Chapter 10: Brownian motion Flashcards
graphs of Brownian motion:
jagged movements
shown examples, for graphs of different times each pic has 1/5 of next. For T unit of time on the x-axis and root(T) units of space on the y-axis.
E[|X_T|]≈ root(T) as T tends to infinity
As we scale out we see convergence to a limit, random, continuous time stochastic processes.
THM 10.2.1
Brownian motion
There is a stochastic process (Bt) such that:
- The paths of (Bt) are continuous.
- For any 0 ≤ u ≤ t, the random variable Bt −Bu is independent of σ(Bv ; v ≤ u).
- For any 0 ≤ u ≤ t, the random variable Bt −Bu has distribution N(0,t−u).
1* Bt is a random continuous funct
2* movement between time t and u is indep f what happened before
(2 points in time indep of past)
Definition 10.2.2
STANDARD Brownian motion
If B0 = 0 we say that (Bt) is a standard Brownian motion.
WE DENOTE Bt as STANDARD BM: We write (Bt) for a standard Brownian motion, and Ft = σ(Bu ; u ≤ t) for its generated(natural) filtration. We work over the filtered space (Ω,F,(Ft),P)
Bt
We write (Bt) for a standard Brownian motion, and Ft = σ(Bu ; u ≤ t) for its generated filtration. We work over the filtered space (Ω,F,(Ft),P)
Brownian motion notes:
Bt (standard)
- B_t -B_0 ~N(0,t-0) so..
- Putting u = 0 into the second property and noting that B0 = 0, we obtain that the distribution of Brownian motion at time t is Bt ∼ N(0,t).
- B_t -B_0 ~N(0,t-0)
- E[Bt] = 0 for all t
- B^n_t ∈ L^1 for all n ∈ N. (*)
if Z ∼ N(0,t) then E[Z^2] = t.
so Brownian motion…
(*)
if Z ∼ N(0,t) then E[Z^2] = t.
Hence E[B^2_t ] = var(B_t) = t
for all t
if Z ∼ N(µ,σ^2) then
E[e^Z]
(*)
if Z ∼ N(µ,σ2) then
E[e^Z] = e^{µ+(1/2)σ^2}
E[e^{B_t}]
*
E[e^{B_t}] = e^{t/2}
B_t^n ∈ L^p for all p ≥ 1
B_t^n ∈ L^p for all p ≥ 1
⇔
E[|B_t|^p] less than infinity for all t,p
Brownian motion and heat equation
Temp @ position x @ time t is u(t,x), t ∈ [0,∞) time and x ∈R space.
Suppose at t=0 temp at point x is f(x)=u(0,x)
heat equation is
∂u/∂t = ∂^2u/∂x^2
with the initial condition
u(0,x) = f(x)
If we start Brownian motion from x ∈ R, then Bt ∼ x + N(0,t) ∼ N(x,t), so we can write down its probability density function
φ_{t,x}(y) =
(1 /√2πt) exp[−(y−x)^2/2t]
LEMMA 10.3.1
A sol of the heat equation
φt,x(y) satisfies the heat equation
φ_{t,x}(y) =
(1 /√2πt) exp[−(y−x)^2/2t]
If we start Brownian motion from x ∈ R, then Bt ∼ x + N(0,t) ∼ N(x,t), so we can write down its probability density function
Proof 10.3.1
φt,x(y) satisfies the heat equation
φ_{t,x}(y) =
(1 /√2πt) exp[−(y−x)^2/2t]
Because if u(t,x) solves so does u(t,x-y)
for any value of y. Assume y=0 and need to show that
φt,x(0) =
(1 /√2πt) exp[−(x^2/2t] satisfies ∂u/∂t = ∂^2u/∂x^2. Partial differentiation: chain and product rule
∂ φ/∂ t= (-0.5/√(2πt^3))exp(-x^2/2t)+ (1/√2πt)(-x^2(-1)/2t^2)exp(-x^2/2t)
1 √2π x2 2t5/2 − 1 t3/2exp−x2 2t and ∂φ ∂x = (1/√2πt)(−2x/2t) exp(−x2^/2t).
= −x/√2πt^3 exp(−x2^/2t).
so as
∂2φ/∂x2
= −1 √2πt3 exp−x2 2t+ −x √2πt3 −2x 2t
exp(−x2^/2t).
=
(1/√2π)[( x^2/2t^{5/2}) −
1/t^{3/2}] exp(−x2^/2t).
Hence, ∂φ/∂t = ∂^2φ/∂x^2 .
Lemma 10.3.2
We can use Lemma 10.3.1 to give a physical explanation of the connection between Brownian motion and heat diffusion. We define
w(t,x) = E_x[f(B_t)
That is, to get w(t,x), we start a particle at location x, let it perform Brownian motion for time t, and then take the expected value of f(Bt).
w(t,x)
w(t,x) = E_x[f(B_t)
satisfies
heat equation
∂u/∂t = ∂^2u/∂x^2
and the initial condition
u(0,x) = f(x)
PROOF
Lemma 10.3.2
w(t,x) = E_x[f(B_t)
satisfies
heat equation
∂u/∂t = ∂^2u/∂x^2
and the initial condition
u(0,x) = f(x)
w(t,x)= E_x[f(B_t)] = integral from -infinity to infinity of [ f(y)φ_{t,x}(y)].dy
*by looking in reverse to see how much heat there was initially
E_x[f(B_t)
start BM at x and consider packets of heat moving with BM
- f represents initial cond: how muh heat initially u(0,x)=f(x)
- B_t is heat varying ranf back in time
- E_x approx average heat effect over lots of heat packets
Consequently u=u
PROOF:
B_0=x, So w(0,x)= E_x[f(B_0) = E_x[f(x)] = f(x) and hence initial cond satisfied
integral from -∞ to ∞ of
[f(y) (∂/∂t)(φt,x(y))]dy
integral from -∞ to ∞ of
[f(y) (∂^2/∂x^2)(φt,x(y))]dy
=
(∂^2/∂x^2) integral from -∞ to ∞ of
f(y)φ_{t,x}(y)dy
=
∂^2w/∂x^2
BM symmetry note
Normal dist x ~ N(0,σ^2) is symmetric about 0, in the sense that −Z also has the distribution N(0,σ2). This symmetry about 0 is also present in Brownian motion.
Lemma 10.4.1
SELF-SYMMETRY OF BM
The stochastic process W_t = −B_t is a standard Brownian motion.
PROOF
Lemma 10.4.1
SELF-SYMMETRY OF BM
The stochastic process W_t = −B_t is a standard Brownian motion.
Check Wt = −Bt satisfies the three defining properties of Brownian motion.
- By the first property, Bt is almost surely continuous, hence −Bt is also almost surely continuous. Also, for 0 ≤ u ≤ t we have Wt −Wu = −(Bt −Bu).
- Since, by the second property, Bt −Bu is independent of Fu, so is Wt −Wu, and we have σ(Wv ; v ≤ u) = σ(−Wv ; v ≤ u) = σ(Bv ; v ≤ u) = Fu. Thus Wt −Wu is independent of σ(Wv ; v ≤ u).
- Lastly, if Z ∼ N(0,t) then, using the symmetry of normal random variables, −Z ∼ N(0,t), so we have Wt −Wu = −(Bt −Bu) ∼ N(0,t−u) by the third property.
Hence, all three properties also hold for (Wt), so Wt is a Brownian motion. Since W0 = −B0 = 0, we have that (Wt) is a standard Brownian motion.
LEMMA 10.4.2
NON DIFFERENTIABILITY
Let t ∈ [0,∞). Almost surely, the function t → Bt is not differentiable at t.
P[u →B_n is diff at t]=0
PROOF
LEMMA 10.4.2
NON DIFFERENTIABILITY
Let t ∈ [0,∞). Almost surely, the function t → Bt is not differentiable at t.
P[u →B_n is diff at t]=0
Proof:
Recall defn of deriv
f’(t)= limit as h tends to 0 of [ (f(t+h) -f(t))/h]
2) of BM and scaling properties of normal dist
(B_{t+h} - B_t ) /h ~ B_h /h ~ X/√h.
where X ∼ N(0,1).
(N(0,t+n-t))
Note that X is positive half the time and negative half the time (and P[X = 0] = 0).
(probabilities 0.5)
Hence, as h → 0, we obtain that
[X/ √h ]-a.s→ X_∞ =
{∞ with probability 1/2,
{−∞ with probability 1/2.
from lemma 6.1.2 almost sure convergence implies convergence in dist, so this limit also holds in dist. Since (B_{t+h} - B_t ) /h has same dist as X/√h., it converges in dist to X_∞
Consider event E={Bt is differentiable at t} when it occurs (B_{t+h} - B_t ) /h converges to a finite quanitity as h tends to 0. However shown that the limit X_∞ with P[X_∞ ∈{∞,−∞}] =1, so prob that limit has finite quantity =0.
P[E]=0
LEMMA 10.4.3 BM
Relationship to martingales
Brownian motion is a martingale.
It is enough to look at the case (Bt) of standard Brownian motion, since adding and subtracting a deterministic constant does not change if a process is a martingale.
Since Bt ∼ N(0,t) we have var(Bt) < ∞, which implies that Bt ∈ L1.
((B_t) is adapted and (F_t) is because its the natural filtration)
Since the filtration (Ft) is the generated filtration of Bt, is immediate that Bt is adapted. Lastly, for any 0 ≤ u ≤ t we have
E[Bt|Fu] = E[Bt −Bu|Fu] +E[Bu|Fu] = E[Bt −Bu] + Bu = Bu. Here, we use the properties of Brownian motion: Bt − Bs is independent of Fu and
E[Bt] = E[Bu] = 0.
LEMMA 10.4.4
B^2_t −t
B^2_t −t is a martingale
(B_t is F_t measurable so B^2_t -t is, so (B^2_t -t ) adapted to (F_t))
Proof: Since Bt ∼ N(0,t) we have var(Bt) < ∞, which implies B2 t ∈ L1.
Hence also B^2_t −t ∈ L1. Since B2 t −t is a deterministic function of Bt, we have that Mt is adapted to the generated filtration of Bt. Lastly, for 0 ≤ u ≤ t,
E[B^2_t −t|Fu] = E[(Bt −Bu)^2 + 2BtBu −B^2_u|Fu]−t = E[(Bt −Bu)2|Fu] + 2BuE[Bt|Fu]−B2 u −t = E[(Bt −Bu)2] + 2B2 u −B2 u −t = (t−u) + B2 u −t
= B^2_u −u
as required. Here we use that Bt−Bu is independent of Fu, along with both (10.1) and Lemma 10.4.3.
(cant E(B^2_t | F_t), can E[(B_t - B_n)^2 | F_n])
