1
$\begingroup$

I cannot understand one step of a proof in Theorem 32 of the first Chapter of Preliminaries of Protter. enter image description here I cannot follow how the Optional stopping theorem was applied n times here to yield $P(A) \prod f_{t_j-t_{j-1}}(u_j)$. I see that we can use the independent increments of the Levy Process to simplify $f$'s

In other words I cannot show how

$$E\{ \mathcal{1}_A \prod \frac{M_{T+t_j}^{u_j}}{M_{T+t_{j-1}}^{u_j}}\}=P(A)$$ Can you please help me out. I would be grateful?

1 Answers 1

2

Start with the case $n=1$: You need to show that $$ E\left\{1_A{M^{u_1}_{T+t_1}\over M^{u_1}_{T+t_0}}{f_{T+t_1}(u_1)\over f_{T+t_0}(u_1)}\right\}=E\left\{1_A{M^{u_1}_{T+t_0}\over M^{u_1}_{T+t_0}}{f_{T+t_1}(u_1)\over f_{T+t_0}(u_1)}\right\}=E\left\{1_A f_{t_1-t_0}(u_1)\right\}. $$ The first equlity above follows from the optional stopping theorem at time $T+t_0$, using that both $T$ and $M^{u_1}_{T+t_0}$ are $\mathcal F_{T+t_0}$-measurable. The second equality is because, for $t=T(\omega)$, $$ \eqalign{ f_{t+t_1}(u_1) &=E\{e^{iu_1X_{t+t_1}}\}\cr &=E\{e^{iu_1(X_{t+t_1}-X_{t+t_0})}e^{iu_1X_{t+t_0}}\}\cr &=E\{e^{iu_1(X_{t+t_1}-X_{t+t_0})}\}E\{e^{iu_1X_{t+t_0}}\}\cr &=f_{t_1-t_0}(u_1)f_{t+t_0}(u_1).\cr } $$ In the general case work backwards in time, using optional stopping at $T+t_{n-1}$, then at $T+t_{n-2}$, etc.

  • 0
    Thank you very much John . I am grateful. If you might have a few moments could you look at this too? http://math.stackexchange.com/questions/2098662/doubt-in-the-proof-of-right-continuity-of-the-completed-levy-filtration – 2017-01-15
  • 0
    I have two more questions: Why does showing that the expectation of the product equals the product of expectations (which is usually a consequence of independence) imply independence. I mean we need to show it every element in the sigma algebra generated by $Y_t$. What is the role played by the $(u_1,u_2 \dots u_n)$ and taking exponential in proving this? I know I am asking too many questions and I would understand if you cant answer them. Thank you very much – 2017-01-16