(ω)S_{1}^{(0)}(t) + ξ_{1}^{(1)}(ω)S_{1}^{(1)}(t).

13. Could you indicate any similarities and distinctions in two methods of simulation of the Brownian motion (see pages 19–20) ?

14. Let a random vector (Y_{1}, Y_{2}) has the density function

f(y_{1}, y_{2}) = μ^{2}e^{−μy_{2}}, 0 < y_{1} < y_{2},

and zero otherwise. Show that the random variables X_{1} = Y_{1} and X_{2} = Y_{2} − Y_{1} are independent and identically distributed with exponential E(μ) distribution.

15. Let {Y (t), t ≥ 0} be defined as on page 24. Prove that

P (Y (t) = n) = P (Y (t) ≤ n) − P (Y (t) ≤ n − 1).