3 Itô Calculus Flashcards

Question

Lemma 3.2.4. For Y ∈ Hₜˢᵗᵉᵖ we have E[|I(Y )|²] (second moment) E[I(Y )]

Answer 1

For Y ∈ H_T ˢᵗᵉᵖ we have E[|I(Y )|²] = E[ ∫₀T|Y (s)|² ds] E[I(Y )] = 0. (RV ∫₀1 Yₙ(t) dW(s) has finite second moment) (MEAN is 0: RV as sum of RV x increment of brownian motion, centred RVs)

Answer 2

Answer 3

(Y_n(t))_ t in [0,1] in Hₜˢᵗᵉᵖ I: Hₜˢᵗᵉᵖ → L₂(Ω) (Yₙ(t))₎ₜ ᵢₙ [₀,₁] → I(Y)= ∫₀¹Y(s) dW(s) its a function that takes a simple process and gives a random variable stochastic process integrated wrt BM operator with properties

Answer 4

if we multiply the step with a real we also get a step we can define uniquely from disjoint sets

Answer 5

Considering: Y(t) ²= :=Σᵢ₌₁ⁿ Yₜ_ᵢ² **1**_[tᵢ₊₁,tᵢ] this is because the intervals are pairwise disjoint so only i=j are 1 if I fix omega this is just a step function so for each omega i can find the Riemann integral ∫₀¹Y(s) ² dW(s) = Σᵢ₌₁ⁿ Yₜ_ᵢ² (tᵢ₊₁-tᵢ) which is again a RV then I take the expectation. Note that: E[∫₀¹Y(s) .dW(s))²] = ‖∫₀¹|Y(s) dW(s)‖² L₂ = ‖ (Y(s))_t i [0,1] ‖²_L₂(Ωx[0,1]) RHS: = E[ ∫₀¹|Y (s)|² ds]= ∫_Ω ∫₀¹(Y (s))² ds dP As the integral operator takes a stochastic process and gives a RV, norm of the size of object same as norm of the image (ito's isometry)

Answer 6

PROOF: 1) Y ∈ H_T ˢᵗᵉᵖ so Y(t) = :=Σᵢ₌₁ⁿ Yₜ_ᵢ **1**_[tᵢ₊₁,tᵢ] where Yₜ_ᵢ is F_t measurable and bounded Consider E[(∫₀¹Y (s) dW(s) )²]= E[ (Σᵢ₌₁ⁿ Yₜ_ᵢ(W(tᵢ₊₁)-W(tᵢ)) ²] = E[Σᵢ,ⱼ₌₁ⁿ Yₜ_ᵢ(W(tᵢ₊₁)-W(tᵢ)) Yₜ_ⱼ(W(tⱼ₊₁)-W(tⱼ))] = (diagonal elements taken out) E[Σᵢⁿ Y²ₜ_ᵢ(W(tᵢ₊₁)-W(tᵢ))² + Σᵢ≠ⱼⁿ 2Yₜ_ᵢ(W(tᵢ₊₁)-W(tᵢ)) Yₜ_ⱼ(W(tⱼ₊₁)-W(tⱼ))] = E[Σᵢⁿ Y²ₜ_ᵢ(W(tᵢ₊₁)-W(tᵢ))² + E[ Σᵢ≠ⱼ 2Yₜ_ᵢ(W(tᵢ₊₁)-W(tᵢ)) Yₜ_ⱼ(W(tⱼ₊₁)-W(tⱼ))] For i≠j: WLOG i

Answer 7

In the context of a filtration F on a probability space, if something is F_tᵢ -measurable, then it is also F_tⱼ -measurable for tᵢ≤tⱼ The intuition behind this is that F_tᵢ represents all the information available up to time tᵢ , and if something is measurable with respect to this information, then it's also measurable with respect to all the information available at a later time tⱼ , where tⱼ≥tᵢ In other words, as time progresses, more information becomes available, so anything measurable at t_ᵢ will also be measurable at any later time tⱼ.

Answer 8

For Y ∈ H_T ˢᵗᵉᵖ let us define for t ∈ [0, T] I(Y, t) = ∫₀ᵗ Y (s) dW(s) := I(Y **1**_[0,t))

Answer 9

The process (I(Y, t))_t∈[0,T] (integral_[0,t] Y(s) dW(s)) is a continuous martingale with respect to F. proof: see notes (F-wiener process, filtration (W(t))_ t in [0,T]

Answer 10

missed out, seen similar in a proof? Y ∈ H_T ˢᵗᵉᵖ we have E[ sup_ₜ∈[₀,ₜ] |I(Y,t)|² ] ≤ 4E[ ∫₀T|Y (s)|² ds proof: This is a direct consequence of Doob’s inequality. Indeed, since the stochastic integral is a continuous martingale, we have by Theorem 2.1.12 and Lemma 3.2.4 E[ sup_ₜ∈[₀,ₜ] |I(Y,t)|² ] ≤ 4E[|I(Y,T)|²] = 4E[ ∫₀T|Y (s)|² ds

Answer 11

say we take a linear function on x in reals T(x) T:R to R if its linear form ax |T(x)|=|x| in our case isometry crucial for our stochastic integral defn

Answer 12

E[∫₀¹ Y (s) dW(s)] = E[Σᵢⁿ Yₜ_ᵢ(W(tᵢ₊₁)-W(tᵢ)] = Σᵢⁿ E[Yₜ_ᵢ(W(tᵢ₊₁)-W(tᵢ)] = Σᵢⁿ E[E[(Yₜ_ᵢ(W(tᵢ₊₁)-W(tᵢ)) | Fₜ_ᵢ]] = ΣᵢⁿE[Yₜ_ᵢ E[((W(tᵢ₊₁)-W(tᵢ)) | Fₜ_ᵢ]] = ΣᵢⁿE[Yₜ_ᵢ E[((W(tᵢ₊₁)-W(tᵢ))]] = 0 as E[((W(tᵢ₊₁)-W(tᵢ)) =0

Answer 13

If I take a simple stochastic process I can define the stochastic integral which is a random var with mean 0 finite second moment which is given by the integral of the square of the process

Answer 14

If Y ∈ Hˢᵗᵉᵖ : then Integral ∫₀¹Y (s) dW(s) ∈ L₂(Ω) norm ‖∫₀¹ Y(s) dW(s) ‖² L₂ = E[(∫₀¹Y (s) dW(s) )² ] = E[∫₀¹Y(s)² ds ] and E[∫₀¹Y (s) dW(s) ]=0

Answer 15

GIVEN WITHOUT THE PROOF H= {(Y(t))_t∈[0,1] |Y is adapted and E[∫₀¹Y(s)² ds <∞ } (1) take all stochastic processes Y that are adapted to a filtration ie that for all y ∈ [0,1] its F_t measurable ie. a RV Y(t) only depends on the values of brownian motion at time t ie the sigma-algebra generated by BM σ(W(s), s ≤ t) (2) satisfy that exp of square integrated = is finite we can think of it as the Riemann integral. Is it well defined, its a lebesgue integral but we consider it as one (dw). So these are not simple anymore H doesnt look like a step function

Answer 16

We say that a stochastic process Y : Ω × [0, T] → R is in the class H_T , if it is adapted to the filtration F and it satisfies E[∫₀T |Y (s)|²ds] < ∞.

Answer 17

Y_t conditionally on F_t is not a RV but behaves like a constant. This is what adaptedness is. So if we only consider those that are adapted we have this

Answer 18

E[|I(Y )|²] = E[ ∫₀T|Y (s)|² ds] because it shows that if we see H_Tˢᵗᵉᵖ as a subspace of L_2(Ω × R+), then the stochastic integral I : H_Tˢᵗᵉᵖ → L_2(Ω) is an isometry.

Answer 19

Let (Y(t))t∈[0,1] ∈H Then there exists a sequence Yₙ(t))t∈[0,1]∈ Hˢᵗᵉᵖ for n ∈N s.t. limit n to infinity E[∫₀¹ (Yₙ(s)-Y(s))² .ds]=0

Answer 20

Let Y∈H (That means Y is a: stochastic process which is adapted and doesnt depend on values of BM from the future, only past and it has a finite expectation of integral of square)

Answer 21

If Y ∈H_T then there exists Yₙ ∈ H_T ˢᵗᵉᵖ, for n ∈ N such that lim_{n→∞} E[ ∫₀T |Y (s) − Yₙ(s)|²ds] = 0

Answer 22

Let Y ∈ H and let (Yₙ)ₙ₌₁∞ ⊂ H_T ˢᵗᵉᵖ be the sequence from Lemma 3.2.9. Then there exists a random variable Z ∈ L_2(Ω). such that lim_{n to ∞} ‖I(Yn) − Z‖ = 0.

Answer 23

define for function integral of funct considering sequence of funct converging to this funct and our integral will also converge to integral of funct we don't rigourously prove the lemma For the family of functions we consider for a particular omega

Answer 24

the limit of n,m to infinity of |a_n-a_m| tends to 0 if for all epsilon there exists N s.t for all n,m >= N |a_n-a_m|

Answer 25

Lim_{n,m tends to infinity} of ‖X_n-X_m‖_L₂ = Lim_{n,m tends to infinity} of (E[(X_n-X_m)²])¹/² =0 Thus Lim_{n,m tends to infinity} of (E[(X_n-X_m)²]) =0 thus there exists a RV Z In L₂(Ω) s.t ‖Z - X_n‖_L₂ s.t Lim n to infinity [ E[ (X_n-Z)²] =0

Answer 26

Suppose we have a process in H: Y ∈ H Then by lemma, there exists Yₙ ∈ H_T ˢᵗᵉᵖ (simple) s.t lim_{n→∞} E[ ∫₀¹|Y (s) − Yₙ(s)|²ds] = 0 (this also implies its cauchy...) I look at ∫₀¹ Yₙ(s)dW(s) and we show it is a cauchy sequence... Thus there exists Z in L₂(Ω) s.t E[ (∫₀¹ Yₙ(s)dW(s) -Z)²] converges to 0 as n tends to infinity and this limit will be our value we use

Answer 27

∫₀¹ Yₙ(s)dW(s) and we show it is a cauchy sequence ∫₀¹ Yₙ(s)dW(s) - ∫₀¹ Yₘ(s)dW(s) (in L_2 norm??) E[(∫₀¹ Yₙ(s)dW(s) - ∫₀¹ Yₘ(s)dW(s))²] (stochastic integral is linear) = E[(∫₀¹ (Yₙ(s)-Yₘ(s))dW(s))²] it is isometry, expectation of the square ... =E[(∫₀¹ (Yₙ(s)-Yₘ(s))²ds)] by triangle inequality ≤ 4E[∫₀¹ (Yₙ(s)-Y(s))²ds] + 4E[∫₀¹ (Y(s)-Yₘ(s))²ds] both tend to 0 as n,m tends to infinity means the sequence of stochastic integrals is cauchy in L2 which is complete thus it converges Limit is a RV Z

Answer 28

(Ω,F,P) filtration F = {Fₜ}_ ₜ in [0,1] BM (W(t))_t in [0,1], F-wiener process our goal was to construct stochastic integral ∫₀T Y(s) dW(s) we started looking at simple processes STEP 1 (Y(t))_{t∈ [0,1]} ∈ H_T ˢᵗᵉᵖ Y(t) =Σᵢ₌₁ⁿ Yₜ_ᵢ **1**_[tᵢ,tᵢ₊₁] (t) THUS ∫₀T Y(s) dW(s):= Σᵢ₌₁ⁿ Yₜ_ᵢ (W(tᵢ₊₁)-W(tᵢ,)) ∈ L₂(Ω) which is a RV in L_2 which we saw had some properties STEP 2 This was for a simple process, then we extended this to approximate Consider class H_T , if it is adapted to the filtration F and it satisfies E[∫₀T |Y (s)|²ds] < ∞. If we take a Y in H_T then there exists a sequence of simple processes Yₙ ∈ H_T ˢᵗᵉᵖ (simple) s.t lim_{n→∞} E[ ∫₀¹|Y (s) − Yₙ(s)|²ds] = 0 we used this to construct the stochastic integral for Y By seeing the integrals of sequence form a cauchy sequence (RVs) converging in L_2

Answer 29

**linear operator**, if Y₁, Y₂ ∈ Hₜ and a₁, a₂ ∈ R, then we have that I(a₁Y₁ + a₂Y₂) = a₁I(Y₁) + a₂I(Y₂) **Exp = 0** E[∫₀T Y(s) dW(s)]=0 **var =** E[ (∫₀T Y(s) dW(s))² ] = E[ ∫₀T Y(s)² d(s)] we will see (∫₀ᵗ Y(s) dW(s))_(t in [0,T]) is a martingale

Answer 30

**linear operator**, if Y₁, Y₂ ∈ Hₜ and a₁, a₂ ∈ R, then we have that I(a₁Y₁ + a₂Y₂) = a₁I(Y₁) + a₂I(Y₂) **Exp = 0** E[∫₀T Y(s) dW(s)]=0 **var =** E[ (∫₀T Y(s) dW(s))² ] = E[ ∫₀T Y(s)² d(s)]

Answer 31

(Yₙ(t))₎ₜ ᵢₙ [₀,₁] → I(Y)= ∫₀TY(s) dW(s) (stochastic process) → (random variable) I want to consider now (for little t) ∫₀ᵗ Y(s) dW(s) for each t i have a RV so I consider a family of RVs

Answer 32

a family of RVs which is a stochastic process

Answer 33

TRUE BUT if i take a stochastic process (Yₙ(r))₎_r ᵢₙ [₀,T] ∈ H_T then I have stochastic process (**1**_[0,t] Y(r))_ r ᵢₙ [₀,T] ∈ H_T Note: This new stochastic process is equal to the stochastic process Y when r is in [0,t] if r>t then its 0 = {Y(r) if r <=t {0 r>t So trajectories in this case look like: From a simple process which is (continuous on the left) becomes the same but only for time <=t * because it is ∈ H_T we can integrate and value ∫₀ᵗ Y(s) dW(s) = ∫₀T **1**_[0,t] (s) Y(s) dW(s)

Answer 34

(1) This stochastic process is F adapted: So for any time t the stochastic integral, ∫₀ᵗ Y(s) dW(s) is F_t measurable (2) Continuous P(t → ∫₀ᵗ Y(s) dW(s) is continuous)= 1 So always cont funct for each t its a rv for each omega its a function of time and it turns out this is continuous

Answer 35

Consider (Ω,F,P) filtration F = {Fₜ}_ ₜ in [0,T] then stochastic process (M(t))ₜ in [0,T] is a martingale w.r.t filtration F if (1)M(t) is F_t measurable for all t in [0,T] (ADAPTED) (2) INTEGRABLE in(moment 1?), finite E[|M(t)|] < infinity (3) E[M(t)|F_S]=M(s)?? (for any time t, the conditional expectation of the next value given all past values up to time t is equal to the value at time t) E[Xₜ₊₁|X₁,X₂...Xₜ]=Xₜ third property is important:

Answer 36

Considering conditional expectations: projection geometric interpretation projection of M(t) in the space of F_S measurable Random variables F_S measurable variables are random variables which at time s aren't random variables anymore This condition tells us that if t is in the future and s is the present the projection of M(t) to the present (projection is the best possible approximation we can get), the approximation of RV in the future is best estimate is the value is the value in the present M(s). Iterating this shows it would be constant?

Answer 37

Let Mₜ be a martingale with respect to a filtration {Fₜ}ₜ≥₀ let S be a stopping time. Conditional exp: E[M(t)∣Fₛ ]=M(S) This property states that the conditional expectation of M(t) given the information up to time S (i.e., the sigma-algebra F_S ) is equal to the value of the martingale M at time S. This property essentially means that at any stopping time S, the current value of the martingale M provides the best estimate for its future value. In other words, knowing the current value of the martingale M provides no additional information about its future evolution beyond what is already known up to time S

Answer 38

E.g if X is the price of a stock I claim that the price of the stock is adapted to filtration F_t measurable then it means if t is the present then the value is not random (i know its current price) tomorrows price is random, but waiting for a day its not random anymore, This is what adapted means. If i make a conjecture about stock tomorrow, best option is to go for the price of stock today

Answer 39

E[E[M(t)|F_S]]=E[M(s)] implies E[M(t)]=E[M(s)] constant as i can choose s=0 as cond exp becomes exp

Answer 40

TRUE (∫₀ᵗ Y(s) dW(s))_ t in [0,T] is an F-martingale

Answer 41

TRUE (∫₀ᵗ Y(s) dW(s))_ t in [0,T] is an F-martingale using Y=1 stochastic integral becomes ∫₀ᵗ dW(s) =∫₀T **1**_[0,t] (s)dW(s)) W(t) - W(0) = W(t)

Answer 42

We only consider Y in H_Tˢᵗᵉᵖ for now, its true for H_T: Checking the three properties of a martingale We know by defn: Y(r)= Σᵢ₌₁ⁿ Yₜ_ᵢ **1**_[tᵢ,tᵢ₊₁] (r) where Yₜ_ᵢare Fₜ_ᵢ-measurable (*) (1) Stochastic integral ∫₀ᵗ Y(s) dW(s)) =∫₀T **1**_[0,t] (r) Y(r) dW(s)) multiplying (*) by **1**_[0,t](r) gives simple process, which we know how to integrate Y(t)**1**_[0,t](r) = Σᵢ₌₁ⁿ Yₜ_ᵢ **1**_[tᵢ^t,tᵢ₊₁^t] (r) ∫₀t Y(s) dW(s):= ∫₀T Y(t)**1**_[0,t](r) . dW(s) = Σᵢ₌₁ⁿ Yₜ_ᵢ (W(tᵢ₊₁^t)-W(tᵢ^t)) Which is a stochastic process dep on t We want to show this is a martingale: the finite sum of martingales is a martingale so we show: Yₜ_ᵢ (W(tᵢ₊₁^t)-W(tᵢ^t)) (1) is F_t measurable for all t in [0,T] (ADAPTED): W is a BM adapted to filtration Yₜ_ᵢ is Fₜ_ᵢ-measurable, I want it to be F_t measurable for ttᵢ: Yₜ_ᵢ is Fₜ_ᵢ-measurable but now Fₜ_ᵢ⊂Fₜ W(tᵢ^t) will be Fₜᵢ^ₜ-measurable W(tᵢ₊₁^t) will be Fₜᵢ₊₁^ₜ-measurable Fₜᵢ^ₜ⊂Fₜ Fₜᵢ₊₁^ₜ ⊂Fₜ So all terms are Fₜ-measurable( subtract and multiply still Fₜ-measurable) (2) E[| Yₜ_ᵢ (W(tᵢ₊₁^t)-W(tᵢ^t)) |] < infinity? E[| Yₜ_ᵢ (W(tᵢ₊₁^t)-W(tᵢ^t)) |] Yₜ_ᵢ uniformly bounded, finite in L_infinity by defn so we can take it out ≤ ‖Yₜ_ᵢ‖L∞ E[| (W(tᵢ₊₁^t)-W(tᵢ^t)) |] by holders inequality(exercise for norm p

Answer 43

Y(r)= Σᵢ₌₁ⁿ Yₜ_ᵢ **1**_[tᵢ,tᵢ₊₁] (r) (*) multiplying (*) by **1**_[0,t](r) gives Y(t)**1**_[0,t](r) = Σᵢ₌₁ⁿ Yₜ_ᵢ **1**_[tᵢ,tᵢ₊₁] (t)**1**_[0,t](r) as the product of two characteristic functions is the characteristic function of the intersection(min ^): Y(t)**1**_[0,t](r) = Σᵢ₌₁ⁿ Yₜ_ᵢ **1**_[tᵢ^t,tᵢ₊₁^t] (r) which is a simple process

Answer 44

sum subtract composition multiplication still measurable

Answer 45

Let Y, Z ∈ H_T and a, b ∈ R i) almost surely . for all t ∈ [0, T] (ii) For all t ∈ [0, T] E[] =0 (iii) (Itô’s isometry) For all t ∈ [0, T] E[]=E[] (iv) (I(Y, t))t≥0 is a martingale with respect to F. (v) Doob’s inequality holds,

Answer 46

Define m(t)= Yₜ_ᵢ (W(tᵢ₊₁^t)-W(tᵢ^t) CASE 1: s

Answer 47

is also a martingale t' fixed t is random

Answer 48

Y ∈ H_T, then E[sup_{T ∈ T} |∫₀ᵗ Y(s).dW(s)|²] ≤4E[∫₀T Y(s)².ds] (without the supremum and with terminal time capital T and without 4 we would have the isometry, but instead we take the supremum of all times and we have this bound)

Answer 49

True we saw only in class H-step

Answer 50

random variable which has 0 mean and variance = square of the expectation of the integral

Answer 51

filtration **F**={Ƒₜ}ₜ∈[₀,T] generated by BM Ƒₜ = σ(W⁻¹(A), A∈B(R)) inverse image of Borel sets If a random variable is measurable w.r.t. F_t is means that if we know all the values of BM W(s) for all s up to time t then in principle we know values so no longer random var

Answer 52

If M(t) is an **F**-Martingale continuous s.t E|M(t)|²] <∞ then there exists a unique process in the class Y ∈ H_T s,t, M(t)= M(s) + ∫₀ᵗ Y(s) dW(s))

Answer 53

every stochastic integral is a martingale If we have a martingale which has the property which is continuous in time and square integrable then we can define integral Y s.t M(t) is the stochastic integral of Y wrt. W

Answer 54

Let G be the augmentation of the natural filtration of W. Let (M(t))_t≤T be a continuous, martingale with respect to G such that M(T) ∈ L_2(Ω). Then, there exists a unique (Y (t))_t≤T ∈ H_T , adapted to G, s.t for each t ∈ [0, T], with probability one M(t)= M₀ + ∫₀ᵗ Y(s) dW(s))

Answer 55

If we take a RV Z which is integrable and any filtration Fₜ. (1) adapted to filtration: M(t) is Fₜ-measurable E[Z|Fₜ] by defn as conditional on Fₜ (2)integrable: E[|E[Z|Fₜ]|] E[E[|Z||Fₜ])] =E[|Z|] <∞ which we know is finite (you should know this!!!: convex functions and conditional expectations |E[Z|Fₜ]|≤ E[|z||Fₜ]) We also know E[E[X|G]]=E[X] (3)

Answer 56

conditional expectation def: cond expectation is a g-measurable function and E[**1**_AX]=E[**1**_A E[X|G]] for all A in G so we choose an A s.t this works: A=Ω

Answer 57

we extended to class H now I want to extend to a larger class S

Answer 58

A map 𝜏 : Ω → [0, +∞] is called **stopping time** **with respect to a given filtration F** if for all t ∈ [0, T] we have {𝜏 > t} ∈ Fₜ * 𝜏 is a RV which takes non-neg values *it has the property that for event {𝜏 > t}= {ω∈Ω: 𝜏(ω)>t}∈ Fₜ for all t ≥0 so RVs 𝜏 for which map as above for which the event that {𝜏>t} is in Fₜ for all t ≥0 called stopping time related to gambling: strategy to get out when do we stop!

Answer 59

If F_tᵢ is flow of information F_s⊆ F_t ----s----------t-----> time line: increasing flow of information recall that a RV that is F_t at time t is not longer random event {𝜏 > t}= {ω∈Ω: 𝜏(ω)>t}∈ Fₜ for all t ≥0 means that * event is Fₜ measurable *and at time t we know whether 𝜏 > t or not as event ∈ Fₜ *say tau describes a random time an event happens: we want to know at time t if the event has happened and event {𝜏 > t} means it has not happened yet up to time t! *describes mathematically only by looking at information from the past I want to determine whether this event corresponding to time tau has happened or not

Answer 60

F_tᵢ -measurable, then it is also F_tⱼ -measurable for tᵢ≤tⱼ When we say X is F_tᵢ -measurable, it means that X is determined by the information available at time tᵢ . So, X is independent of any information that becomes available after time tᵢ . Now, if tᵢ ≤t_jt i ≤t_j , then the sigma-algebra F_tᵢ is a subset of F_t_j , meaning that the information available at time tᵢ is also available at time t_j . Therefore, any random variable that is F_tᵢ -measurable is also F_t_j -measurable for t_i≤t_j In other words, if X depends only on information available up to time tᵢ , then it automatically depends on information available up to time t_j whenever tᵢ ≤t_j .

Answer 61

diagram: y=b level stochastic process X path may cross b map t to X(t) as a function of time (0, ifninity) I look at set for which X(t)>b {t ∈ [0, T] : X(t) ≥ b} take infimum: first time that stochastic process hits the level b is τ_b(w_1) τ_b := inf{t ∈ [0, T] : X(t) ≥ b},..... Now this is random as if I have a different path the w will be different( τ_b(w_2) I CLAIM ITS A STOPPING TIME:(i suggest you check on your own, dont want to spend too much time) in general if you have ADAPTED PROCESSES and you have STOPPING TIME of this form (first time process hits a level) underlying process is adapted and so its a stopping time: We need to show: for all t ∈ [0, T] that {𝜏 > t} ∈ Fₜ: (the largest value X can take is something strictly less than b on interval [0,t] for t less than 𝜏) {𝜏 > t}={sup_{s≤t} {X(s)} t} ∈ F_t. Hence τ_b is indeed an F-stopping time. An example of why in gambling you might tell yourself the first time you hit £100 I get out of here, the most probable is that you lose all your money (dep on how much you start with) tau can take value +infinity, you have to wait that long?

Answer 62

MINIMUM OF TWO STOPPING TIMES IS A STOPPING TIME not discussed: Let 𝜏₁ and 𝜏₂ be stopping times WRT filtration F then {𝜏₁ >t} {𝜏₂>t} ∈ F_t ∀t ∈ [0, T]. {𝜏₁ ∧ 𝜏₂ > t} = {𝜏₁ > t and 𝜏₂ > t} = {𝜏₁ > t} ∩ {𝜏₂ > t} which is in F_t , because σ-algebras are closed under (countable and thus also finite) intersection. We have shown that {τ1 ∧ τ2 > t} ∈ F_t for all t ∈ [0, T]. Therefore τ1 ∧ τ2 is indeed an F-stopping time

Answer 63

SPECIFICALLY GONE THROUGH????? **1**_[0,𝜏(w)] (t) is a stochastic process (depends on w and on t) looks like: 1 for t less than or equal to 𝜏(w), 0 for t>𝜏(w) I claim its adapted and thus F_t measurable for all t>=0 fix t>=0: **1**_[0,𝜏(w)] (t)= {1 t ≤𝜏(w) {0 t>𝜏 so to check if F_t measurable suffices to check: that the inverse image of the set (-infinity,a) as this generates a Borel sigma algebra. checking {**1**_[0,𝜏(w)] (T) >a} in F_t through the inverse map (**1**_[0,𝜏])⁻¹ (-∞,a) can only take specific values (**1**_[0,𝜏] (t))⁻¹(B) = {ω ∈ Ω : 1[0,τ(ω))(t) ∈ B} = {Ω if 0, 1 ∈ B {∅ if 0, 1 ∈/ B {{t < τ} if 1 ∈ B, 0 ∈/ B {{t < τ}^c if 0 ∈ B, 1 ∈/ B Since Ft is a σ-algebra: Ω, ∅ ∈ Ft . Furthermore since τ is a stopping time: {τ > t} ∈ Ft . So since σ-algebras are closed under complementation, we also have that {τ > t}^c ∈ F_t. Therefore **1**_[0,𝜏] (t))⁻¹(B) ∈ Ft ∀B ∈ B([0, ∞]) hence adapted? F_t measurable some more info in exercise notes

Answer 64

Solution: See Exercise 3.4.2 and Exercise 3.4.3. missed? The stopping time from the exercise above it is call the first exit time of X(t) from (a, b).

Answer 65

We know class Hₜˢᵗᵉᵖ⊆ H_T ⊆S_T We denote by S_T the class of all F-adapted processes Y : Ω × [0, T] → R such that T ≥ 0, almost surely ∫_[0,T] |Y (s)|^2 ds < ∞. S_T= {(Y)_{t ∈[0,T]} ¦ adapted and P(∫_[0,T] |Y (s)|^2 ds < ∞)=1 } This is a larger class than H_T, instead of the probability we want expectation of integrand to be finite We want to extend the stochastic integral to integrands from the class S_T .

Answer 66

taking process in H_T means expectation inf integrand finite, meaning almost surely ie probabilit finite =1 thus it is also a process in S_T

Answer 67

FALSE take any RV X>0 E[X]< ∞ implies P(X<∞)=1 If the probability that X is not finite>0 means that RV attains infinity with some positive probability which means the expectation has to be infinite so if a expectation is finite then RV Is finite almost surely

Answer 68

Let Y ∈ H_T and let 𝜏 be a stopping time bounded by T. Then, almost surely I(Y, τ ) = I(**1**[0,𝜏]Y ) = I(**1**_[0,𝜏)Y ) ----------------------------------------------------

Answer 69

Y ∈ S_T and n ∈N definite 𝜏_n = inf {t≥0 ¦ ∫_ [0,t] (Y(s))^2 .ds ≥ n} infimum of times of which the integral crosses the level n starts from 0, non negtive, increasing process (sketch not jagged strictly increasing) Y is adapted thus integral is adapted thus this 𝜏_n is a stopping time thus it is adapted I claim Y_n belongs to H: so we started with something in s and truncated by using indicator function Y_n ∈ H_T we have to check 2 things 1)adapted checked the first time integral hits 𝜏_n we know Y by def is adapted and therefore 𝜏_n is adapted and so Y_n is adapted 2) expectation of integral of Y_n ^2 is finite which we check E[ ∫₀T|Y (s)|² ds =E [ ∫₀T **1**_[0,𝜏_n ](S)Y (s)|² ] =E [ ∫_[0, t^𝜏_n ] Y (s)².dS ] =E[n] less than or equal than n so finite So we have shown belongs in H and we can take the stochastic integral we have that ||cinverges to 0 in probability more details in notes I won't insist on that not a martingale if in S is a local martingale

Answer 70

(Y ∈ H_T means Y is adapted and that integration E[ ∫₀T|Y (s)|² ds is finite < infinity:) (𝜏 stopping time means {t<𝜏} is in F_t we checked already **1** is adapted new process **1**_[0,𝜏)(s)Y(s) Y is adapted as by defn thus **1**_[0,𝜏)(s)Y(s) is adapted and **1**_[0,𝜏)(s)Y(s) is F_s measurable for all s>=0 E[ ∫₀T|**1**_[0,𝜏)(s)Y(s) |² ds= E[ ∫_{0, 𝜏^T} (Y(s))² ds note upper limit with min{t,𝜏} is random, 𝜏 is random, cand bring in Expectation through to integral like if it was deterministic time , but as limit is random we can't but integrating something positive ≤ E[ ∫_{0, T} (Y(s))² ds

Answer 71

true, shown prev Since **1**_[0,𝜏] Y ∈ H_T it means i can take the STOCHASTIC INTEGRAL OF IT So set M(t) = ∫_[0,t]Y(s) dW(s) Then we have M(𝜏) = ∫_[0,T]**1**_[0,𝜏](s) Y(s) dW(s) DANGER this is not obvious: true for all t but as upper limit random as 𝜏 random we wouldnt expect this also this equals ∫_[0,𝜏] Y(s) dW(s) (not in notes)

Answer 72

Answer 73

For Y ∈ HT we have E " sup t∈[0,T] Z t 0 Y (s) dW(s) # ≤ 3E Z T 0 |Y (s)| 2 ds1/2 .

Answer 74

A process X : Ω × [0, T] → R is a local martingale if there there exists a sequence of stopping times (τ_n)∞ n=1 satisfying (i) for any n ∈ N, P(τn ≤ τn+1) = 1 (ii) for all ω ∈ Ω, τ_n(ω) = T for all n large enough. (iii) the for each n ∈ N, the process (X(t ∧ τ_n))t∈[0,T] is a martingale. Such a sequence of stopping times is called localising sequence for (X(t))t∈[0,T] . For Y ∈ ST it easy easy to see that the sequence of stopping times defined in (3.10) is a localising sequence for (I(Y, t))t∈[0,T] .

Answer 75

only briefly discussed We want to show that for any ε > 0, we have limit as n tends to infinity of P( sup_t≤T | ∫_[0,t] Y_n(s) dW(s)- ∫_[0,t] Y(s)dW(s) |≥ε ) =0 For this it suffices to show that for any ε > 0 and for any δ > 0, lim sup_n to infinity P( sup_{t≤T} | ∫_[0,t] Y_n(s) dW(s)- ∫_[0,t] Y(s)dW(s) |≥ε ) ≤δ. carries on to apply.... notes

Answer 76

THM 3.2.15contained within H_t H_T:all square integrable adapted processes y∈Hₜˢᵗᵉᵖ has property that: ( ∫₀ᵗ Y(s)dW(s))_t∈[0,t] **is a random variable** with 1) E[ ∫₀ᵗ Y(s)dW(s)]=0 2) E[ ∫₀ᵗ Y(s)dW(s)]^2 = E( ∫₀ᵗ (Y(s))^2 .ds < infinity (finite defn for y in H_T) also it's linear: the integral of elements from H_T can be split up as a sum of integrals 3) the process is a martingale also extended this to a class S are adapted processes s.t the probability that the RV integral_[0,T] is finite with probability one, local martingale Setting m(t)= integral_[0,t] Y(s)dW(s)

Answer 77

E[sup_{t<= T} |( ∫₀T Y(s) dW(s)|] <= cE[ ∫₀T Y(s)^2 ds] (if Y wasn't in H_T would be infinite) This implies that if I had Yⁿ converges to y in the sense E [ ∫₀T (Yⁿ(s)-Y(s))^2 ds] converges to 0 as n tends to infinity THEN stochastic integrals converge: ∫₀ᵗ Yⁿ(s)dW(s) converges to ∫₀ᵗ Y(s)dW(s) in the sense E [sup_(t ≤T) |∫₀ᵗ Y(s) dW(s) - ∫₀ᵗ Yⁿ(s)dW(s)|] converges to 0 as n tends to infinity

Answer 78

if y in S_T P(sup_{t ≤T}| ∫₀ᵗ Y(s)dW(s) |≥ ε)≤P( ∫₀T (Y(s))^2 ds ≥δ) + δ/ε² relates the stochastic integral and integral of the square and so if Yⁿ converges to y in the sense ∫₀T (Yⁿ(s)-Y(s))ds converges to 0 in probability as n tends to infinity ∫₀T (Yⁿ(s) dW(s) converges ∫₀T Y(s))dW(s) in the sense sup_{t ≤T}| ∫₀T (Yⁿ(s)dW(s)- ∫₀T Y(s))dW(s)| tends to 0 in probability

Answer 79

If Y in H_T then the process m(t)= ∫₀ᵗ Y(s).dW(s) is an F martingale Y in S_T (m(t))_{t in [0,T]} is not a martingale in general If E[( ∫₀T (Y(s))^2.ds)^0.5]< infinity (finite) then (m(t))_{t in [0,T]} is a martingale In general Y in S_T is a local martingale (m(t))_{t in [0,T]} : there exists a sequence of stopping times we used Yⁿ(t)= **1**_[0,𝜏_n](t) Y(t) 𝜏_n= inf{ ∫₀T ((Y(s))^2 . ds >n} first time the stochastic process crosses the level n gives Yⁿ in H_T also Yⁿ(t)=Y(t) for t in [0,𝜏_n] so m(t ^ 𝜏_n) = ∫_[0,t ^ 𝜏_n]Y(s) dW(s) = ∫_[0,t ^ 𝜏_n]Yⁿ(s) dW(s) but as this is the stochastic integral of somthin tht belongs in H_t then it is a martingale for each n thus m(t) is a local martingale

Answer 80

If Y is in S ONLY NOT IN H_T: not true If Y in H_T :We know that the expectation of the absolute value is finite E[| ∫₀ᵗY(s)|] < infinity so given Y in H_T we use this condition

Answer 81

d/dt(g(f(t))) = g'(f(t))f'(t) IF f has differential f(t) =f(0) +∫_[0,t] a(S) dS g(f(t)) = g(f(0)) + ∫_[0,t] g'(f(s))s(s) ds not involving the derivative if g df(g=f'(g(t)?? .dt

Answer 82

iff g'(t) =a(t) notation dg(t) =a(t).dt dg/dt=a

Answer 83

probability space (Ω,F,P) filtration F={F_t}_{t in [0,T]} BM wrt this filtration F-Wiener process (W(t))_{t in [0,T]} Stochastic process (X(t))_{t in [0,T]} is an **ito process** if there exists processes (A(t))_{t in [0,T]} in D_T and (B(t))_{t in [0,T]} in S_T (stochastic integrals s.t the integral is finite) X(t) =X(0)+ ∫_[0,t] A(s).ds + ∫_[0,t] B(s) dW(s) when B =0 the function is differentiable wrt t

Answer 84

X has Ito differential or stochastic differential dX(t) = A(t)dt + B(t) dW(t) just notation short hand for the equation

Answer 85

Let (X(t))_{t in [0,T]} be an Itô process A in D_T B in S_T X(t) =X(0)+ ∫_[0,t] A(s).ds + ∫_[0,t] B(s) dW(s) Let f be in c^2 (twice differentiable) Then composition (f(x(t)))_{t in [0,T]} is also an Ito process which satisfies f(x(t)) =f(x(0))+ ∫_[0,t] f'(X(s))A(s).ds + ∫_[0,t] f'(X(s)) B(s) dW(s) + 0.5 ∫_[0,t] f''(X(s))(b(s))^2 ds includes the Ito correction term we can group together and write in terms of stochastic differentials (also if B =0 usual chain rule)

Answer 86

Integral X(t) =X(0)+ ∫_[0,t] A(s).ds + ∫_[0,t] B(s) dW(s) can write as: dX=A(t)dt + B(t)dw(t) then f(x(t)) =f(x(0))+ ∫_[0,t] f'(X(s))A(s).ds + ∫_[0,t] f'(X(s)) B(s) dW(s) + 0.5 ∫_[0,t] f''(X(s))(b(s))^2 ds du(x(t)) = u'(x(t))A(t) + 0.5u"(X(t))(B(t))^2 .dt + u'(X(t))B(t).dW(t) if we group we have a dt term and a dw term showing its an ito process

Answer 87

H_T martingales S_T local martingales

Answer 88

S_T contains stochastic processes that are F-adapted for this class we have the property that elements have P( integral_[0,T] X(s)^2 .ds

Answer 89

there exists a process a in class D_T and process B in class S_T s.t X(t) =X(0)+ ∫_[0,t] A(s).ds + ∫_[0,t] B(s) dW(s) when B =0 the function is differentiable wrt t DETERMINISTIC INTEGRAL +STOCHASTIC INTEGRAL

Answer 90

consider X(t) =X(0)+ ∫_[0,t] A(s).ds + ∫_[0,t] B(s) dW(s) dX=A(t)dt + B(t)dw(t) write the table (dt)^2 =0 dW(t).dt =0 (dW(t))^2 = dt whenever we see these we write YOU ONLY NEED TO REMEMBER THIS **dX(t)=u'X(t)d(X(t) + 0.5u"(d(X(t))^2** links to taylors expansion first deriv dX +0.5 second deriv (dX)^2

Answer 91

these will be VERY VERY important for the exam please also check the coursework and FEEDBACK and check the solutions, this is important for the exam too <3 already I gave you three exercises, cannot complain

Answer 92

recovering the formula: dX(t)=u'(X(t))**d(X(t)** + 0.5u"(d(X(t))^2 =u'(X(t)**A(t)dt+ B(t)dW(t)** +0.5u"(X(t))(**A(t)dt+ B(t)dW(t)**)^2 =u'(X(t)**A(t)dt+ B(t)dW(t)** +0.5u"(X(t))(**(A(t)dt)^2+ (B(t)dW(t))^2 +2A(t)B(t)dt dW(t)**) using the table =u'(X(t)A(t)dt+ u'(X(t)) B(t)dW(t)+0.5u"(X(t))((B(t))^2 dt as we needed

Answer 93

product is also differentiable differential form (fg)' = f'g+g'f f(t)g(t)=f(0)g(0)+ ∫_{0,t}f'(s)g(s).ds +∫_{0,t}f(s)g'(s).ds integration by parts and product rule

Answer 94

is an ito process and we have the equality d(X(t)Y(t))= X(t)dY(t) + Y(t)dX(t) + dX(t)dY(t) convenient to write in terms of differentials consider the differential for X Aₓ(t) dt+Bₓ(t)dW(t) differential for Y Aᵧ(t)dt+Bᵧ(t). dW(t) replacing d(X(t)Y(t))= X(t)dY(t) + Y(t)**Aₓ(t) dt+Bₓ(t)dW(t)** + dX(t)**Aᵧ(t)dt+Bᵧ(t). dW(t)**+ (**Aᵧ(t)dt+Bᵧ(t). dW(t)**)(**Aₓ(t) dt+Bₓ(t)dW(t)**) write the table (dt)^2 =0 dW(t).dt =0 (dW(t))^2 = dt =(**BₓBᵧ + (X(t)Aᵧ(t)** + Y(t)Aₓ(t) )dt+ ( X(t)Bᵧ(t) +Y(t)Bₓ(t)).dW(t) first part integration by parts

Answer 95

1) this equality means that I want to show that Z is an ito process and it has a stochastic differential recalling that ito process has the form dX(t)=A(t)dt +B(t)dW(t) dZ(t) = σZₜ dW(t), means A=0 B=σZₜ Consider process X(t)= **σW(t) -0.5σ² t** (exp argument) Is X an ito process? Can I write of the form above or equivalently of the form: X(t)= X(0) +∫₀ᵗ A(s).ds + ∫₀ᵗ B(S) dW(s) =0 +∫₀ᵗ -0.5σ² ds + ∫₀ᵗ σ dW(s) (YESSS) so X is an ito process writing as: dX(t)=(-0.5σ²) dt +σdW(t) Then using itos formula we have u(X(t)): exp(X(t)) Z(t)=u(X(t)) thus dZ= du(X(t) = u'(X(t))dX(t) + 0.5u"(X(t)) (dX(t))² substituting dZ= u'(X(t))**(-0.5σ²) dt +σdW(t)** + 0.5u"(X(t)) (**(-0.5σ²) dt +σdW(t)**)² from the table =-0.5σ²u'(X(t))dt +σu'(X(t))dW(t) + 0.5u"(X(t))σ² dt as u'=u as exponential = σu'(X(t))dW(t) =σZ(t) dW(t) we want to check: IC Z(0)=1 means Z(0)= exp(σW(0)-0.5σ²(0))=exp(0)=1 as required

Answer 96

2)not gone through? seems standard enough look through notes next one similar? important next one

Answer 97

dZ= du(X(t) = u'(X(t))dX(t) + 0.5u"(X(t)) (dX(t))² replacing from the table

Answer 98

firstly Z(0)=1 Sufficient to show its an ito process and then that the integral dW(t) part is in H_T to show it is an F martingale We use ito's formula: eᵗ/²cos(W(t)) cos part is a smooth function of brownian motion and thus expect to be an ito process and eᵗ/² expect to be ito process, thus product will also be. ITO PROCESSES ARE THOSE WHICH ARE DETERMINISTIC INTEGRAL + STOCHASTIC INTEGRAL we also know that if the stochastic integral is in the class H_T it is a martingale dX(t)=A(t)dt +B(t)dW(t) X(t)= X(0) +∫₀ᵗ A(s).ds + ∫₀ᵗ B(S) dW(s) STEP show ito processes Y(t)=eᵗ/² X(t)= cos(W(t)) *dY(t)= 0.5eᵗ/² dt (we use **itos formula** for dX: u(x)=cosx X(t)=W(t) dX(t)=u'X(t)d(X(t) + 0.5u"d(X(t))^2))) *dX(t)= -sin(W(t)) dW(t) -0.5cos(W(t)) **(dW(t))²** = -sin(W(t)) dW(t) - 0.5cos(W(t)) dt STEP with the differentials of X and Y we find using **itos product rule** d(XY)= X(t)dY(t) + Y(t)dX(t) +dX(t)dY(t) = 0.5 cos(W(t))eᵗ/² dt+ eᵗ/²[-sin(W(t)) dW(t) - 0.5cos(W(t)) dt] +0.5eᵗ/² dt [-sin(W(t)) dW(t) - 0.5cos(W(t)) dt] = 0.5 cos(W(t))eᵗ/² dt - eᵗ/²sin(W(t)) dW(t) - eᵗ/²0.5cos(W(t)) dt using the table =- eᵗ/²sin(W(t)) dW(t) so dZ(t)= - eᵗ/²sin(W(t)) dW(t) so Z(t)= **1**-∫₀ᵗeᵗ/²sin(W(t)) dW(t) so we have show Z(t)= constant +stochastic integral We know the stochastic integral: if the element inside the integral is inside H_T it is a martingale STEP show this 1)process (X(t))_{t in [0,T]} in H_T if by defn F adapted, F_t measurable for all t : By defn W(t) is F_t measurable for all t in [0,T] since W is adapted map composed with smooth continuous function is also F_t measurable -eᵗ/²sin(W(t)) f(x)=-eᵗ/²sinx f:R to R continuous and therefore f(W(t))borel measurable also 2) check exp is finite E [∫_[0,T] X(s)² .ds ] = E[∫_[0,T] eᵗ/²sin(W(t))² .dt] ≤E[∫_[0,T] exp(T/2).1dt deterministic integral =exp(T/2)T < infinity so -0.5eᵗ/²sin(W(t)) dW(t) )_{t in [0,T]} in H_T so -∫₀ᵗeᵗ/²sin(W(t)) dW(t) is an F-martingale so Z(t)= 1 -∫₀ᵗeᵗ/²sin(W(t)) dW(t) is also an F-martingale THIS WAS A SPECIFIC EXERCISE know very well!!

Answer 99

The higher dimension ito's formula: I just want you to know the formula itself DEFINITELY YOU SHOULD KNOW ITOS FORMULA SHOULD KNOW PRODUCTS RULE HOW TO USE AS IN PREV EXERCISE you should probably know this higher dim, you don't need to remember this, it easy to remember, I will not ask for it but its nice to remember as makes it easier to derive the product one

Answer 100

dX=A(t)dt + B(t)dw(t) W(t) Ω to R X(t) : Ω to R

Answer 101

want to consider now stochastic process W(t): Ω to R^{d_0} X(t) Ω to R^{d} d_0 dimensional wiener process: [missing line defining the process W(t) = (W₁(t), W₂(t), ..., W_{d₀}(t))???] where each component Wᵢ(t) is an R-valued F wiener process, W_1,...W_{d_0 }independent independent brownian motions in vector, give D_0 dimensional random variable as time evolves stochastic process with values in R^{d_0}

Answer 102

want to consider now stochastic process W(t): Ω to R^{d_0} X(t) Ω to R^{d} Process X(t) =(X_1(t),...,X_D(t)) ^T (in general D_0 not equal to D} W=(w_1,...,W_d_0) each component is adapted and there exists Aᵢin Dᵢ Bᵢⱼ in S_T where i=1,..d, j=1,...d_0 s.t each component has this form Xᵢ(t)= Xᵢ(0) +∫₀ᵗ Aᵢ(s).ds + Σ_{j=1, d_0}BᵢⱼdWⱼ(s) true for all i=1,...,d For fixed i resembles ito process with d_0 brownian motions, might have multiple sources derivative in finance depending on multiple stocks

Answer 103

(X_1(t)) (...) (X_d(t)) + integral_{0,t} (A_1(s)) (...) (A_d(s)).ds + (B_11(s)....B1d_0(s)) ...... (B_d1(s)....B_dd_0(s)) (dW_1(s)) (...) (dW_d_0 (s)) vectors and matrices or **x**+ integral_{0,t}**A**.ds + integral_{0,t} **B** dW(s) noise values in R^d_0 output with values in R^d

Answer 104

Aᵢin Dᵢ Bᵢⱼ in S_T where i=1,..d, j=1,...d_0 s.t Xᵢ(t)= Xᵢ(0) +∫₀ᵗ Aᵢ(s).ds + Σ_{j=1, d_0}BᵢⱼdWⱼ(s) true for all i=1,...,d then with function u from R^d to R [missing du(X(t))] =Σ_{i=1 to d} u_{x_i}dX_i (t) + 0.5Σ_{i,j=1 to d} u_{xᵢxⱼ}(X(t)) dXᵢ(t)dXⱼ(t) where u_{x_i} is the partial derivative of u wrt x_i ith component (if d =1 only one partial derivative as expected) also this is the non rigourous version of the one you SHOULD REMEMBER of saying that...[missing]

Answer 105

df(Xₜ)=∂ₓᵢ f(Xₜ) dXᵢ(t)+0.5∂²ₓᵢₓⱼf(Xₜ) dXᵢ(t)dXⱼ(t) partial derivs which help to remember the multi dimensional formula ALSO dtdWⱼ(t)=dWᵢ(t)dWⱼ(t)=(dt)²=0 i≠j if i=j (dWᵢ(t))² =dt I will not ask you in the exam for the multidimensional formula but there is an exercise we will look at..

Answer 106

(we saw prev that ito processes with purely stochastic part and no deterministic part are martingales because stochastic integrals are martingales) (first look:looks like an ito process as has dt term and dw term, here set d=1?) Step 1: i=1 u(x)=exp(x) or he used f(x)=exp(x) (derivatives of f all equal due to exp funct) X(t) = ct +Σ_{ⱼ=1 to d} aⱼWⱼ(t)) dX(t)=c dt + Σ_{ⱼ=1 to d} aⱼ.dW(t) This is exactly the n dim form of ito formula for i=1 Aᵢin Dᵢ Bᵢⱼ in S_T where i=1,..d, j=1,...d_0 s.t Xᵢ(t)= Xᵢ(0) +∫₀ᵗ Aᵢ(s).ds + Σ_{j=1, d_0}BᵢⱼdWⱼ(s) Z(t) = exp(ct +Σ_{ⱼ=1 to d} aⱼWⱼ(t)) = f(X(t)) Thus dZ(t)=dF(X(t)) Then by itos lemma.? formula **= f'(x(t))dX(t) +0.5f"(X(t))(dX(t))²** (as derivs of f(X(t)) equal f(X(t)) they equal Z(t)) =Z(t)dX(t) +0.5Z(t) (dX(t))² subbing in dX(t) =Z(t)(**c dt + Σⱼ₌₁ᵈ aⱼ .dW(t)** +0.5Z(t) (**c dt + Σⱼ₌₁ᵈ aⱼ .dW(t)**)² using table values BUT WITH MULTI DIM PART i=j =Z(t)(**c dt + Σⱼ₌₁ᵈ aⱼ .dW(t)**+0.5Z(t)Σᵢ₌₁ᵈaᵢ² dt =[Z(t)(c + 0.5Σᵢ₌₁ᵈaᵢ²) ] dt + Σⱼ₌₁ᵈ aⱼ .dW(t) So to choose such a c that gives 0 deterministic part c + 0.5Σᵢ₌₁ᵈaᵢ² =0 c=-0.5Σᵢ₌₁ᵈaᵢ² ;(to show the sum of the stochastic integral is a martingale see prev exercise) Then: dZ(t)=Z(t)(Σⱼ₌₁ᵈaⱼ²dWⱼ(t) If aⱼZ(t) is in the class H_T then each integral ∫₀ᵗaⱼZ(t) dWⱼ(s) is a martingale thus sum Σⱼ₌₁ᵈ∫₀ᵗaⱼZ(t) dWⱼ(s) is a martingale THUS It suffices to show (Z(t))_{t in [0,T]} in H_T for this we need to check it's adapted: Because if Z(t)= exp(ct +Σ_{i=1 to d} aᵢWᵢ(t)) is F_t measurable as if we fix t Wⱼ(t) are F_t measurable continuous function add constant composition of continuous functions is F_t measurable still 2) Need to check also that E[integral_[0,**T**] Z(s)^2 .ds]< infinity E[ integral_[0,T][exp(ct +Σ_{i=1 to d} aᵢWᵢ(s)]^2 .ds] =E[ integral_[0,T][exp(2ct +2Σ_{i=1 to d} aᵢWᵢ(s)] .ds] by properties of exponential =E[ integral_[0,T][exp(2ct) Π{i=1 to d} exp(2aᵢWᵢ(s))] .ds] less than or equal to exp(2|c|T)integral_[0,T] E[Π{i=1 to d} exp(2aᵢWᵢ(s))] .ds]] by independence of W_i's exp(2|c|T)integral_[0,T] Π{i=1 to d} E[exp(2aᵢWᵢ(s))] .ds] =exp(2|c|T)integral_[0,T] Π{i=1 to d} E[exp(√s2aᵢWᵢ(s)√s)] .ds] we know Wᵢ(t)√t) has distribution N(0,1) =exp(2|c|T)integral_[0,T] Π{i=1 to d} E[exp(√s2aᵢWᵢ(s)√s)] .ds] =exp(2|c|T)integral_[0,T] exp( Σⱼ₌₁ᵈ√s2aⱼ) .ds] using E[exp(aZ)]=exp(a^2/2) exp(2|c|T) exp(2√T Σⱼ₌₁ᵈaⱼ) T < infinity Thus if we choose c as above dZ(t)=Z(t)(Σⱼ₌₁ᵈaⱼ²Wⱼ(t) with Z(t)aⱼ in H_T and thus Z is a martingale In the case of level 3: multi dimensional case not in exam But if i would only deliver what would be asked in the exam I would have finished the module earlier!

Answer 107

exp(a^2/2) then W_j/ square root(t) ~N(0,1)

Answer 108

first deriv wrt time second wrt space We use itos formula/lemma du(t,W(t))= ∂ₜu(t,W(t))dt + ∂ₓu(t,W(t))dW(t)+0.5 ∂ₓₓ²u(t,W(t))dt = [∂ₜu(t,W(t)) + 0.5∂ₓₓ²u(t,W(t))]**dt** + ∂ₓu(t,W(t))**dW(t)** first part partial diff eq=0 so du(t,W(t)) =∂ₓu(t,W(t))**dW(t)** means in integral form u(t,W(t))=u(0,W(0))+ ∫₀ᵗ ∂ₓu(s,W(s))**dW(s)** because its brownian motion W(0)=0 u(t,W(t))=u(0,0)+ ∫₀ᵗ ∂ₓu(s,W(s))**dW(s)** taking expectations we have required E[u(t,W(t))]= u(0,0) as first is deterministic and expectation of stochastic integral is 0 PROVIDED the integral term is in the class H_T: 1) W(t) is F_t measurable, u is smooth, ∂ₓ continuous in x so it is also borel in x, thus the composition of these is F_t measurable 2) E[∫_[0,**T**][∂ₓu(t,W(t))]^2 .dt] ≤E[∫_[0,**T**][C^2 .dt] = TC^2

Answer 109

in general connection between PDEs and ito processes apply itos formula for derivs in space and time if funct satisfies PDE involving these terms we use this also we will see we can represent sols using ito processes using expectations instead of solving approximate underlying process

Answer 110

Suppose we have an ito process which looks like this CANT SEE THE BOARD formula tells you how __evolves if function dep on x function also depends on time f(X(t) we can see from higher dimensions this is actually two dimensional setting components y_1=x and y_2 =t I have a function of two variables f(t,X(t))=g(Y(t) g(y):= g(y₁,y₂)=f(y_1,y_2) Then I know from multi dimensional formula differentials df(t,X(t)= dg(Y(t)) from itos formula using partial derivatives = ∂_y₁g(Y(t))dY¹(t)+∂_y₂g(Y(t))dY²(t)+0.5∂²_y₂y₂g(Y(t))(dY²)² partial deriv of g wrt y_1 is the partial deriv of f wrt first variable, here its the time variable second derivative of g wrt second variable is the same as the derivative of f wrt to the second variable, called x = ∂ₜf(t,X(t))dt+∂ₓf(t,X(t))dX(t)+0.5∂²ₓₓf(t,X(t))(B²(t))dt (when i=1 then product means is 0, same with 1. Only case 2 survives) one case: second partial derivative of f wrt y_2 so df(t,X(t)) = [∂ₜf(t,X(t))+∂ₓf(t,X(t))A(t)+0.5∂²ₓₓf(t,X(t))(B²(t))]dt +∂ₓf(t,X(t))B(t)dW(t) if you dont have time dependence you have the same without the first term here and in this case the dt part usual ito formula without t so extra term if time dependence partial deriv of f wrt t USUAL ITO FORMULA +EXTRA TERM

Answer 111

for all A ∈ F_T , P(A) = 0 if and only if Q(A) = 0 there exist two probability measures that equal measure probability 0 for the exact same events probability measure might be different for any two non zero values (thm 5.1.1)

Answer 112

in S_T: integrals for stochastic integrals, meaning i can integrate wrt W Z(t) exists because of this fact note Z(T)>=0 since non negative we can take the expectation; we don't know if finite but for non neg RV can be taken Brownian motion wrt measure?? Define P❋(A):= E[**1**_A Z(T)] with expectation wrt measure p which defines function P❋:F to [0,+infinity] Girsanov’s theorem tells us the following... Assume that (Z(t))_{t in[0,T]} is a martingale wrt filtration (F_t)_{t in [0,T]}

Answer 113

Assuming (Z(t))_{t in[0,T]} is a martingale wrt filtration (F_t)_{t in [0,T]} THEN i)P❋ is a probability measure on(Ω, F) and it is equivalent to P. (ii) The process (W∗(t))t∈[0,T] defined is an F-Wiener process on (Ω, F, P∗)

Answer 114

equivalent probability measures statistical properties of W∗ in general P(W∗(t) in A) not equal to P∗(W∗(t) in A) probabilities not necessary the same meaning expectations and variance is not the same but thm tells us that under the P∗ measure, W∗is a brownian motion: indep increments gaussian distribution satisifies BM properties!

Answer 115

says that If E[exp(0.5 ∫₀ᵗ |B(s)|^2 ds] < ∞. Then (Z(t))_{t∈[0,T]} is a martingale with respect to F. ----------------------- for example if we have process B which satisfies the exponential of this integral is finite then we have Z(t) martingale In particular if B is a stochastic process which is bounded by constant C then of course finite as its an exp of an ito process we can use itos formula to express as ito process, dt parts cancel so stochastic integral which is a martingale if in H_T? ----------- notes: If E[exp(0.5 Σ_{i=1,d} |B_i(s)|^2 ds] < ∞.

Answer 116

Corollary 3.8.3. Assume that B ∈ Sd×1_ T is bounded, that is, there exists a constant C such that for all i = 1, ..., d and all (ω, t) ∈ Ω × [0, T] we have |B_i(ω, t)| ≤ C. Then, the conclusions of Theorem 3.8.1 hold.