1 Homework 9
to be handed in: May 18, 2016
Note: always check the web for the latest adaptations in blue in LN and BN.
Exercise 1 (3 pts) LN Exercise 3.11 Exercise 2 (2,5 pts)
LN Exercise 3.12 (for part (iv) see also Exercise 3).
Exercise 3 (2 pts)
Consider again the situation of Exercise 2. Let x be a holding point, with P{σx> t} = e−αxt, 0 < αx< ∞.
i) Show that there exists a probability distribution q on (E, E ), such that Px{Xσx ∈ B, t < σx < ∞} = e−αxtq(B).
Hint: use conditioning on FtX+.
ii) If all points of E are holding points, then show that there exists a transition kernel q : E × E → [0, 1], such that
Px{Xσx ∈ B, t < σx< ∞} = e−αxtq(x, B).
Exercise 4 (2,5 pts) LN Exercise 3.13.
Remark Suppose that all points of E are holding points, with σx = exp(αd x), x ∈ E. The Markov process spends an exponentially distributed amount of time in a state, then jumps to the next state according to the kernel q, again spends an exponentially distributed amount of time in the new state, etc.
This means that the corresponding Markov process is a Markov jump process of the following form. Let T1, . . . , be an i.i.d. collection of exp(1)-distributed random variables.
Note that Tn/c is exp(c) distributed, c > 0.
Let Y be an (E, E )-valued Markov chain with transition kernel q (this exists, by a char- acterisation lemma approach).
Define by Jn=Pn−1
k=0Tk/αYk be the time of the n-th jump. Put Xt= x ⇐⇒ Yn= x, Jn≤ t < Jn+1.
Then X is a Markov process (w.r.t. the natural filtration). Note that absorption points can be incorporated in this framework as well!
