- #1

mafra

- 10

- 0

## Homework Statement

A Wiener Process W(t) is a stochastic process with:

W(0) = 0

Trajectories almost surely continuous

Independent increases, that means, for all t1 ≤ t2 ≤ t3 ≤ t4, we have (W(t2) - W(t1)) is independent of (W(t4) - W(t3))

For t ≤ s, (W(s) - W(t)) follows a normal centered law of variance (s-t).

## Homework Equations

a) The process W(t) is markovian?

b) For s,t > 0, what is the law of the random variable W(s) + W(s+t)?

c) For u,v,w > 0, calculate E[W(u)W(u+v)W(u+v+w)]

d) Calculate the autocovariance function of the process exp(-t)W(exp(2t))

## The Attempt at a Solution

**a) The process W(t) is markovian?**

A process is markovian if:

P[X(tn) ∈ Bn | X(tn−1) ∈ Bn−1, ... , X(t1) ∈ B1] = P(X(tn) ∈ Bn | X(tn−1) ∈ Bn−1)

Where Bi (I believe) is the Borel σ-algebra of the real numbers.

By the data given:

P[W(t2) - W(t1) ∈ B1 | W(t4) - W(t3) ∈ B1] = P[W(t2) - W(t1) ∈ B1

For all t1 ≤ t2 ≤ t3 ≤ t4

I don't know how to follow from now on

The answer given is yes.

**b) For s,t > 0, what is the law of the random variable W(s) + W(s+t)?**

W(s) is gaussian and W(s+t) is gaussian, so W(s) + W(s+t) is gaussian

Mean[W(s) + W(s + t)] = Mean[W(s)] + Mean[W(s+t)]

Mean[W(s) - W(t)] = 0

Mean[W(s)] = Mean[W(t)] which means that the mean is constant

Mean[W(0)] = 0 → Mean[W(s)] = 0 → Mean[W(s) + W(s+t)] = 0

Var[W(s) - W(t)], t ≤ s = s - t

For t = 0:

Var[W(s) - W(0)] = s - 0

Var[W(s)] = s

σ² = Var[W(s) + W(s+t)]

σ² = E{[W(s) + W(s+t)]*[W(s+T) + W(s+t+T)]}

σ² = E[W(s)W(s+T) + W(s)W(s+t+T) + W(s+t)W(s+T) + W(s+t)W(s+t+T)]

σ² = Var[W(s)] + E[W(s)W(s+t+T)] + E[W(s+t)W(s+T)] + Var[W(s+t)]

σ² = 2s + t + E[W(s)W(s+t+T)] + E[W(s+t)W(s+T)]

W(s+t+T) = [W(s+t+T) - W(s+T)] + W(s+t)

E[W(s)W(s+t+T)]

E{[W(s+t+T) - W(s+T)]*W(s)} + E[W(s+t)W(s)]

t1=0, t2=s, t3=s+T, t4=s+t+T, t1 ≤ t2 ≤ t3 ≤ t4 (Not sure of what I'm doing here, since T could be negative)

E[W(s+t+T) - W(s+T)]*E[W(s)] + s

s

σ² = 3s + t + E[W(s+t)W(s+T)]

I tried some similar substitutions but that doesn't seem to work in this last expectation

The given answer is σ² = 4s + 1

**c) For u,v,w > 0, calculate E[W(u)W(u+v)W(u+v+w)]**

E[W(u)W(u+v)W(u+v+w)]

W(u) = w1

W(u+v) = w2

W(u+v+w) = w3

∫ ∫ ∫ w1*w2*w3*p(w1,w2,w3) dw3 dw2 dw1

p(w1,w2,w3) = p(w3|w2,w1)*p(w2|w1)*p(w1)

Markovian process, w2 > w1 → p(w3|w2,w1) = p(w3|w2)

∫ ∫ ∫ w1*w2*w3*p(w3|w2)*p(w2|w1)*p(w1) dw2 dw2 dw1

∫ w1 ∫ w2 ∫ w3*p(w3|w2) dw3 p(w2|w1) dw2 p(w1) dw1

∫ w3*p(w3|w2) dw3 = E[w3|w2=w2]

E[w3-w2] = 0

E[w3] = E[w2]

E[w3|w2=w2] = E[w2|w2=w2] = w2

∫ w3*p(w3|w2) dw3 = w2

∫ w1 ∫ w2² p(w2|w1) dw2 p(w1) dw1

∫ w2² p(w2|w1) dw2 = Rw[w2|w1]

Rw[w2|w1] = var[w2|w1] + E[w2|w1=w1]²

Rw[w2|w1] = v + w1²

∫ w1*(v + w1²) p(w1) dw1

∫ w1*v p(w1) dw1 + ∫ w1³ p(w1) dw1

v*mean[w1] + ∫ w1³ p(w1) dw1

v*mean[w1] = 0

∫ w1³*p(w1) dw1 = 0 because w1³*p(w1) is an odd function

E[W(u)W(u+v)W(u+v+w) = 0

The result matches the given answer, but I don't know if there are any other paths or if I all the resolution is correct

PS: Calculations for conditional variance var[w2|w1]

P[W(u+v) - W(u)] = N(0, v)

P[W(u+v) - W(u)] = P[W(u+v) - W(u) | W(u) - W(0) = a]

P[W(u+v) - W(u) | W(u) = a] = N(0,v)

P[W(u+v) | W(u) = a] = N(a,v) - Variance doesn't change with summed constants like W(u) is here

**d) Calculate the autocovariance function of the process exp(-t)W(exp(2t))**

Here I could do it just by assuming beforehand that the process was Wide-Sense Stationary and the conditional probabilities were all gaussian

X(t) = exp(-t)W(exp(2t))

As before, the mean rests in zero, so the covariance is E[X(0)E(T)], T≥0

X(0) = x1

X(T) = x2

∫∫ x1*x2*p(x1,x2) dx1 dx2

∫∫ x1*x2*p(x1|x2)*p(x2) dx1 dx2

∫ x1 ∫ x2*p(x2|x1) dx2 p(x1) dx1

p(x) = p(exp(-t)*W(exp(2t)))

p(W(exp(2t))) = N(0,exp(2t))

mean(a*x) = a*mean(x), variance(a*x) = a²*variance(x)

p(x) = p(exp(-t)*W(exp(2t))) = N(0,1)

p(x2|x1) = p( X(t2) | X(t1) = x1 )

t2 = T, t1 = 0:

p( X(T) | X(0) = x2 )

p( exp(-T)*W(exp(2T)) | W(1) = x1 )

p(W(exp(2T)) | W(1) = x1) = N(x1, exp(2T) - 1)

p( exp(-T)*W(exp(2T)) | W(1) = x1 ) = N(x1*exp(-T), 1 - exp(-2T))

p(x1) = N(0,1)

p(x2|x1) = N(x1*exp(-T), 1 - exp(-2T))

∫ x1 ∫ x2*p(x2|x1) dx2 p(x1) dx1

∫ x2*p(x2|x1) dx2 = E[x2|x1] = x1*exp(-T)

∫ x1²*exp(-T)*p(x1) dx1

exp(-T) ∫ x1²*p(x1) dx1 = exp(-T)*Var(x1) = exp(-T)

The answer is exp(-|T|), but again, I don't know how to show that the covariance is symmetrical