Fine I'll do it. Conditioning on the Z's you can pull the first n Y's out of the expectation...
Now use your distribution of the Y's
So your conditional expectation is
Hi !
Looks like it's the same as last year : I can't wait for the next class to solve a problem ^^'
We've studied martingales for only 3 hours already, so I'm still very new to it.
So we have a sequence of iid rv's such that and
where
Let ... and
Let's consider the filtration
I have to show that is a positive martingale.
3 points to show :
1/ is -measurable
2/ is integrable
3/
Point 1/ is okay.
Point 2/ is okay.
Point 3/ is totally unknown Which method to use ?
Thanks for any help/hints
Fine I'll do it. Conditioning on the Z's you can pull the first n Y's out of the expectation...
Now use your distribution of the Y's
So your conditional expectation is
easy?
Everything is easy when you know what you're doing.
I used to say certain problems were easy (and also some were hard)
in class and that would annoy the students who didn't realize the difference.
But I don't know why they have the condition p<q here.
Maybe that's the next step.
That we expect more failures than successes.
This then goes to infinity.
and I saw the unedited post, lol.
I've found out that the cow sees everything.
I have exams to grade next week, if anyone is interested.
Hmmm no, I would've said obvious or direct. Once you've done enough of conditional expectation of course.
Yep, we had to prove then that goes to -infinity. And if you're interested, the following questions are :But I don't know why they have the condition p<q here.
Maybe that's the next step.
2. Let
Consider the martingale (it's true because Tk is a bounded stopping time and Zn is a martingale)
Then show that
(I think this is where the condition over p and q is the most important actually)
3. Finally show that
Huh ? which undedited post ?and I saw the unedited post, lol.
I've found out that the cow sees everything.