Lecture 4 : Second-Order Calculus: Continuity, Differentiation, Integration of the Process. Karhunen-Loeve Expansion.

Second-Order Calculus: Continuity

Here we consider second-order processes {X(t), t ∈ T } with T ⊂ R. We are going to develop a theory in which it would be possible to talk about continuity, differentiation and integration of the process.

Definition 1. A stochastic process {X(t), t ∈ T } is said to be continuous in mean square iff

E(X(t + h) − X(t))^{2} → 0 as h → 0 ∀t ∈ T.

It is easy to see that mean square continuity implies stochastic continuity. Indeed (proof is the same as for Chebyshev inequality; for simplicity we suppose that a random variable X(t + h) − X(t) possesses density f),

E(X(t + h) − X(t))^{2} = ∫ u^{2}f(u)du ≥ ∫_{{u: |u|≥ε}}u^{2}f(u)du ≥ ε^{2} ∫_{{u: |u|≥ε}}f(u)du =

ε^{2}P (|X(t + h) − X(t)| > ε) ⟹ P (|X(t + h) − X(t)| > ε) ≤ \frac{E(X(t + h) − X(t))^{2}}{ε^{2}}

for every ε > 0.

Dependence between different notions of continuity for stochastic processes:

all sample paths continuity⟹mean square continuity⟹stochastic continuity.

If to remind the example of a stochastic process which is stochastically continuous but has all sample paths discontinuous (see Lectures 1–2, page 12), the same process will be also mean square continuous (see Exercise 1).

An example of a stochastic process which is stochastically continuous but is not mean square continuous is given in Exercise 2.

Characterization of Mean Square Continuity

Theorem 1. Assume m(t) to be a continuous function on T. Then the process {X(t), t ∈ T } is continuous in mean square iff the covariance function K is continuous at (t, t) for every t ∈ T.

Proof. It is easy to see that the process {X(t), t ∈ T } is continuous in mean square iff the process {X(t) − m(t), t ∈ T } is continuous in mean square. But the second process has the same covariance function K(s, t) as the first one and zero mean function. Therefore, we may assume further that m(t) = 0.

Let K be continuous at (t, t). Then as h → 0,

E(X(t + h) − X(t))^{2} = K(t + h, t + h) − K(t + h, t) − K(t, t + h) + K(t, t) → 0.

Conversely, let {X(t), t ∈ T } be continuous in mean square. Hence,

E(X(t + h) − X(t))^{2} → 0 as h → 0,
E(X(t + h′) − X(t))^{2} → 0 as h′ → 0.

Thus,

|K(t + h, t + h′) − K(t, t)| ≤ |K(t + h, t + h′) − K(t + h, t)| + |K(t + h, t) − K(t, t)| =

|E (X(t + h)(X(t + h′) − X(t))) | + |E (X(t)(X(t + h) − X(t))) | ≤

\sqrt{E(X(t + h)^{2})}\sqrt{E(X(t + h′) − X(t))^{2}} + \sqrt{E(X(t)^{2})}\sqrt{E(X(t + h) − X(t))^{2}} → 0.

Lemma 1. The covariance function K is continuous at (s, t) for every (s, t) ∈ T × T iff it is continuous at (t, t) for every t ∈ T.

Proof. Of course, we need only to prove that if K is continuous at (t, t) for every t ∈ T then it is continuous at (s, t) for every (s, t) ∈ T × T. Again assume m(t) = 0. If K is continuous at any point (t, t) then according to Theorem 1

E(X(s + h) − X(s))^{2} → 0 as h → 0,
E(X(t + h′) − X(t))^{2} → 0 as h′ → 0.

Thus, applying the same ”standard” reasoning as during the proof of Theorem 1,

|K(s + h, t + h′) − K(s, t)| → 0.

It is useful now to rewrite Theorem 1 and Lemma 1 for the covariance function of a stationary in the wide sense stochastic process as well.

Examples

1. Both the Poisson process and Brownian motion are continuous in mean square.

2. A second-order stochastic process {X(t), t ≥ 0} having independent increments and the property P (X(0) = 0) = 1 is continuous in mean square iff both m(t) and B(t) = E((X(t) − m(t))^{2} are continuous functions on T.

3. Let Z_{1}, Z_{2} be independent identically distributed random variables having normal N(0, σ^{2}) distribution. Define the stochastic process {X(t), t ∈ R} as X(t) = Z_{1} cos(λt) + Z_{2} sin(λt), where λ is a fixed real number. The process is continuous in mean square since, as it was already calculated (see Lecture 3, Example 4, page 7), K(s, t) = σ^{2} cos(λ(t − s)) – continuous function on R^{2}.

4. Let X_{t}, t ∈ T ⊂ R, be independent identically distributed random variables having normal N(0, 1) distribution and consider a stochastic process {X(t), t ∈ T }, where X(t) = X_{t}, t ∈ T. Then

K(s, t) = \{ \begin{matrix}1, & s = t \\ 0, & s ≠ t \end{matrix} .

The process is not continuous in mean square since the covariance function is not continuous.

Second-Order Calculus: Differentiation

Definition 2. A stochastic process {X(t), t ∈ T } is said to be differentiable (in mean square) iff

E (\frac{X(t + h) − X(t)}{h} − X′(t))^{2} → 0 as h → 0 ∀t ∈ T

for some second-order stochastic process X′(t).

Theorem 2. Assume m(t) to be a differentiable function on T. Then the process \{X(t), t ∈ T \} is differentiable iff there exists \frac{∂^{2}K(s,t)}{∂s∂t} at points (t, t) for every t ∈ T.

Proof. It is easy to see that the process {X(t), t ∈ T } is differentiable iff the process {X(t) − m(t), t ∈ T } is differentiable. But the second process has the same covariance function K(s, t) as the first one and zero mean function. Therefore, we may assume further that m(t) = 0.

It is worth to remind that

\frac{∂^{2}K(s, t)}{∂s∂t} = lim_{h→0} lim_{h′→0} \frac{K(s + h, t + h′) − K(s + h, t) − K(s, t + h′) + K(s, t)}{hh′}.

Let {X(t), t ∈ T } be differentiable. Applying the ”standard” reasoning, we have as h, h′ → 0,

|\frac{K(t + h, t + h′) − K(t + h, t) − K(t, t + h′) + K(t, t)}{hh′} − E(X′(t))^{2}| → 0. (1)

Conversely, let \frac{∂^{2}K(s,t)}{∂s∂t}|_{(t,t)} exists, that is

lim_{h→0} lim_{h′→0} \frac{K(t + h, t + h′) − K(t + h, t) − K(t, t + h′) + K(t, t)}{hh′}

exists and is equal, say, to C. Then as h, h′ → 0,

E (\frac{X(t + h) − X(t)}{h} − \frac{X(t + h′) − X(t)}{h′})^{2} → C − C − C + C = 0.

The last relation implies*

E (\frac{X(t + h) − X(t)}{h} − X′(t))^{2} → 0 as h → 0

for some X′(t).

Corollary 1. If m(t) = 0 and there exists \frac{∂^{2}K(s,t)}{∂s∂t} at points (t, t) for every t ∈ T, then the process is differentiable and EX(s)X′(t) = \frac{∂K(s,t)}{∂t}, EX′(s)X′(t) = \frac{∂^{2}K(s,t)}{∂s∂t}.

Proof. Differentiability of the process has already been proved in Theorem 2. Then, applying the ”standard” reasoning, we get as h → 0

|\frac{K(s, t + h) − K(s, t)}{h} − EX(s)X′(t)| → 0 ⟹ \frac{∂K(s, t)}{∂t} = EX(s)X′(t),

and

|\frac{EX(s + h)X′(t) − EX(s)X′(t)}{h} − EX′(s)X′(t)| → 0,

or equivalently,

| \frac{\frac{∂K(s+h,t)}{∂t} − \frac{∂K(s,t)}{∂t}}{h} − EX′(s)X′(t)| → 0 ⟹ \frac{∂^{2}K(s, t)}{∂s∂t} = EX′(s)X′(t).

Examples.

1. Neither Brownian motion nor the Poisson process are differentiable. Indeed, consider a second-order stochastic process with covariance function K(s, t) = min{s, t}, s ≥ 0, t ≥ 0. Suppose, for a moment, there exists a second-order stochastic process X′(t) such that

E (\frac{X(t + h) − X(t)}{h} − X′(t))^{2} → 0 as h → 0 ∀t ≥ 0.

Taking (1) with h = h′ and K(s, t) = min{s, t}, we get

|\frac{t + h − t − t + t}{h^{2}} − E(X′(t))^{2}| = |\frac{1}{h} − E(X′(t))^{2}| → 0 as h → 0,

and this is a contradiction.

2. Consider the stochastic process from Example 3 on page 3. We have K(s, t) = σ^{2} cos(λ(t − s)), s, t ∈ R, and therefore

\frac{∂^{2}K(s, t)}{∂s∂t} = σ^{2}λ^{2} cos(λ(t − s)), s, t ∈ R.

Indeed, it is easy to see that in this case

X′(t) = −Z_{1}λ sin(λt) + Z_{2}λ cos(λt),

and its covariance function is precisely \frac{∂^{2}K(s,t)}{∂s∂t}. The process X′(t) is again Gaussian.

Second-Order Calculus: Integration

Let [a, b] ⊂ T be a finite interval and g : [a, b] → R. We define ∫_{a}^{b} g(t)X(t)dt as follows.

Let Δ : a = t_{0} < t_{1} < … < t_{n} = b be a partition of [a, b], with |Δ| = max_{1≤k≤n} |t_{k} − t_{k−1}|.

Define

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

If E (I(Δ) − I)^{2} → 0 as |Δ| → 0 for some random variable I, then we say that g(t)X(t) is integrable (in mean square) on [a, b] and write

I = ∫_{a}^{b} g(t)X(t)dt.

It is possible to show that if m and g are continuous on [a, b] and K is continuous on [a, b] × [a, b], then g(t)X(t) is integrable on [a, b] (arguing as at the end of page 4).

Lemma 2. If m(t) = 0, g(t) and h(t) are continuous on [a, b] and K(s, t) is continuous on [a, b] × [a, b], then

E [∫_{a}^{b} g(s)X(s)ds ∫_{a}^{b} h(t)X(t)dt] = ∫_{a}^{b} ∫_{a}^{b} g(s)h(t)K(s, t)dsdt,

E [X(s) ∫_{a}^{b} h(t)X(t)dt] = ∫_{a}^{b} K(s, t)h(t)dt, E [∫_{a}^{b} g(s)X(s)ds] = E [∫_{a}^{b} h(t)X(t)dt] = 0.

Proof. It is quite similar for all formulas. That is why we show only how to prove the first one. Let

I(Δ) = ∑_{j=1}^{n} g(s_{j})X(s_{j})(s_{j} − s_{j−1}), J(Δ′) = ∑_{k=1}^{n} h(t_{k})X(t_{k})(t_{k} − t_{k−1}),

I = ∫_{a}^{b} g(s)X(s)ds, J = ∫_{a}^{b} h(t)X(t)dt.

Since E (I(Δ) − I)^{2} → 0, E (J(Δ′) − J)^{2} → 0 as |Δ|, |Δ′| → 0, applying the ”standard” reasoning we get |E (I(Δ)J(Δ′)) − E(IJ)| → 0. But (usual Riemann sum)

E (I(Δ)J(Δ′)) → ∫_{a}^{b} ∫_{a}^{b} g(s)h(t)K(s, t)dsdt,

proving the first statement.

Karhunen-Loeve Expansion

Theorem 3. Let {X(t), t ∈ T } be a second-order stochastic process with zero mean and continuous covariance K. Then on [a, b] ⊂ T it admits an orthogonal expansion of the form

Z(t) = ∑_{k=1}^{∞} Z_{k}e_{k}(t),

where Z_{k} = ∫_{a}^{b} X(t)e_{k}(t)dt are orthogonal random variables with EZ_{k} = 0, EZ_{k}^{2} = λ_{k} > 0, while λ_{k} and e_{k}(t) are such that

∫_{a}^{b} K(s, t)e_{k}(t)dt = λ_{k}e_{k}(s), s ∈ [a, b], ∫_{a}^{b} e_{j}(t)e_{k}(t)dt = \{ \begin{matrix}1, & j = k \\ 0, & j ≠ k \end{matrix} .

The series converges to X(t) uniformly in t, in other words,

E (X(t) − ∑_{k=1}^{n} Z_{k}e_{k}(t))^{2} → 0 as n → ∞

uniformly for t ∈ [a, b].

Example.

Let us find the Karhunen-Loeve expansion of Brownian motion {W (t), t ≥ 0} on the interval [0, 1]. We have K(s, t) = min{s, t} and need to find λ_{k} and e_{k}(t) such that

∫_{0}^{1} min{s, t}e_{k}(t)dt = ∫_{0}^{s} te_{k}(t)dt + s ∫_{s}^{1} e_{k}(t)dt = λ_{k}e_{k}(s), s ∈ [0, 1].  (2)

We may differentiate with respect to s to obtain

se_{k}(s) − se_{k}(s) + ∫_{s}^{1} e_{k}(t)dt = ∫_{s}^{1} e_{k}(t)dt = λ_{k}e′_{k}(s), s ∈ [0, 1].  (3)

Differentiate again to obtain −e_{k}(s) = λ_{k}e′′_{k}(s), s ∈ [0, 1]. As it is known, the general solution of this differential equation is

e_{k}(s) = A sin(λ_{k}^{−1/2}s) + B cos(λ_{k}^{−1/2}s), s ∈ [0, 1].

Set s = 0 in (2) to obtain e_{k}(0) = 0. Thus B = 0. Set s = 1 in (3) to obtain e′_{k}(1) = 0. Thus, cos(λ_{k}^{−1/2}) = 0 ⟹ λ_{k}^{−1/2} = (k − 1/2)π, k ≥ 1 ⟹ λ_{k} = (k − 1/2)^{−2}π^{−2}, k ≥ 1 ⟹ e_{k}(t) = \sqrt{2} sin((k − 1/2)πt), k ≥ 1. Finally,

W (t) = \sqrt{2} ∑_{k=1}^{∞} Z_{k} sin((k − 1/2)πt), t ∈ [0, 1],

where {Z_{k}}_{k≥1} are independent random variables having N(0, λ_{k}) distribution, k ≥ 1.

Questions and Exercises

1. 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]. Show (without calculation of the covariance) that this process is mean square continuous.

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

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

where ξ is a random variable having uniform distribution in [0, 1], while f(t) = t for t ∈ Q – the set of all rational numbers, and zero otherwise. Show (without calculation of the covariance) that this process is not mean square continuous but stochastically continuious.

3. Do Exercise 1 through the calculation of the covariance and using Theorem 1. Is the process a stationary in the wide sense ?

4. Do Exercise 2 through the calculation of the covariance and using Theorem 1. Is the process a stationary in the wide sense ?

5. Check, whether the process {X(t), t ∈ [0, 1]} from Exercise 1 is differentiable.

6. Calculate the mean and variance of a random variable ∫_{0}^{1} W (t)dt, where {W (t), t ≥ 0} is a Brownian motion. Can we state that it has a normal distribution ?

7. Calculate the mean and variance of a random variable ∫_{0}^{π} X(t)dt, where {X(t), t ≥ 0} is a stochastic process from Example 3 on page 3.

8. Find the correlation between W (t) and ∫_{0}^{1} W (t)dt, where {W (t), t ≥ 0} is a Brownian motion.

9. Repeat calculation made in Example on page 7 and find the Karhunen-Loeve expansion of Brownian motion {W (t), t ≥ 0} on the interval [0, a].

10. Find the Karhunen-Loeve expansion of a second-order stochastic process with covariance function K(s, t) = st on the interval [0, 1].

11. Find the Karhunen-Loeve expansion of a second-order stochastic process with covariance function K(s, t) = min{s, t} − st on the interval [0, 1].