Chapter 7 - Brownian motion & martingales

Question 1

Definition of a Weiner process (Standard Brownian Motion)

Answer 1

Wt , t>=0 is a Weiner if:
i) W0 = 0
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ii) Wt has continuous sample paths
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iii) For any 0<= s<= t, the increment Wt-Ws is normally distributed, Wt-Ws ~N(0, t-s)
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iv) Wt has independent increments that is for any sequence of times 0<=t1<=2< …. <tn , we have that increments Wtn- Wtn-1, …, Wt3 - Wt2, Wt2-Wt1 are indpendent random variables [this implies Markovian]
(Alternatively Wt-Ws is independent of Wu for u<s)

Question 2

Brownian motion

Answer 2

ii) Wt has continuous sample paths
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iv) Wt has independent increments that is for any sequence of times 0<=t1<=2< …. <tn , we have that increments Wtn- Wtn-1, …, Wt3 - Wt2, Wt2-Wt1 are indpendent random variables [this implies Markovian]
(Alternatively Wt-Ws is independent of Wu for u<s)
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~N (μ(t-s), σ^2 (t-s)). [i.e std b.m when μ=0, σ=1, Z0=0]

Question 3

What is the relationship between a Brownian Motion & STD Brownian Motion?

Answer 3

Zt= Z0 + σWt +μt (diffusion coeff σ and drift μ)

Question 4

Covariance of a Weiner Process

Answer 4

Cov(Ws, Wt)
=E[(Ws-E[Ws])(Wt-E[Wt])]
=E(WsWt - E(Ws)Wt - WsE(Wt) + E(Ws)E(Wt))
=E[Ws(Ws+ (Wt-Ws))]… since E[Wt]=E[Ws]=0
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By independence of increments,
Cov(Ws, Wt) = E[Ws^2] +E[Ws]E[(Wt-Ws)]
=Var[Ws] + 0 = s
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in general Cov(Xs, Xt) = min{s,t}

Question 5

Geometric Brownian Motion

Answer 5

St= e^Zt where Zt = Z0 + σWt+ μt.
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Therefore St ~ logN (Z0 +μt, σ^2t)
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Hence logSt~ N (Z0 +μt, σ^2t)

Question 6

Martingales

Answer 6

def: stoc proc whose expected value in the future is = to its current value. (martingale represents a ‘fair game’ where a player is neither expected to win or lose in the future.
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E[XT/Ft] = Xt for t<T
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filtration Ft represents everything that can be known up to & including time t

Question 7

When is a stochastic process Xt called a martingale with respect to filtration?

Answer 7

• E[|Xt|] < ∞ for all t
  -E[Xt|Fs] = Xs for all s<=t

Question 8

Supermartingale

Answer 8

E[Xt|Fs] <= Xs
(either has a negative or 0 drift)

Question 9

Submartingale

Answer 9

E[Xt|Fs] >= Xs
(either has a positive or 0 drift)

Question 10

Show that a Weiner process is a martingale with respect to its natural filtration, noting that E[|Wt|] <∞ since Wt < ∞ almost surely

Answer 10

E[Wt|Fs^w] = E[Ws + (Wt-Ws) |Fs^w]
=E[Ws|Fs^w] + E[(Wt-Ws) |Fs^w]
Since the increments are independent & Wt-Ws ~N(0, t-s):
E[Wt|Fs^w] = Ws

Question 11

Is Wt^2 -t a margingale?

Answer 11

E[Wt^2 -t|Fs^w] = E[(Ws + (Wt-Ws))^2|Fs^w] - t
= E[Ws^2/Fs^w] +2E[Ws(Wt-Ws)|Fs^w] + E((Wt-Ws)^2\Fs^w] - t
= Ws^2 -s
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Yes since E[Wt^2 -t|Fs^w]= Ws^2 -s, this is a martingale

Question 12

Is exp(λWt -0.5(λ^2)*t) a martingale?

Answer 12

E[exp(λWt -0.5(λ^2)t) |Fs^w] = E[exp(λ(Ws + (Wt-Ws))-0.5(λ^2)(s+(t-s))|Fs^w]
= exp(λWs -0.5(λ^2)s)* E[exp(λ(Wt-Ws) -0.5(λ^2)(t-s)|Fs^w]
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E[exp(λ(Wt-Ws) -0.5(λ^2)(t-s)|Fs^w] = E(exp(Yt)]
=exp(E[Yt] + 0.5Var(Yt))
= exp(-0.5(λ^2)(t-s) +0.5(λ^2)(t-s) ) = 1 … followed from MGF of a normal distribution
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Since E[exp(λWt -0.5(λ^2)t) |Fs^w] =exp(λWs -0.5(λ^2)*s), it means this is a martingale.