Actuarial Outpost > MFE Why dZ(t) x dZ(t) = dt ??
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#31
07-23-2012, 12:36 PM
 Arthur Kade Member Join Date: May 2009 Location: Philadelphia, PA Studying for my SAG card Favorite beer: whatever other folks are buying Posts: 4,248

Quote:
 Originally Posted by david849 would that imply (dZ(t))^3 = ħsqrt(dt)*dt?
Yes, except that you don't keep higher order terms.
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#32
07-23-2012, 12:37 PM
 jas66Kent Member SOA Join Date: May 2012 Location: Canterbury, UK Studying for ST9 and SA2 Favorite beer: Corona :) Posts: 8,310

You obviously have little understanding of Calc in general. But hey, what else is new.

In case you didn't know, the stochastic PDE can be re-written in terms of integration.

I suggest you look into that before acting like an ass.
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#33
07-23-2012, 12:51 PM
 truonda Member SOA Join Date: Apr 2012 Studying for MLC College: Waterloo Alumni Posts: 186

For the purpose of this exam you do not need to know why dZ(t) x dZ(t) = dt; just accept it as a fact.
#34
07-23-2012, 01:18 PM
 zywen Join Date: Jul 2011 College: unversity of waterloo Posts: 7

it is because the quadratic variation of brownian motion ----> t
#35
07-23-2012, 01:36 PM
 Numerical Neuroticism Member SOA Join Date: Jan 2011 Studying for FAP 6-8 and FA Posts: 363

Quote:
 Originally Posted by david849 would that imply (dZ(t))^3 = ħsqrt(dt)*dt?
I think (dZ(t))^3 = (dZ(t))^2 * dZ(t) = dt * dZ(t) = 0
#36
07-23-2012, 01:37 PM
 zywen Join Date: Jul 2011 College: unversity of waterloo Posts: 7

Quote:
 Originally Posted by Numerical Neuroticism I think (dZ(t))^3 = (dZ(t))^2 * dZ(t) = dt * dZ(t) = 0
correct!
#37
07-23-2012, 01:46 PM
 jas66Kent Member SOA Join Date: May 2012 Location: Canterbury, UK Studying for ST9 and SA2 Favorite beer: Corona :) Posts: 8,310

Quote:
 Originally Posted by AAABBBCCC yes I suspect that people actually dont understand what dz(t)^2=dt "means".
You can also use the cross product to prove that dz(t)*dz(t) = dt
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Quote:
 John Kenneth Galbraith "The modern conservative is engaged in one of man's oldest excuses in moral philosophy; that is, the search for a superior moral justification for selfishness."
#38
07-23-2012, 02:16 PM
 AAABBBCCC Member Join Date: Feb 2012 Posts: 1,329

$\sum_{i=1}^{2^n}(Z(i2^{-n}t)-Z((i-1)2^{-n}t))^2 \rightarrow t$ almost surely. Doesnt work along any partition.

Last edited by AAABBBCCC; 07-23-2012 at 02:39 PM..
#39
07-23-2012, 02:37 PM
 AAABBBCCC Member Join Date: Feb 2012 Posts: 1,329

You have to think of dz(t)^2 as something you can integrate against, but it is a stochastic integral. Basically says

$\int g(t)dz(t)^2=\int g(t)dt$ as a stochastic integral, with g anything reasonable and perhaps stochastic. So when you "prove" Ito's Lemma, you can use a taylor series expansion and get rid of any higher powers of dz(t) beyond 2 and convert dz(t)^2 into dt. IT explains why you have the Ito correction term 1/2 g''(X(t))dt in Ito's Lemma.
#40
02-20-2013, 06:37 PM
 algebrat Member Join Date: Nov 2012 Location: Ork Studying for GED Favorite beer: Scieoclean Posts: 1,688

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