BiggieB
Joined: Feb 19, 2005
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 Posted:
Oct 04, 2006 - 18:07 |
 
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MrB
Joined: Jul 05, 2006
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Oct 04, 2006 - 18:08 |
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7 and a bit, by my maths
But then, I studied Theology |
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CC_Blood
Joined: Aug 02, 2003
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Oct 04, 2006 - 18:15 |
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I suggest looking up Girsanov's theorem and Smith's "Wealth of Nations" |
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Plorg
Joined: May 08, 2005
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Oct 04, 2006 - 18:17 |
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Oh that wacky Girsanov and his theorems...
I remember as a kid hearing stories about that crazy old hermit,
throwing slide rules and textbooks at passersby while ranting unintelligibly.
Ah good times. |
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sk8bcn
Joined: Apr 13, 2004
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Oct 05, 2006 - 12:22 |
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I will look at it during job. I propose we share our guessings there. Chat ports are blocked at job, I can't jump in there. |
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sk8bcn
Joined: Apr 13, 2004
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Oct 05, 2006 - 15:46 |
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altough you already solved it, I started by ex1 to warm up and found out a=1/beta
I have an idea for 2a, but need to investigate a bit further.
Imo
you'd have to try to Zt=E(Z t-1 | Ft) and move on until you reach s, and use the advice given to make simplification.
Yet it's just a thought, a path I will try to follow. Don't hesitate to post Biggie, even if you find. I wanna work a bit on my maths.
BTW, I discover how much I love maths. I currently do informatics (what sucks, and I suck in this field), and yes, I am very weird, but I do love maths.... |
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Alf115
Joined: Aug 17, 2005
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Oct 05, 2006 - 16:13 |
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2/3. The answer is always 2/3. |
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SchwipSchwap
Joined: Jul 02, 2005
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 Posted:
Oct 05, 2006 - 16:17 |
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Exercise 2a) is a bit strange.
According to the text this new equivaltent probability measure is defined on the sigma-algebra F_t. But according to the definition it's defined on the sigma-algebra F. So which one is true?
Sk8bcn: I think your idea won't work since t>=0 and not t element of N, so it's continuous. |
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Cloggy
Joined: Sep 23, 2004
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Oct 05, 2006 - 16:21 |
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Oh come on guys. Stop messing with skate.....
The answer is 42, as it always has been. |
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SchwipSchwap
Joined: Jul 02, 2005
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Oct 05, 2006 - 16:23 |
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No seriously...this is martingale in continuous time. Sk8bcn's idea does not work for those martingales.  |
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Panda_
Joined: Jul 14, 2004
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Oct 05, 2006 - 16:24 |
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There is many french coach on this one.
I like matchs too. Except when it comes to statistics/probabilities (which i only support the bloodbowl one guess why). |
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"Rien ne sert de partir a point, il vaut mieux courir." |
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sk8bcn
Joined: Apr 13, 2004
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Oct 05, 2006 - 16:49 |
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| SchwipSchwap wrote: | | No seriously...this is martingale in continuous time. Sk8bcn's idea does not work for those martingales. |
While I do not say that my idea works (I yet struggle on the question), I affirm that t belongs to N.
F is a filtration, and a filtration is defined by:
a filtration is a (suite croissante de tribus) Fn, n>=0 so FnCFn+1, for n belonging to N.
BTW, I currently try to find out if I cannot use those points in the problem.
FsCFt so C=C1 U C2 with C1€Fs and C2€Ft\{Fs}
and I didn't use the fact P tilda is a proba on F
To SchwipSchwap: There is no bug. Ft is a sigma algebra. but we define the probability on the "tribu" F, the "biggest one". Each Ft are included in F |
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sk8bcn
Joined: Apr 13, 2004
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Oct 05, 2006 - 16:50 |
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| Panda_ wrote: | | There is many french coach on this one. |
cheers, we are so great! |
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sk8bcn
Joined: Apr 13, 2004
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| Posted:
Oct 05, 2006 - 17:06 |
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I think I get closer to the answer:
we should use the following properties:
if X independant with G then E[X|G]=E[X]
if Y is G mesurable, E[XY|G]=YE[X|G] and E[Y|G]=Y |
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SchwipSchwap
Joined: Jul 02, 2005
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Oct 05, 2006 - 17:13 |
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| sk8bcn wrote: | F is a filtration, and a filtration is defined by:
a filtration is a (suite croissante de tribus) Fn, n>=0 so FnCFn+1, for n belonging to N.
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You can also define a filtration for t>=0.
I really think you are wrong here and we have to work in continuous time. |
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