1 Homework 4
to be handed in: March 23, 2016
Note: always check the web for the latest adaptations in blue in LN and BN.
We will give you the option of selecting exercises according to your interest. You should select exercises worth in total at most 10 points. If you hand in exercises worth more than 10 points, then this will therefore not be beneficial to your grade.
Exercise 1 (3 pts) LN Exercise 1.30 Exercise 2 (1.5 pts) LN Exercise 1.31.
Exercise 3 (2.5 pts) LN Exercise 2.15.
Exercise 4 (3 pts) BN Problem 7.4.
Exercise 5 (4 pts)
Prove Lemma 1.6.15 (this week’s version, see website).
We sketch part of the proof below first, and indicate which sketched parts you will have to prove on the way. Fix t ≥ 0.
For progressive measurablilty of Xτ w.r.t. (Ft∧τ)t, one has to show for every A ∈ E (remind: Xt takes values in (E, E ))
B(A) := {(s, ω) ∈ [0, t] × Ω | Xsτ(ω) = Xτ (ω)∧s(ω) ∈ A} ∈ B([0, t] × Ft∧τ. Let s ≤ t and define
Cs,t=n
C ∩ {[0, t] × {t ∧ τ ≥ s}} | C ∈ B([0, t]) × Fso . Then Cs,t is a σ-algebra, and Cs,t⊂ B([0, t]) × Ft∧τ. Prove both statements.
Next, split B(A) according to whether t ∧ τ (ω) ≤ s or t ∧ τ (ω) > s. For the resulting event where additionally t ∧ τ (ω) > s you could apply a ‘separation’ trick.
If you search on internet, you will be probably clever enough to find a proof. However, the one(s) I found, use a division of the interval [0, t] into equally spaced sub-intervals. That argument is not allowed!
The statements concerning the shifted process are then simple to prove.
