- #1
mafra
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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